# How do you use the integral test to determine the convergence or divergence of #Sigma 1/n^3# from #[1,oo)#?

The series:

is convergent.

Given the series:

we can choose as test function:

Calculating the integral we find:

Thus the integral is convergent and then the series is proven to be convergent as well.

By signing up, you agree to our Terms of Service and Privacy Policy

To use the integral test to determine the convergence or divergence of the series Σ 1/n^3 from n = 1 to infinity, we need to compare it with the corresponding improper integral.

The integral test states that if f(x) is a continuous, positive, and decreasing function for x ≥ 1, and if the series Σ f(n) converges if and only if the integral ∫_1^∞ f(x) dx converges.

In this case, f(x) = 1/x^3. This function is continuous, positive, and decreasing for x ≥ 1.

Now, we evaluate the improper integral:

∫_1^∞ (1/x^3) dx = lim(a→∞) ∫_1^a (1/x^3) dx

= lim(a→∞) [-1/(2x^2)] from 1 to a

= lim(a→∞) [-1/(2a^2)] + 1/2

= 0 + 1/2

= 1/2

Since the integral ∫_1^∞ (1/x^3) dx converges (equals 1/2), by the integral test, the series Σ 1/n^3 from n = 1 to infinity also converges.

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1-2+3-4+...+n(-1)^(n-1)#?
- What does the alternating harmonic series converge to?
- How do you determine if #a_n=(1-1/8)+(1/8-1/27)+(1/27-1/64)+...+(1/n^3-1/(n+1)^3)+...# converge and find the sums when they exist?
- The integral #int_0^a (sin^2x)/x^(5/2)dx# converges or diverges ?
- How do you find #lim sin(2x)/x# as #x->0# using l'Hospital's Rule?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7