[Math] How to calculate the infinite sum of this double series

Question

I'm calculating this double sum:
$$
\sum _{m=1}^{\infty } \sum _{k=0}^{\infty } \frac{(-1)^m}{(2 k+1)^2+m^2}
$$
I know the answer is
$$
\frac{ \pi \log (2)}{16}-\frac{\pi ^2}{16}
$$
which can be verified by numerical calculations. I used the Taylor expansions of $log (1+x)$ and $arcsin (x)$ at x=1 to replace $log (2)$ and $\pi$. I got
$$
\sum _{m=1}^{\infty } \sum _{k=0}^{\infty } \frac{(2 m+1) (-1)^{k+m}}{4 m(2 k+1) (2 m-1)}
$$
I tried to play with the dummy variables but failed.
Any idea?

Answer 1

Consider for any real number $s>1$ the double series $$ \sum_{m=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^m}{((2k+1)^2+m^2)^s}. \tag{1} $$ This double series converges absolutely. Moreover for any given $m$, the inner series over $k$ converges to some $C(m,s)$, say, which for any given $s \ge 1$ is a monotone decreasing function of $m$. Combined with (the idea behind the) alternating series test, it is not hard to justify that the series in (1) tends to our desired sum as $s\to 1^+$.

Now let $R(n)$ denote the number of ways of writing $n$ as $a^2+b^2$ with $a$ and $b$ integers; note $R(1)=4$ since $1=1^2+0^2= (-1)^2+0^2=0^2+1^2=0^2+(-1)^2$. It is well known that $$ \sum_{n=1}^{\infty} \frac{R(n)}{n^s} = 4\zeta(s) L(s,\chi_{-4}), $$ and that $$ \sum_{n=1, \text{n odd}}^{\infty} \frac{R(n)}{n^s} = 4 \Big(1-\frac{1}{2^s}\Big) \zeta(s)L(s,\chi_{-4}). $$ Here $L(s,\chi_{-4})=1/1^s-1/3^s+1/5^s-1/7^s+\ldots$ is the Dirichlet $L$-function for the character $\pmod 4$.

Now let us return to the sum in (1). Write $(2k+1)^2+m^2=n$. If $n$ is odd, then $m$ is necessarily even, and this number is counted in (1) a total of $R(n)/8$ times; the only exceptions are when $n$ is an odd square, where the solutions $(2k+1)^2+0^2$ are not counted. So the contribution of the odd numbers $n$ to (1) is $$ \frac{1}{8} \sum_{n \text{odd} } \frac{R(n)}{n^s} - \frac{1}{2} \sum_{k=0}^{\infty} \frac{1}{(2k+1)^{2s}}= \frac{1}{2} \Big(1-\frac{1}{2^s}\Big) \zeta(s)L(s,\chi_{-4}) - \frac 12 \Big(1-\frac{1}{2^{2s}}\Big) \zeta(2s). $$

Now consider the contribution of $n\equiv 2\pmod 4$ to (1). These are the terms with $m$ odd, and they appear with sign $-1$. Here each $n$ appears $R(n)/4$ times. So these terms give $$ -\frac{1}{4} \sum_{n\equiv 2\pmod 4} \frac{R(n)}{n^s} = -\frac{1}{4} \frac{1}{2^s} \sum_{\ell \text{ odd} }\frac{R(\ell)}{\ell^s} = -\frac{1}{2^{s}}\Big(1-\frac{1}{2^s}\Big)\zeta(s) L(s,\chi_{-4}), $$ upon writing $n=2\ell$ with $\ell$ odd, and using here $R(n)=R(\ell)$.

Thus our sum in (1) equals $$ \Big(\frac12 -\frac{1}{2^s}\Big) \Big(1-\frac{1}{2^s}\Big) \zeta(s) L(s,\chi_{-4}) - \frac 12 \Big(1-\frac{1}{2^{2s}}\Big) \zeta(2s). $$ Now let $s\to 1^+$ and use $\zeta(2)=\pi^2/6$, $L(1,\chi_{-4})=\pi/4$ (Gregory's formula), and $(1/2-1/2^s) \zeta(s) \to (\log 2)/2$ (zeta has a simple pole with residue $1$ at $1$). The claimed identity follows.