Alternating series, summation from n=1 to infinity (1/(n+1)^3

Question

∑ from n=1 to ∞ 1/(n+1)³

Would someone solve please..

is it convergent or divergent ??

And how did you find it please ?

Answer 1

We can use the integration test to solve this. Let's take the limit as b approaches infinity and do the integral from 1 to b of 1/(n+1)^3.

The integral of 1/(n+1)^3 is -1/(2*(n+1)^2) [Using the power rule].

Then we will use our limits to finish the integral. Keeping in mind our limit of b, we get:

-1/(2*(b+1)^2) - [-1/(2*(1+1)^2] = -1/(2*(b+1)^2) + 1/(2*(2)^2)

The first term is equal to 0 because as b approaches infinity this term quickly goes to 0

The second term is 1/8

Because our integral gave us a clear answer (i.e. not infinity or does not exist), we can say that this series converges. Please note, it does not approach 1/8, the answer from our integral does not indicate to what number does the series converge. The answer to our integral only tells us if we get a number, then the series must converge.

Please let me know if this helps!