Double summation

Evaluate the sum

$\mathcal{S} = \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2-j^2}{\left ( n^2+j^2 \right )^2}$

Solution

Lemma: It holds that

$\sum_{m=1}^{\infty} \frac{1}{\sinh^2 m \pi} = \frac{1}{6} - \frac{1}{2\pi}$

Proof: Consider the function $\displaystyle f(z)=\frac{\cot \pi z}{\sinh^2 \pi z}$ and let us it integrate over the following contour $\gamma$

By the residue theorem it follows that

$\oint \limits_{\gamma} f(z) \, \mathrm{d}z = 2 \pi i \sum \mathfrak{Res}_{z_k} f(z)$

For the residues we have

$\begin{matrix} \displaystyle \mathfrak{Res}_{z=n}{\frac{\cot\pi z}{\sinh^2\pi z}} &= & \displaystyle \frac{1}{\pi\sinh^2\pi n} \\\\ \displaystyle \mathfrak{Res}_{z=ni}{\frac{\cot\pi z}{\sinh^2\pi z}} &= &\displaystyle \frac{1}{\pi\sinh^2\pi n} \\\\ \displaystyle \mathfrak{Res}_{z=0}{\frac{\cot\pi z}{\sinh^2\pi z}} & = &\displaystyle -\frac{2}{3\pi} \end{matrix}$

The integrals along the sides vanish; hence:

$-2i\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2\pi x}=2\pi i\left(-\frac{2}{3\pi}+\frac{4}{\pi}\sum_{n=1}^\infty\frac{1}{\sinh^2\pi n}\right)$

and since

$\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2\pi x}=\left[ \frac{1}{\pi}\tanh\pi x \right]_{-\infty}^\infty=\frac{2}{\pi}$

the result follows. $\blacksquare$

Back to the problem. We have successively:

$\begin{align*} \mathcal{S} &= \frac{1}{2} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \left ( 1 + \frac{j^2}{n^2} \right ) \\ &=\frac{1}{2}\sum_{j=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \prod_{n=1}^{\infty} \left ( 1 + \frac{j^2}{n^2} \right ) \\ &=\frac{1}{2} \sum_{j=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \frac{\sinh \pi j }{\pi j} \\ &=\frac{1}{2} \sum_{j=1}^{\infty} \left ( \frac{1}{j^2} - \frac{\pi^2}{\sinh^2 \pi j} \right ) \\ &= \frac{\pi}{4} \end{align*}$

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One thought on “Double summation”

Tolaso says:

July 9, 2020 at 1:09 am

A related series is the following:

$\begin{align*} \sum_{n=1}^{\infty}\frac{1}{\sinh^2 n \pi }&=4\sum_{n=1}^{\infty}\frac{e^{-2\pi n}}{(1-e^{-2\pi n})^2} \\ &=4\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}k e^{-2\pi n k} \\ &=4\sum_{k=1}^{\infty}k\left(\sum_{n=1}^{\infty}e^{-2\pi n k}\right) \\ &=4\sum_{k=1}^{\infty}\frac{k}{e^{2\pi k}-1} \end{align*}$

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