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How can we prove the following formula? $$\sum_{k=1}^{\infty}\frac{1}{(mk^2-n)^2}=\frac{-2m+\sqrt{mn}\pi\cot\left(\sqrt{\frac{n}{m}}\pi\right)+n\pi^2\csc^2\left(\sqrt{\frac{n}{m}}\pi\right)}{4mn^2}$$

What is the general method for finding sums of the form $\sum\limits_{k=1}^{\infty}\frac{1}{(mk^2-n)^\ell}, \ell\in\mathbb{N}$?

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  • $\begingroup$ I would first rewrite everything in terms of the quantity $x=\sqrt{n/m}$ and make precise the domain of $x$. Then perhaps transform twice the LHS into a sum over every $k$ in $\mathbb Z$. Then a click might occur... $\endgroup$ – Did Oct 3 '13 at 6:50
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Start with the infinite product expansion

$$\frac{\sin z}{z} = \prod_{k=1}^{\infty}\left(1 - \frac{z^2}{k^2\pi^2}\right)$$

Taking logarithm, substitute $z$ by $\pi\sqrt{x}$ and differentiate with respect to $x$, we find $$ \sum_{k=1}^{\infty} \frac{1}{k^2 - x} = -\frac{d}{dx} \left[ \sum_{k=1}^{\infty}\log\left(1 - \frac{x}{k^2}\right)\right] = -\frac{d}{dx} \left[ \log\left(\frac{\sin(\pi \sqrt{x})}{\pi \sqrt{x}}\right) \right] $$ Differentiate both sides with respect to $x$ for $\ell - 1 $ more times and then divide by $-(\ell-1)!$, we get in general:

$$ \sum_{k=1}^{\infty} \frac{1}{(k^2 - x)^\ell} = -\frac{1}{(\ell-1)!} \frac{d^{\ell}}{dx^{\ell}} \left[ \log\left(\frac{\sin(\pi \sqrt{x})}{\pi \sqrt{x}}\right) \right] $$ In the case $\ell = 2$, the RHS simplifies to $$-\frac{1}{2x^2} + \frac{\pi}{4x}\left( \frac{1}{\sqrt{x}}\cot(\pi\sqrt{x}) + \pi \csc(\pi\sqrt{x})^2 \right)$$ Substitute $x$ by $\frac{n}{m}$ will give you the formula you have for $\ell = 2$. Formula for other $\ell$ can be obtained by taking corresponding number of derivatives.

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    $\begingroup$ You have proven that $$ \sum_{k=1}^{\infty} \frac{1}{k^2 - x} = -\frac{d}{dx} \left[ \log\left(\frac{\sin(\pi \sqrt{x})}{\pi \sqrt{x}}\right) \right] $$ Is there exist a similar formula for $\sum_{k=1}^{\infty} \frac{1}{k^3 - x}?$ $\endgroup$ – user91500 Oct 3 '13 at 8:12
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    $\begingroup$ @Artin I'm not aware of formula like that for $\sum_{k=1}^{\infty} \frac{1}{k^3-x}$. One can probably cook up one by first performing a partial fraction decomposition of $\frac{1}{k^3-x}$ wrt $k$ and then uses the infinite product expansion of gamma function $\frac{1}{\Gamma(z)} = z e^{\gamma z}\prod_{k=1}^{\infty} (1+\frac{z}{k}) e^{-\frac{z}{k}}$. However, the end result will involve derivatives of gamma function at complex points. It doesn't look very useful for practical purposes. $\endgroup$ – achille hui Oct 3 '13 at 12:45
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This sum may be evaluated by considering the following contour integral in the complex plane:

$$\oint_C dz \frac{\pi \cot{\pi z}}{(m z^2-n)^2}$$

where $C$ is a rectangular contour that encompasses the poles of the integrand in the complex plane, up to $z=\pm \left ( N +\frac12\right)$, where we consider the limit as $N \to\infty$.

