Question

Find
\[\sum_{n=1}^\infty \frac{n}{(2n-1)^2(2n+1)^2}\]

Answer 1

begin with partial sums of the series

Answer 2

I have been trying to find the partial sums by hand, but I couldn't.

Answer 3

Hope this helps
http://www.wolframalpha.com/input/?i=sum_ {n%3D1}^\infty+\frac{n}{%282n-1%29^2%282n%2B1%29^2}

Answer 4

" http://www.wolframalpha.com/input/?i=sum_ {n%3D1}^\infty+\frac{n}{%282n-1%29^2%282n%2B1%29^2}"

Answer 5

http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+n%2F%28%282n-1%29^2%282n%2B1%29^2%29

Answer 6

\[\sum_{n=1}^k \frac{n}{(2n+1)^2(2n-1)^2}=\frac{1}{8}\sum_{n=1}^k\frac{1}{(2n-1)^2}-\frac{1}{(2n+1)^2}=\frac{1}{8}(1-\frac{1}{\biggl (2k+1)^2}\biggr )\]