Using the integral test, how do you show whether #sum 1/n^3# diverges or converges from n=1 to infinity?
Check that
The integral converges, therefore the series converges. (We have not found what the series converges to.)
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To determine whether the series ( \sum_{n=1}^{\infty} \frac{1}{n^3} ) converges or diverges, we can use the integral test.

Define the function ( f(x) = \frac{1}{x^3} ).

Check if ( f(x) ) is continuous, positive, and decreasing for ( x \geq 1 ).

Integrate ( f(x) ) from 1 to ( \infty ):
[ \int_{1}^{\infty} \frac{1}{x^3} , dx ]
 Evaluate the integral. If the integral converges, then the series also converges. If the integral diverges, then the series diverges.
[ \int_{1}^{\infty} \frac{1}{x^3} , dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^3} , dx = \lim_{b \to \infty} \left( \frac{1}{2x^2} \right) \bigg_{1}^{b} ]
[ = \lim_{b \to \infty} \left( \frac{1}{2b^2} + \frac{1}{2} \right) ]
[ = \frac{1}{2} ]
Since the integral ( \int_{1}^{\infty} \frac{1}{x^3} , dx ) converges (to ( \frac{1}{2} )), by the integral test, the series ( \sum_{n=1}^{\infty} \frac{1}{n^3} ) also converges.
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The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the seriesTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) convergesTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges ifTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(fTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if andTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(nTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only ifTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) fromTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if theTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from nTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improperTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n =To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integralTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 toTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫fTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinityTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x)To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity alsoTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dxTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also convergesTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. ConverselyTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 toTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if theTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integralTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ convergesTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral divergesTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
ForTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then theTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the seriesTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the seriesTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series alsoTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also divergesTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/nTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In thisTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3),To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this caseTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), theTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case,To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the correspondingTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we haveTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding functionTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have theTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function isTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the seriesTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(xTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) =To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/nTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/xTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3),To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), andTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3.To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and weTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. ThisTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we wantTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This functionTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want toTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuousTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compareTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous,To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare itTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positiveTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it toTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive,To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to theTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, andTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integralTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasingTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing onTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on theTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the intervalTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/xTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1,To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3)To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dxTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now,To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, weTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 toTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluateTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improperTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integralTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/xTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/xTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dxTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx =To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dxTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx fromTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dxTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2)To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3)To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * xTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dxTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluatedTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated fromTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2xTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 toTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinityTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2)To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] fromTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity =To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 toTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) *To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ =To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/inTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinityTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2*To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2*∞To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)] To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2*∞^To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2*∞^2To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2*∞^2)]To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2*∞^2)] To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2*∞^2)]  [To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2*∞^2)]  [1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2)To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2*∞^2)]  [1/(To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) *To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0 To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] =To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) =To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2) To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
SinceTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since theTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integralTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) =To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/xTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
SinceTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3)To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since theTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dxTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integralTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx convergesTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equalsTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/xTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3)To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dxTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2),To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx fromTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), byTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by theTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integralTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 toTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral testTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinityTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test,To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, theTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity convergesTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the seriesTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, theTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the seriesTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/nTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3)To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/nTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) alsoTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/n^To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) also converTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/n^3To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) also convergesTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/n^3)To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) also converges.To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/n^3) alsoTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) also converges. ThereforeTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/n^3) also converTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) also converges. Therefore, theTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/n^3) also convergesTo determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) also converges. Therefore, the seriesTo determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/n^3) also converges.To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) also converges. Therefore, the series ∑To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/n^3) also converges.To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) also converges. Therefore, the series ∑(To determine whether the series ∑(1/n^3) converges or diverges, we can use the integral test.
The integral test states that if f(x) is a positive, continuous, and decreasing function for all x ≥ 1, and if ∫(f(x)) dx from 1 to infinity converges, then the series ∑(f(n)) from n = 1 to infinity also converges. Conversely, if the integral diverges, then the series also diverges.
In this case, we have the series ∑(1/n^3), and we want to compare it to the integral ∫(1/x^3) dx from 1 to infinity.
∫(1/x^3) dx = ∫(x^(3)) dx = (1/2) * x^(2) evaluated from 1 to infinity = [(1/2) * (1/infinity^2)]  [(1/2) * (1/1^2)] = (1/2)  (1/2) = 1
Since the integral ∫(1/x^3) dx from 1 to infinity converges (it equals 1), by the integral test, the series ∑(1/n^3) also converges.To determine whether the series ∑(1/n^3) converges or diverges using the integral test, we need to compare it with the integral of the corresponding function.
The integral test states that if a continuous, positive, and decreasing function f(n) is defined on the interval [1, ∞), then the series ∑f(n) converges if and only if the improper integral ∫f(x) dx from 1 to ∞ converges.
For the series ∑(1/n^3), the corresponding function is f(x) = 1/x^3. This function is continuous, positive, and decreasing on the interval [1, ∞).
Now, we evaluate the improper integral ∫(1/x^3) dx from 1 to ∞:
∫(1/x^3) dx = [ 1/(2x^2) ] from 1 to ∞ = [1/(2∞^2)]  [1/(21^2)] = 0  (1/2) = 1/2
Since the integral ∫(1/x^3) dx converges (equals 1/2), by the integral test, the series ∑(1/n^3) also converges. Therefore, the series ∑(1/n^3) converges.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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