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I ran across another challenging and interesting series, and I am wondering if someone could shed some light on how to evaluate it.

$$ \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(n^{2}+k^{2})^{2}}=\zeta(2)\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{2}}-\zeta(4)$$

This has turned out to be rather challenging. At first glance, I thought it may be somewhere along the lines of the famous $$ \sum_{n=1}^{\infty}\frac{1}{n^{2}+k^{2}}=\frac{\pi}{2k}\coth(\pi k)-\frac{1}{2k^{2}}$$ that is often seen in Complex Analysis.

So, I ran the first sum through Maple and it gave me:

$$ \sum_{n=1}^{\infty} \sum_{k=1}^{\infty}\frac{1}{(n^{2}+k^{2})^{2}}=\frac{{\pi}^{2}}{4}\sum_{n=1}^{\infty}\frac{\coth^2(\pi n)}{n^{2}}+\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^{3}}-\frac{{\pi}^{2}}{4}\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^{4}}$$

Of course, two of these are the very familiar $\zeta(2), \;\ \zeta(4)$. I managed to evaluate $\displaystyle \sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^{3}}=\frac{7{\pi}^{3}}{180}$ using Complex Analysis.

The one that has given me the fit is $\displaystyle\sum_{n=1}^{\infty}\frac{\coth^{2}(\pi n)}{n^{2}}$.

This evaluates to $\displaystyle\frac{2}{3}K+\frac{19{\pi}^{2}}{180}$. But, how?.

I tried a Complex Analysis method, but had trouble finding the residues at $ni$, which are the zeroes of $\sinh^{2}(\pi n)$ .

Anyone have any good ideas on how to evaluate the original sum at the top or even just this 'coth-squared' one?. Complex analysis or other wise. I thought maybe a clever use of Fourier would work, but maybe not.

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    It wasn't until I got to the end of the question that I realised "CA" meant "complex analysis"... also I personally think it's unhelpful to put "interesting" in your title – I already assume that your question is interesting, else why would you post it?2012-09-16
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    Ha. I thought CA meant Computer Algebra.2012-09-16
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    See http://meta.math.stackexchange.com/questions/5082/should-we-retitle-posts-whose-titles-contain-interesting.2012-09-16
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    Sorry, I did not realize that was such an issue.2012-09-16

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