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

Question

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!

Answer 1

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

Answer 2

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 $B_{2n}$.

Answer 3

Integrate the function $$g(z) = \frac{z^{4n+1}e^{iz}}{\cosh(z) + \cos(z)} $$ around the contour $[-\sqrt{2} \pi N, \sqrt{2} \pi N] \cup \sqrt{2} \pi Ne^{i[0, \pi]}, $ where $N$ is a positive integer.

This is the same approach that I used in this answer to show the $n=0$ case $$\sum_{k=0}^{\infty} \frac{2k+1}{e^{(2k+1)\pi }+1} = \frac{1}{24}.$$

The integral vanishes on the semicircle as $N \to \infty$ through the positive integers for the same reasons explained in my previous answer.

So we have

$$ \begin{align} &\int_{-\infty}^{\infty} \frac{x^{4n+1}e^{ix}}{\cosh(x) + \cos(x)} \, \mathrm dx \\&= 2 \pi i \left(\sum_{k=0}^{\infty}\operatorname{Res} \left[g(z), \frac{(2k+1) \pi (1+i)}{2} \right] + \sum_{k=0}^{\infty}\operatorname{Res} \left[g(z), \frac{(2k+1) \pi (-1+i)}{2} \right]\right) \\ &= 2 \pi i \left(\sum_{k=0}^{\infty} \lim_{z \to \frac{(2k+1) \pi (1+i)}{2} }\frac{z^{4n+1}e^{iz}}{\sinh(z) - \sin(z)} + \sum_{k=0}^{\infty} \lim_{z \to \frac{(2k+1) \pi (-1+i)}{2} }\frac{z^{4n+1}e^{iz}}{\sinh(z) - \sin(z)} \right) \\ &= \small 2 \pi i \left( \frac{\pi }{2} \right)^{4n+1} (-1)^{n} 2^{2n} \left(\sum_{k=0}^{\infty} \frac{(2k+1)^{4n+1}(1+i) i e^{-(2k+1) \pi /2}}{(-1+i)\cosh \left(\frac{(2k+1) \pi }{2} \right)} \color{red}{-} \sum_{k=0}^{\infty} \frac{(2k+1)^{4n+1}(-1+i) i e^{-(2k+1) \pi /2}}{ (1+i) \cosh \left(\frac{(2k+1) \pi }{2} \right)}\right) \\&= 2 \pi i \left( \frac{\pi }{2} \right)^{4n+1} (-1)^{n} 2^{2n} \left( \sum_{k=0}^{\infty} \frac{(2k+1)^{4n+1}e^{-(2k+1) \pi/2}}{\cosh \left(\frac{(2k+1) \pi }{2}\right)}+ \sum_{k=0}^{\infty} \frac{(2k+1)^{4n+1}e^{-(2k+1) \pi/2}}{\cosh \left(\frac{(2k+1) \pi }{2}\right)} \right) \\ &= \frac{(-1)^{n}\pi^{4n+2}}{2^{2n-1}} i \sum_{k=0}^{\infty} \frac{(2k+1)^{4n+1} e^{- (2k+1) \pi /2}}{\cosh \left(\frac{(2k+1)\pi }{2} \right)} \\ &= \frac{(-1)^{n}\pi^{4n+2}}{2^{2n-2}} i \sum_{k=0}^{\infty} \frac{(2k+1)^{4n+1}} {e^{(2k+1)\pi }+1}. \end{align} $$

Then equating the imaginary parts on both sides of the equation, we have

$$ \begin{align} \sum_{k=0}^{\infty} \frac{(2k+1)^{4n+1}}{e^{(2k+1)\pi}+1} &= \frac {(-1)^{n} 2^{2n-2}}{\pi^{4n+2}} \int_{-\infty}^{\infty} \frac{x^{4n+1} \sin(x)}{\cosh(x)+\cos(x)} \, \mathrm dx \\& = \frac{(-1)^{n} 2^{2n-1}}{\pi^{4n+2}} \int_{0}^{\infty} \frac{x^{4n+1} \sin(x)}{\cosh(x)+\cos(x)} \, \mathrm dx \\ &= \frac{(-1)^{n} 2^{2n}}{\pi^{4n+2}} \Im \int_{0}^{\infty} x^{4n+1} \sum_{m=1}^{\infty} (-1)^{m-1} e^{(-1+i)mx} \, \mathrm dx \\ &= \frac{(-1)^{n} 2^{2n}}{\pi^{4n+2}} \Im \sum_{m=1}^{\infty} (-1)^{m-1} \int_{0}^{\infty} x^{4n+1} e^{(-1+i)mx} \, \mathrm dx \\ &\overset{\spadesuit}{=} \frac{(-1)^{n} 2^{2n}}{\pi^{4n+2}} \Im \sum_{m=1}^{\infty} (-1)^{m-1} \frac{(4n+1)!}{\left((1-i)m\right)^{4n+2}} \\ &= \frac{(4n+1)!}{2 \pi^{4n+2}} \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^{4n+2}} \\ &= \frac{(4n+1)!}{2 \pi^{4n+2}} \eta(4n+2) \\ &\overset{\clubsuit}{=} \frac{(4n+1)!}{2 \pi^{4n+2}} \frac{B_{4n+2} \pi^{4n+2}}{(4n+2)!} \left(2^{4n+1}-1 \right) \\ &= \frac{B_{4n+2}}{8n+4} \left(2^{4n+1}-1 \right). \end{align}$$

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$\spadesuit$ $\int_{0}^{\infty} x^{a}e^{-sx} \, \mathrm dx = \frac{\Gamma(a+1)}{s^{a+1}}, \quad (a >-1, \ \Re(s) >0) $

$\clubsuit$ https://en.wikipedia.org/wiki/Dirichlet_eta_function#Particular_values