# Property of sum $\sum_{k=1}^{+\infty}\frac{(2k+1)^{4n+1}}{1+\exp{((2k+1)\pi)}}$

Is it true that for all $n\in\mathbb{N}$, \begin{align}f(n)=\sum_{k=1}^{+\infty}\frac{(2k+1)^{4n+1}}{1+\exp{((2k+1)\pi)}}\end{align} is always rational. I have calculated via Mathematica, which says \begin{align}f(0)=\frac{1}{24},f(1)=\frac{31}{504},f(2)=\frac{511}{264},f(3)=\frac{8191}{24}\end{align} But I couldn't find the pattern or formula behind these numbers, Thanks for your help!

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What is the origin of this sum? –  vesszabo Oct 18 '12 at 14:42
@vesszabo Emm..My friend asked me via forum, I still wonder what was his intention to study this series. –  Golbez Oct 18 '12 at 14:49
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## 1 Answer

This series appears in Apostol's book "Modular Functions and Dirichlet Series in Number Theory" p.25 according to (8) from MathWorld with the result (if your series starts with $k=0$) : $$f(n)=\frac {2^{4n+1}-1}{8n+4}\,B_{4n+2}$$ with $B_n$ a Bernoulli number.

UPDATE: Apostol's book may be consulted here and the theorem $13.17$ is the proof of the classical relation between $\zeta(2n)$ and $B_{2n}$.

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Great!(+2) if it was possible. –  vesszabo Oct 18 '12 at 17:39
@vesszabo: Glad you liked it! For other results of this kind see 'zeta constant' and Vepstas' paper 'On Plouffe's Ramanujan Identities'. –  Raymond Manzoni Oct 18 '12 at 20:17
I know Bernoulli numbers and it was suspicious that the sum has relation with B numbers (because in the denominator there is exp function), but I couldn't find it. Thanks for the links. –  vesszabo Oct 19 '12 at 8:31
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