# Find the sum of this series

1. Jul 16, 2010

### jwxie

1. The problem statement, all variables and given/known data

Determine whether the series converges or diverges. For convergent series, find the sum of the series.

sima (k=1, infinity), (2k +1) / ((k^2) (k+1)^2 )

2. Relevant equations

3. The attempt at a solution
Well, the kth test for divergence said this series has limit ak = 0 because if we simplify the leading terms, we have 2k / c*k^4, which is 1/k^3, this is a p-series. We know that for p > 1, the p-series will converge.

But how do you find the sum?
Partial fraction expanision seems not a good choice?

(2k +1) / ((k^2) (k+1)^2 ) = A/k + B/k^2 + C/(k+1) + D/(k+1)^2
->>>>
(2k+1) = A(k)(k+1)^2 + B(k+1)^2 + C(k^2)(k+1) + D(k^2)

We can ignore anything that has power higher than 1, so after mulitiplication, we have
2k + 1 = Ak + 2Bk + B

it's clear that b = 1
2k = k(a+2b)
2k = k(a+2)
2 = a + 2
a = 0

so are we left with the expansion 1/k^2 ? which doesn't make sense to me...

well anyhow, even if i have a compute error, this is not a telescopic series, no positive terms will cancel. How do I compute the sum then?

2. Jul 16, 2010

### Staff: Mentor

So the k-th term test for divergence tells us nothing.

3. Jul 16, 2010

### jwxie

wait. kth term test said if lim ak =/= 0, it must diverges. if it is zero, it may or may not diverge.

So i just that to begin with. Now I need a different approach to find the sum. It converges to 1. I mean I can approximate it goes goes to 1 if I look at the pattern numbers. But definitely I can't be sure.

How would I compute it?

4. Jul 16, 2010

### Staff: Mentor

Yes, lim ak = 0, which means that the k-th term test does not apply. You can't just simplify the leading terms, and this series is not a p-series.
???
Carry this out. Find A, B, C, and D.
No, this is not how it works. On the right side you are going to have terms up to degree 3. The coefficients of the k3 and k2 terms have to be zero, but you can't just ignore them.
If you complete your partial fractions work, you will find that this is a telescoping series.

And this tells us exactly nothing.