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Mathematics 68 Online
OpenStudy (watchmath):

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

OpenStudy (anonymous):

begin with partial sums of the series

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (watchmath):

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

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