Using the integral test, how do you show whether #sum 1/n^3# diverges or converges from n=1 to infinity?

Question

Isabella Ackerman · Accepted Answer

Check that #1/x^3# is (eventually) decreasing, then see whether #int_1^oo 1/(x^3) dx# diverges or converges.

If your grader or teacher won't let you claim that it is obvious that #1/x^3# is a decreasing function on #[1, oo]#, then point out that for #f(x) = 1/x^3#, we get #f'(x) = -3/x^4# which is always negative, so #f# is always decreasing. #int_1^oo 1/(x^3) dx = lim_(brarroo) int_1^b 1/(x^3) dx# # = lim_(brarroo) (-1/(2x^2)]_1^b)# # = lim_(brarroo) (-1/(2b^2) + 1/2)# # = 1/2# The integral converges, therefore the series converges. (We have not found what the series converges to.)

Axel Alvarez · Answer

undefined To determine whether the series $ \sum_{n=1}^{\infty} \frac{1}{n^3} $ converges or diverges, we can use the integral test.

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

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

3. Integrate $ f(x) $ from 1 to $ \infty $:

$$ \int_{1}^{\infty} \frac{1}{x^3} \, dx $$

4. 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.

Emma Aldridge · Answer

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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.

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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.

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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.

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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.

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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.

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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/(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) * (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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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/(2*1^2)]
             = 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.