We note here that we assume that the ratio $n/m$ is not the square of an integer. Now, the contour integral is zero because the individual integrals along each piece of the contour cancel. On the other hand, the contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand. Working this out, we find that

$$\sum_{k=-\infty}^{\infty} \frac{1}{(m k^2-n)^2} = -\sum_{\pm}\operatorname*{Res}_{z=\pm \sqrt{n/m}} \frac{\pi \cot{\pi z}}{(m z^2-n)^2}$$

Since the pole is a double pole, we have

$$\sum_{\pm}\operatorname*{Res}_{z=\pm\sqrt{n/m}} \frac{\pi \cot{\pi z}}{(m z^2-n)^2} =\sum_{\pm} \frac{\pi}{m^2} \left [\frac{d}{dz} \frac{\cot{\pi z}}{(z\pm \sqrt{n/m})^2} \right ]_{z=\pm\sqrt{n/m}} $$

I assume that the reader can take the derivatives and do the subsequent algebra. I get for the sum

$$-\frac{\pi}{m^2} \frac{m}{2 n} \left [\sqrt{\frac{m}{n}}\cot{\left(\pi \sqrt{\frac{n}{m}} \right)}+ \pi \csc^2{\left(\pi \sqrt{\frac{n}{m}} \right)}\right ] $$

So we now have

$$\sum_{k=-\infty}^{\infty} \frac{1}{(m k^2-n)^2} = \frac{\pi}{2 m n} \left [\sqrt{\frac{m}{n}}\cot{\left(\pi \sqrt{\frac{n}{m}} \right)}+ \pi \csc^2{\left(\pi \sqrt{\frac{n}{m}} \right)}\right ] $$

We now exploit the evenness of the summand; the result is

$$\sum_{k=1}^{\infty} \frac{1}{(m k^2-n)^2} = -\frac{1}{2 n^2} + \frac{\pi}{4 m n} \left [\sqrt{\frac{m}{n}}\cot{\left(\pi \sqrt{\frac{n}{m}} \right)}+ \pi \csc^2{\left(\pi \sqrt{\frac{n}{m}} \right)}\right ] $$

The result to be proven follows.

For the general sum

$$\sum_{k=1}^{\infty} \frac{1}{(m k^2-n)^{\ell}}$$

where $\ell \gt 2$ and $\ell \in \mathbb{Z}$, we take the same approach. The residue is an $\ell-1$ derivative of the integrand.

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    $\begingroup$ You are very fast. $\endgroup$ – Felix Marin Oct 3 '13 at 7:20
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    $\begingroup$ @FelixMarin: Thanks. I am very efficient at odd hours when I should be asleep. $\endgroup$ – Ron Gordon Oct 3 '13 at 7:21
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$$ \sum_{k = 1}^{\infty}{1 \over \left(mk^{2} - n\right)^{2}} = {1 \over m^{2}}\sum_{k = 1}^{\infty}{1 \over \left(k^{2} - \mu^{2}\right)^{2}} = {1 \over 2\mu m^{2}}{{\rm d} \over {\rm d}\mu} \sum_{k = 1}^{\infty}{1 \over k^{2} - \mu^{2}}\,, \qquad \mu^{2} \equiv {n \over m} $$


\begin{align} \sum_{k = 1}^{\infty}{1 \over k^{2} - \mu^{2}} &= \sum_{k = 1}^{\infty}{1 \over \left(k + \mu\right)\left(k - \mu\right)} \\[3mm]&= -\,{1 \over \mu^{2}} + {\Psi\left(\mu\right) - \Psi\left(-\mu\right) \over \mu - \left(-\mu\right)} = -\,{1 \over \mu^{2}} + {1 \over 2}\,{\Psi\left(\mu\right) - \Psi\left(-\mu\right) \over \mu} \\[3mm]&=-\,{3 \over 2\mu^{2}} - {\pi\cot\left(\mu\right) \over 2\mu} \end{align} since $$ \Psi\left(\mu\right) - \Psi\left(-\mu\right) = \Psi\left(\mu\right) - \Psi\left(\mu + 1\right) - \pi\cot\left(\pi\mu\right) = -\,{1 \over \mu} - \pi\cot\left(\pi\mu\right) $$
\begin{align} &\sum_{k = 1}^{\infty}{1 \over \left(mk^{2} - n\right)^{2}} \\[3mm] = &\ \bbox[15px,border:1px dotted navy]{\displaystyle\frac{\csc ^2\left( \sqrt{\frac{n}{m}}\right) \left[\vphantom{\LARGE A}\pi m \sqrt{\frac{n}{m}} \sin \left(2 \sqrt{\frac{n}{m}}\right)-6 m \cos \left(2 \sqrt{\frac{n}{m}}\right)+6 m+2 \pi n\right]}{8 mn^{2}}} \end{align}

${\tt\mbox{I need additional information about}}$ $m$ ${\tt\mbox{and}}$ $n$.

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