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.

Answer 2

Here is a proof based on Hachino's idea. We have agreed it's enough to prove that $$S=\sum_{m=1}^\infty\frac{(-1)^m\tanh(m\pi /2)}{m}=\frac{\log 2-\pi}{4}. $$ Plug in the expansion $$\tanh(m\pi /2)=\frac{1-e^{-m\pi }}{1+e^{-m\pi }}=1+2\sum_{n=1}^\infty(-1)^ne^{-nm\pi }. $$ Changing order of summation and using the Taylor series of $\log(1+x)$ gives $$S=-\log 2+2\sum_{n=1}^\infty(-1)^{n+1}\log(1+e^{-n\pi })=\log\left(\frac{\prod_{n=1}^\infty(1+e^{-(2n-1)\pi})^2}{2\prod_{n=1}^\infty(1+e^{-2n\pi})^2}\right).$$ The infinite products can be recognized as theta values. Recall that $$\theta_2(z,q)=2q^{1/4}\cos z\prod_{n=1}^\infty(1-q^{2n})(1+2q^{2n}\cos 2z+q^{4n}), $$ $$\theta_3(z,q)=\prod_{n=1}^\infty(1-q^{2n})(1+2q^{2n-1}\cos 2z+q^{4n-2}). $$ There are lots of conflicting notations for theta functions; I use the conventions of Whittaker and Watson, A course in modern analysis. It follows that $$S=\log\left(e^{-\pi/4}\frac{\theta_3(0,e^{-\pi})}{\theta_2(0,e^{-\pi})}\right). $$ It remains to show that $$\frac{\theta_3(0,e^{-\pi})}{\theta_2(0,e^{-\pi})}=2^{1/4}. $$ This is clear from generalities on elliptic functions. Again in the notation of Whittaker and Watson, the modulus $k$ is given by $$k=\frac{\theta_2(0,q)^2}{\theta_3(0,q)^2}. $$ The value $q=e^{-\pi}$ corresponds to $k=k'$, where $k'$ is the complementary modulus defined by $k^2+(k')^2=1$. Thus, $k=1/\sqrt 2$, which gives the desired result.

If you had a tough childhood, you might worry that I changed the order of summation in series that are not absolutely convergent. You might want to plug in a convergence factor $x^m$, repeat the argument above and prove that both sides of the resulting identity are left continuous at $x=1$.

Answer 3

Well, using WolframAlpha, one has the identity

\begin{equation} \sum_{k=0}^{\infty} \frac{1}{k^2 + a^2} = \frac{\pi a \coth (\pi a) + 1}{2a^2} \end{equation}

which, after separating the sum into odd/even contributions and factorizing the terms, leads to :

Edit : corrected after Hjalmar's remark

\begin{equation} \sum_{k=0}^\infty\frac 1{(2k+1)^2+m^2}=\frac{\pi\tanh(\pi m/2)}{4m} \end{equation}

Answer 4

So I think this can be done, in two steps:

(1) understand sums of the 1/Q(k, m) for certain integral quadratic forms Q; and

(2) express this sum as a linear combination of such sums.

The first part is classical number theory, while the second part is elementary, though in general the answers in (1) will come from double sums over all pairs of integers (positive and negative), obviously omitting (0,0).

The simplest case for (1) is where Q is the square of the distance of the lattice point (k,m) to the origin. This actually diverges ... so while this is all perfectly standard number theory, you need to be taking care that the linear combination cancels the divergences. So the Dedekind zeta function of the Gaussian integers, and related functions, should be invoked.