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!

Solutions Collecting From Web of "Property of sum $\sum_{k=1}^{+\infty}\frac{(2k+1)^{4n+1}}{1+\exp{((2k+1)\pi)}}$"

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 $B_{2n}$.

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
$$(2^{4n+1}-1)\frac{B_{4n+2}}{8n+4}.$$