# An alternating series identity with a hidden hyperbolic tangent

How does one use the inverse Mellin transform to prove that the following identity holds?

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n(e^{n\pi} + 1)} = \frac{1}{8}(\pi - 5\log(2))$$

The identity follows from MO189199 and the penultimate identity on this page (for $x=\frac{1}{2}$).

Scratch-work: I computed the Mellin transform of

$$f(x) = \frac{(-1)^x}{x(e^{x\pi} + 1)}$$

and re-wrote the function in terms of its inverse Mellin Transform as (substituting $x = n$)

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n(e^{n\pi} + 1)} = \frac{1}{2\pi i}\int_{C} (2 \pi)^{1-s} \Gamma(s-1)\big(\zeta(s-1, \frac{1}{2} - \frac{i}{2}) - \zeta(s-1, 1-\frac{i}{2})\big)\zeta(s)ds$$

But I am not altogether confident that this has been computed correctly, and I am still not sure how to find the poles and compute the residues. For example, checking for a residue at $s=0$ gives

$$\frac{1}{8}(\pi - 2\pi i)$$

There seems to be a nontrivial contribution at $s = 1$ and $s = -1$ as well, but already the mathematics has exceeded what I understand; I am especially unclear with regard to how one accounts for residues with respect to the generalized Riemann zeta function $\zeta(s,a)$.

Clarity on using this approach or another to demonstrate the main identity would be appreciated!

• @CAFosta The series certainly converges: Use the Alternating series test. In particular, the series converges to the right-hand side, i.e., $\frac{1}{8}(\pi - 5\log(2))$. My question is how to prove this identity holds (ideally without appealing to the two links in the body of my post!). – Benjamin Dickman Jan 23 '15 at 10:40
• See this. – Mhenni Benghorbal Jan 23 '15 at 11:02
• @MhenniBenghorbal Could you add more details? The scenario here is somewhat different from your linked example insofar as the inverse Mellin transform includes the generalized Riemann zeta function (which has tangled me up in finding poles/computing residues). – Benjamin Dickman Jan 23 '15 at 19:14

Using$$\tanh\left(x\right)=1+2\underset{n=1}{\overset{\infty}{\sum}}\frac{\left(-1\right)^{n}}{e^{2nx}}$$ we have$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n\left(e^{\pi n}+1\right)}=\frac{1}{2}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}\left(1-\tanh\left(\frac{\pi n}{2}\right)\right)}{n}=\frac{1}{2}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n}-\frac{1}{2}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n}\tanh\left(\frac{\pi n}{2}\right)=$$ $$=-\frac{1}{2}\log\left(2\right)-\frac{1}{2}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n}\tanh\left(\frac{\pi n}{2}\right).$$ Now we using the identity$$\tanh\left(\frac{\pi x}{2}\right)=\frac{4x}{\pi}\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{2}+x^{2}}$$ and so we have$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n\left(e^{\pi n}+1\right)}=-\frac{1}{2}\log\left(2\right)-\frac{2}{\pi}\underset{n\geq1}{\sum}\left(-1\right)^{n}\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{2}+n^{2}}=-\frac{1}{2}\log\left(2\right)-\frac{2}{\pi}\underset{k\geq1}{\sum}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{\left(2k-1\right)^{2}+n^{2}}.$$ The last sum can be calculated$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{\left(2k-1\right)^{2}+n^{2}}=\frac{\left(2k-1\right)\pi-\sinh\left(\left(2k-1\right)\pi\right)}{2\left(2k-1\right)^{2}\sinh\left(\left(2k-1\right)\pi\right)}$$ hence$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n\left(e^{\pi n}+1\right)}=-\frac{1}{2}\log\left(2\right)+\frac{1}{\pi}\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{2}}-\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)\sinh\left(\left(2k-1\right)\pi\right)}.$$ Obviously$$\underset{k\geq1}{\sum}\frac{1}{\left(2k-1\right)^{2}}=\frac{1}{8}\pi^{2}$$ and using$$\frac{\pi}{\sinh\left(\pi x\right)}=\frac{2\pi}{e^{\pi x}-e^{-\pi x}}=x\underset{m\in\mathbb{Z}}{\sum}\frac{\left(-1\right)^{m}}{x^{2}+m^{2}}$$ we have$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n\left(e^{\pi n}+1\right)}=-\frac{1}{2}\log\left(2\right)+\frac{1}{8}\pi-\frac{1}{\pi}\underset{k\geq1}{\sum}\underset{m\in\mathbb{Z}}{\sum}\frac{\left(-1\right)^{m}}{\left(2k-1\right)^{2}+m^{2}}$$ and the last sum is (you can find the proof here closed form for a double sum) is$$-\frac{1}{\pi}\underset{k\geq1}{\sum}\underset{m\in\mathbb{Z}}{\sum}\frac{\left(-1\right)^{m}}{\left(2k-1\right)^{2}+m^{2}}=-\frac{\log\left(2\right)}{8}$$ and this complete the proof.
• @BenjaminDickman There is a nontrivial contribution in $1,0$ and $−(2k−1),k=1,2…$ So the residues calculation should be a series but I didn't work on it. – Marco Cantarini Jan 26 '15 at 13:50
• We can also notice that the last sum is related with the number of representations of an integer as a sum of two squares, hence with the Dirichlet function $L(\chi_4,s)$, as pointed by Lucia on MO (mathoverflow.net/questions/189199/…) – Jack D'Aurizio Jan 29 '15 at 13:09
It is quite easy to prove that: $$\sum_{n=1}^{+\infty}\frac{\cos(\pi n)}{n} e^{-nk\pi} = -\log\left(1+e^{-k\pi}\right)$$ hence the original sum equals: $$-\log\prod_{k=1}^{+\infty}\frac{1+e^{-(2k-1)\pi}}{1+e^{-2k\pi}}.$$ The last product is clearly related with the Jacobi triple product. In particular, since the values of the Jacobi theta functions $\vartheta_3$ and $\vartheta_4$ in $q=e^{-\pi}$ are known, the claim simply follows.