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I found this sum in an old math problems book and it asks me to find its closed form. And for the life of me I can't find. Here it is

$$\sum_{p=1}^{\infty}\frac{4^{p+1}\pi^{2p}}{(2p)!}|B_{2p}|\left(\frac{1}{2p+3}+\frac{1}{4p+2}-\frac{3}{4p+4}\right)$$

Any hits or suggestions are welcomed.

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  • $\begingroup$ the first fraction of it make me think of $4cosh(2\pi)$ however I'm not sure how to put the rest with it. What's $|B_{2p}|$? Perhaps writing the partial fraction as on fraction as three different sums might help and try to find three different closed forms of sum? $\endgroup$ – user88595 Nov 10 '13 at 14:53
  • $\begingroup$ $|B_{2p}|$ is the absolute value for the Bernoulli numbers. I've tried splitting it into 3 sums and fiddling around with the bernoulli numbers, but no result. $\endgroup$ – Leonida Nov 10 '13 at 14:57
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    $\begingroup$ The term before the parenthesis is $8\zeta(2p)$, maybe that could lead to something useful? $\endgroup$ – Daniel Fischer Nov 10 '13 at 15:00
  • $\begingroup$ @DanielFischer Good point, I totally missed that. Maybe it helps! $\endgroup$ – Leonida Nov 10 '13 at 15:05
  • $\begingroup$ Seems to be slowly converging to $\approx -0.4$. The 1281-st partial sum is -0.39665. $\endgroup$ – David H Nov 10 '13 at 15:43

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