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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
up vote 6 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
@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

Here is an approach using Mellin transforms to enrich the collection of solutions. We seek to evaluate (assuming that we start the original series at $k=0$ as observed above)

$$f(n) = \sum_{k\ge 1} \frac{(2k-1)^{4n+1}}{1+\exp((2k-1)\pi)},$$ this one started at $k=1$ which corresponds to $k=0$ in the problem statement.

There is a harmonic sum here which we now evaluate by Mellin transform inversion.

Introduce $$S(x) = \sum_{k\ge 1} \frac{((2k-1)x)^{4n+1}}{1+\exp((2k-1)\pi x)}$$ so that we are interested in $S(1).$

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have that $$\lambda_k = 1, \quad \mu_k = 2k-1 \quad \text{and} \quad g(x) = \frac{1}{1+\exp(\pi x)}.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{1}{1+\exp(\pi x)} x^{s-1} dx = \int_0^\infty \frac{\exp(-\pi x)}{1+\exp(-\pi x)} x^{s-1} dx \\= \int_0^\infty \left(\sum_{q\ge 1} (-1)^{q-1} e^{-\pi q x} \right) x^{s-1} dx = \sum_{q\ge 1} (-1)^{q-1} \int_0^\infty e^{-\pi q x} x^{s-1} dx \\= \frac{1}{\pi^s} \Gamma(s) \sum_{q\ge 1} \frac{(-1)^{q-1}}{q^s} = \frac{1}{\pi^s} \left(1 - \frac{2}{2^s}\right)\Gamma(s) \zeta(s).$$

The series that we have used here converges absolutely for $x$ in the integration limits.

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by

$$Q(s) = \frac{1}{\pi^{s+4n+1}} \left(1 - \frac{2}{2^{s+4n+1}}\right)\Gamma(s+4n+1) \zeta(s+4n+1) \left(1 - \frac{1}{2^s} \right) \zeta(s) \\ \text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{(2k-1)^s} = \left(1 - \frac{1}{2^s} \right) \zeta(s)$$ for $\Re(s) > 1.$

To see this note that the base function of the sum is $$\frac{x^{4n+1}}{1+\exp(\pi x)} .$$

The Mellin inversion integral for $Q(s)$ is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero.

As it turns out we only need the contribution from the pole at $s=1.$

We have that $$\mathrm{Res}\left(Q(s)/x^s; s=1\right) = \frac{1}{\pi^{4n+2}} \left(1-\frac{2}{2^{4n+2}}\right) \times (4n+1)! \times \zeta(4n+2)\times \frac{1}{2} \times \frac{1}{x} \\= \frac{1}{\pi^{4n+2}} \frac{2^{4n+2}-2}{2^{4n+2}} \times (4n+1)! \times \frac{(-1)^{(2n+1)+1} B_{4n+2} (2\pi)^{4n+2}}{2\times (4n+2)!} \times \frac{1}{2} \times \frac{1}{x} \\ = (2^{4n+1}-1)\frac{B_{4n+2}}{8n+4} \times \frac{1}{x}.$$

This almost concludes the evaluation the result being the residue we just computed because we can show that $Q(s)/x^s$ with $x=1$ is odd on the line $\Re(s) = -2n$ so that it vanishes and if we stop after shifting to that line the Mellin inversion integral that we started with is equal to the contribution from the pole at $s=1.$

To see this put $s=-2n+it$ to obtain $$\frac{1}{\pi^{2n+1+it}} \left(1 - \frac{2}{2^{2n+1+it}}\right)\Gamma(2n+1+it) \zeta(2n+1+it) \left(1 - \frac{1}{2^{-2n+it}} \right) \zeta(-2n+it)$$ which is $$\frac{1}{\pi^{2n+1+it}} \frac{2^{2n+1+it}-2}{2^{2n+1+it}} \Gamma(2n+1+it) \zeta(2n+1+it) \frac{2^{-2n+it}-1}{2^{-2n+it}} \zeta(-2n+it)$$

Now use the functional equation of the Riemann Zeta function in the following form: $$\zeta(1-s) = \frac{2}{2^s\pi^s} \cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$ to transform the above into $$(2^{2n+1+it}-2) \times \frac{\zeta(1-(2n+1+it))}{2\cos(\pi(2n+1+it)/2)} \frac{2^{-2n+it}-1}{2^{-2n+it}} \zeta(-2n+it)$$ which is $$(2^{2n+it}-1) \times \frac{\zeta(-2n-it))}{\cos(\pi it + \pi(2n+1)/2)} (1-2^{2n-it}) \zeta(-2n+it)$$ which we finally rewrite as $$(2^{2n+it}-1) (1-2^{2n-it}) \frac{(-1)^{n+1}}{\sin(\pi it/2)} \zeta(-2n+it)\zeta(-2n-it)$$ or $$(1-2^{2n+it}) (1-2^{2n-it}) \frac{(-1)^n}{\sin(\pi it/2)} \zeta(-2n+it)\zeta(-2n-it).$$

Among this product of five terms the first two taken together are even as are the last two zeta function terms. The middle sine term is odd in $t$, so the entire product is odd in $t$ and we are done, having obtained the answer $$(2^{4n+1}-1)\frac{B_{4n+2}}{8n+4}.$$

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