12
$\begingroup$

Consider the following series: $$ S = \sum_{\text{odd } n} \sum_{\text{odd } m} \frac{(-1)^{(n+m)/2}}{nm} \frac{\sinh( \pi \sqrt{n^2 + m^2}/2)}{\sinh( \pi \sqrt{n^2 + m^2})} $$ From the physical context, one can argue that the series should converge to $$ S = - \frac{\pi^2}{96}, $$ and numerically calculating the first 10–20 terms of this series seems to show rapid convergence to this value.

Is it possible to prove this statement without appealing to a physical argument? I honestly have no idea how to begin tackling a problem like this.

$\endgroup$
  • $\begingroup$ A 1D version of this sum is : $\frac{\pi}{8}=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)\cosh\frac{(2n+1)\pi}{2}}$ (link) $\endgroup$ – Paul Enta Apr 10 '16 at 21:53
  • 1
    $\begingroup$ Can you explain the physical context of guessing the value for $S$? It would also be great to know where this series comes from. $\endgroup$ – D. S. Park Apr 25 '16 at 16:05
12
$\begingroup$

Here is a solution that I have found while working on other lattice sums. It utilizes a very simple result:

Define $f$ by $$f(x)=\sum_{n=0}^{\infty} (-1)^n (2n+1) e^{-\pi x (n+\frac12)^2}.$$ Then $$\int_0^{\infty} e^{-y x} f(x)dx=\operatorname{sech}\sqrt{\pi y}.\tag{$\star$}$$

The proof is simple - just integrate term by term, and the result follows from the series $$\sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)}{(2n+1)^2+(2x)^2}=\frac{\pi}{4} \operatorname{sech} \pi x.$$

Therefore, $$\large \operatorname{sech} \frac{\pi}{2} \sqrt{n^2+m^2} = \int_0^{\infty} e^{-\frac{\pi}{4} x(n^2+m^2)} f(x)dx.$$ Now put $(2n+1)$ and $(2m+1)$ instead of $n$ and $m$, multiply by $\displaystyle \, \frac{(-1)^{n+m}}{(2n+1)(2m+1)},$ and sum both $n$ and $m$ over the non-negative integers:

$$ 2 S=\sum_{n,m=0}^{\infty} \frac{(-1)^{n+m}}{(2n+1)(2m+1)} \operatorname{sech}\left(\frac{\pi}{2} \sqrt{(2n+1)^2+(2m+1)^2}\right) = \int_0^{\infty} f(x) \left(\sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} \right)^2 dx$$

Here comes the neat part: we notice that $$\frac{d}{dx} \sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} = -\frac{\pi}{4} f(x),$$ and conclude that $$ 2 S = \frac13 \left(-\frac{4}{\pi}\right) \left(\sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} \right)^3\Biggr{|}_0^\infty \\= \frac13 \left(\frac{4}{\pi}\right) \left(\frac{\pi}{4}\right)^3=\frac{\pi^2}{48}.$$

Can you see how nicely it generalizes to higher dimension versions of this sum?

Note: The function $f(x)$ is actually $\eta^3(i x)$ where $\eta$ is the Dedekind eta function (This is a consequence of Jacobi's triple product identity). The result $(\star)$ was obtained by Glasser, along with other beautiful formulas for integrals involving $\eta(i x)$, in his article Some Integrals of the Dedekind Eta-function (2008) (arxiv link). I omitted this detail simply becuase it is not important for this purpose.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ We can generalize this result using the series (with $\theta\in [0,\pi/2]$) $$\displaystyle \sum_{n \geq 1} \frac{(-1)^{n+1}n\sin(\theta n)}{n^2+x^2}=\frac{\pi}{2} \frac{\sinh \theta x}{\sinh \pi x}.$$ For instance, the 2D case is $$\sum_{n,m=1}^{\infty} \frac{(-1)^{n+m}}{n m} \sin(\theta n) \sin(\theta m) \frac{\sinh \theta\sqrt{n^2+m^2}}{\sinh \pi \sqrt{n^2+m^2}} =\frac{\theta^3}{12 \pi}.$$ Note the curious case when $\theta=1$. Again, this easily generalizes to series of higher dimensions. $\endgroup$ – nospoon Jun 4 '16 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy