TITLE: Property of sum $\sum_{k=1}^{+\infty}\frac{(2k+1)^{4n+1}}{1+\exp{((2k+1)\pi)}}$
QUESTION [15 upvotes]: 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!

REPLY [18 votes]: 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
$$(2^{4n+1}-1)\frac{B_{4n+2}}{8n+4}.$$