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As seen in the title I'm interested in a way to evaluate $$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$$ But I'm not sure what to do, I did attempt something but ended up with a wrong result $$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k-\sqrt{2}n\right)\left(k+\sqrt{2}n\right)}$$ then applying partial fraction decomposition \begin{align*} &=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3\left(k-\sqrt{2}n\right)}+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3\left(k+\sqrt{2}n\right)}+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^2n^2}\\ &=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3}\int _0^1x^{k-\sqrt{2}n-1}\:\mathrm{d}x+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3}\int _0^1x^{k+\sqrt{2}n-1}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\ &=-\sum _{n=1}^{\infty }\int _0^1\operatorname{Li}_3\left(-x\right)x^{-\sqrt{2}n-1}\:\mathrm{d}x-\sum _{n=1}^{\infty }\int _0^1\operatorname{Li}_3\left(-x\right)x^{\sqrt{2}n-1}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\ &=\int _0^1\frac{\operatorname{Li}_3\left(-x\right)}{x\left(1-x^{\sqrt{2}}\right)}\:\mathrm{d}x-\int _0^1\frac{x^{\sqrt{2}}\operatorname{Li}_3\left(-x\right)}{x\left(1-x^{\sqrt{2}}\right)}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\ &=\int _0^1\frac{\operatorname{Li}_3\left(-x\right)}{x}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\ &=-\frac{7}{8}\zeta (4)+\frac{5}{4}\zeta (4)\\ &=\frac{3}{8}\zeta (4) \end{align*} Which seems to be a wrong answer due to numerical methods and approximations, what did I do wrong? how else could I approach this problem?

Salva Fernandez
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2 Answers2

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Let $S_1(n)=\sum\limits_{k=1}^{\infty } \frac{\left ( -1 \right )^{k-1}}{ k^2-2n^2} $, then $S=\sum\limits_{n=1}^{\infty }\left (\sum\limits_{k=1}^{\infty } \frac{\left ( -1 \right )^{k-1}}{n^2\left ( k^2-2n^2 \right )} \right )=\sum\limits_{n=1}^{\infty }\frac{S_1(n)}{n^2}$

Let's consider the integral in the complex plane along a big circle of radius $R$ (counter-clockwise) with the center at $(0;0)$:

$\oint=\oint_C\frac{\pi}{\sin\pi z}\frac{1}{z^2-2n^2}dz=2\pi i \sum Res\frac{\pi}{\sin\pi z}\frac{1}{z^2-2n^2}$ inside the circle

It can be shown that $\oint\to0$ as $R\to\infty\,\,\, \Rightarrow$

$\sum Res\frac{\pi}{\sin\pi z}\frac{1}{z^2-2n^2}=\sum\limits_{k=-\infty}^{\infty } \frac{(-1)^k}{k^2-2n^2} +\frac{\pi}{\sqrt2 n\sin(\sqrt2 \pi n)}=0$

$-\sum\limits_{k=-\infty}^{\infty } \frac{(-1)^k}{k^2-2n^2}=2S_1(n)+\frac{1}{2n^2}=\frac{\pi}{\sqrt2 n\sin(\sqrt2 \pi n)}\,\,\,\Rightarrow\,\,\,S_1(n)=\frac{\pi}{2\sqrt2 n\sin(\sqrt2 \pi n)}-\frac{1}{4n^2}\,\,\,\Rightarrow$ $$S=\sum\limits_{n=1}^{\infty }\frac{\pi}{2\sqrt2 n^3\sin(\sqrt2 \pi n)}-\frac{1}{4}\zeta(4)$$

The evaluation of $\sum\limits_{n=1}^{\infty }\frac{1}{n^3\sin(\sqrt2 \pi n)}=-\frac{13 \pi^3}{360 \sqrt{2}}$ is done here: Compute the series $\sum_{n=1}^{+\infty} \frac{1}{n^3\sin(n\pi\sqrt{2})}.$

NB The approach is the same: integration along the closed contour in the complex plane with the special integrand $\biggl(\frac{1}{z^3\sin(\pi z)\sin(\sqrt2\pi z-\pi z)}\biggr)$

Finally, $$S=-\frac{13\pi^4}{360*4}-\frac{\pi^4}{360}=-\frac{17\pi^4}{1440}$$

Svyatoslav
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    Very nice solution. More generally, let $$S(a) = \sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-an^2\right)}$$ then $S(a)$ is a rational multiple of $\pi^4$ for every non-square rational number $a$, for example $$S(19) = \frac{45199 \pi ^4}{12093120} \qquad S(30)=-\frac{13 \pi ^4}{21600} \qquad S(\frac{11}{13}) = \frac{24703 \pi ^4}{411840}$$ – pisco Jul 05 '21 at 12:18
  • @pisco, thank you for providing a broader view of the problem. I didn't even think that the solution could be so nicely generalized... – Svyatoslav Jul 05 '21 at 14:09
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$$\frac 1{n^2\left(k^2-2n^2\right)}=\frac 1{n^2\left(k-n \sqrt 2\right)\left(k+n \sqrt 2\right)}$$ Using partial fracion decomposition $$\frac 1{n^2\left(k^2-2n^2\right)}=\frac{1}{2 \sqrt{2} n^3}\left(\frac{1}{k-\sqrt{2} n}-\frac{1}{k+\sqrt{2} n}\right)$$ $$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k-\sqrt{2} n}=\frac{1}{2} \left(H_{-\frac{n}{\sqrt{2}}}-H_{-\frac{n}{\sqrt{2}}-\frac{1}{2}}\right)$$ $$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k+\sqrt{2} n}=\frac{1}{2} \left(H_{\frac{n}{\sqrt{2}}}-H_{\frac{n}{\sqrt{2}}-\frac{1}{2}}\right)$$ $$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k-\sqrt{2} n}-\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k+\sqrt{2} n}=\pi \csc \left(\sqrt{2} \pi n\right)-\frac{1}{\sqrt{2} n}$$ $$\frac{1}{2 \sqrt{2}}\sum_{k=1}^\infty \frac 1 {n^3}\left(\pi \csc \left(\sqrt{2} \pi n\right)-\frac{1}{\sqrt{2} n}\right)=\frac{1}{2 \sqrt{2}}\left(-\frac{13 \pi ^3}{360 \sqrt{2}}-\frac{\pi ^4}{90 \sqrt{2}}\right)=-\frac{\pi ^4}{360}-\frac{13 \pi ^3}{1440}$$

Claude Leibovici
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