Double series

[Tolaso J Kos]

Prove that:

$$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n-1}}{n^2+m^2}=\frac{\pi^2}{24}+\frac{\pi \log 2}{8}$$

I don't have a solution.

[r9m]

The number of ways of writing an integer as sum of two square integers (both positive and negative), $\displaystyle r_2(n) = 4\sum\limits_{\substack{d |n\\ d \text{ odd }}} (-1)^{\frac{d-1}{2}}$. Then, for $\mathfrak{Re}(s) > 1$, we have:

\begin{align*}\sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{1}{(j^2+k^2)^s} = \sum\limits_{n=1}^{\infty} \frac{r_2(n)}{n^s} &= 4\sum\limits_{\substack{d \ge 1\\d \text{ odd }}} \sum\limits_{n \equiv 0 (d)} \frac{(-1)^{\frac{d-1}{2}}}{n^s} \\&= 4\sum\limits_{\substack{d \ge 1\\d \text{ odd }}} \sum\limits_{m=1}^{\infty} \frac{(-1)^{\frac{d-1}{2}}}{m^sd^s} \\&= 4\sum\limits_{d \ge 1} \frac{(-1)^{d-1}}{(2d-1)^s}\sum\limits_{m=1}^{\infty} \frac{1}{m^s} = 4\beta(s)\zeta(s)\end{align*}

and similarly, $\displaystyle \sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{(-1)^{j+k}}{(j^2+k^2)^s} = \sum\limits_{n=1}^{\infty} (-1)^n\frac{r_2(n)}{n^s} = -4\beta(s)\eta(s)$.

Now we use the fact that $\displaystyle (1+(-1)^j)(1+(-1)^k) = 4$ only if both $j,k$ are even and vanishes otherwise, we have:

\begin{align*}\frac{16}{4^s}\zeta(s)\beta(s) = \frac{4}{4^s}\sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{1}{(j^2+k^2)^s} &= \sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{1+(-1)^j+(-1)^k+(-1)^{j+k}}{(j^2+k^2)^s}\\&= 4\zeta(s)\beta(s) + 2\sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{(-1)^j}{(j^2+k^2)^s}-4\eta(s)\beta(s)\\ \sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{(-1)^j}{(j^2+k^2)^s} &= -2\left(1-\frac{1}{4^{s-1}}\right)\zeta(s)\beta(s) + 2\eta(s)\beta(s) = -\frac{4}{2^{s}}\eta(s)\beta(s)\\ \sum\limits_{j,k \ge 1} \frac{(-1)^{j-1}}{(j^2+k^2)^s} &= -\frac{1}{2}\eta(2s) + \frac{1}{2}\zeta(2s) + \frac{1}{2^{s}}\eta(s)\beta(s)\end{align*}

and our special case $s = 1$, follow from $\displaystyle \eta(1) = \log 2$, $\displaystyle \zeta(2) = \frac{\pi^2}{6}$ . $\displaystyle \eta(2) = \frac{\pi^2}{12}$ and $\displaystyle \beta(1) = \frac{\pi}{4}$.

[Tolaso J Kos]

Nicely done RD. You know days ago I was playing with some differential equations and a series arose, namely this:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \tanh n}$$

Although this series does not have a nice closed form (it is expressed in terms of Theta Jacobi Functions) the series $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \tanh \pi n}$ does though. One has to invoke the Fourier series of $\coth$ , namely the formula:

\begin{equation} \frac{\pi}{n}\operatorname{coth}(\pi n)=\frac{1}{n^2}+2\sum_{m=1}^{\infty} \frac{1}{n^2+m^2} \end{equation}

Hence

\begin{align*} \sum_{n=1}^{\infty}(-1)^{n-1} \sum_{m=1}^{\infty}\frac{1}{n^2+m^2} &\overset{(1)}{=}\sum_{n=1}^{\infty} (-1)^{n-1} \left [ \frac{\pi \coth \pi n}{2n } - \frac{1}{2n^2} \right ] \\ &=\frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}\pi \coth \pi n}{n} - \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}\\ &=\frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}\pi \coth \pi n}{n} - \frac{1}{2}\eta(2) \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \pi \coth \pi n}{n} - \frac{1}{2}\left ( 1-2^{1-2} \right )\zeta(2)\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \pi \coth \pi n}{n} - \frac{\pi^2}{24} \end{align*}

This means that:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \tanh \pi n} = \frac{\pi}{6} + \frac{\log 2}{4}$$

P.S: nospoon at mathematics stack exchange mentioned that there is the following formula

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \operatorname{coth}(n \pi x) = \frac{\pi x}{6}- \frac13 \ln \left( \frac{\vartheta_3(e^{-\pi x}) \vartheta_4(e^{-\pi x})}{2 \vartheta_2^2(e^{-\pi x})}\right)$$

However, I don't have a proof for that. Here $\vartheta_n$ denote Jacobi theta functions.