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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

1 Answer

up vote 5 down vote accepted

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
1  
@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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