# Can I get an independent proof of a closed form of these two related infinite series?

In this issue regarding the Shannon-Whittaker sampling and reconstruction formula (regarding bandlimited reconstruction of periodic discrete-time sampled functions) at the DSP SE, it appears that we have an ancillary result of both:



$$\sum\limits_{m=-\infty}^{\infty} \frac{(-1)^m}{x-mN} \ = \ \frac{\tfrac{\pi}{N}}{\sin\left(\tfrac{\pi}{N} x\right)} \qquad N \in \mathbb{Z}, \ N \text{ odd} \qquad x \in \mathbb{R}$$



$$\sum\limits_{m=-\infty}^{\infty} \frac{1}{x-mN} \ = \ \frac{\tfrac{\pi}{N}}{\tan\left(\tfrac{\pi}{N} x\right)} \qquad N \in \mathbb{Z}, \ N \text{ even} \qquad x \in \mathbb{R}$$

 I tried, but haven't been able to, independently confirm these two mathematical facts except as a consequential bi-product of the above mentioned result.

Can any of you math wiz-bangs derive these two results directly?

I guess I could express it as a single identity:



$$\sum\limits_{m=-\infty}^{\infty} \frac{(-1)^{mN}}{x-mN} \ = \ \frac{\tfrac{\pi}{2N} \left( \cos\left(\tfrac{\pi}{N} x\right) + 1 + (-1)^N \left( \cos\left(\tfrac{\pi}{N} x\right) - 1 \right) \right)}{\sin\left(\tfrac{\pi}{N} x\right)}$$



with $N \in \mathbb{Z}$ and $x \in \mathbb{R}$.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\sum_{m = -\infty}^{\infty}{\pars{-1}^{m} \over x - mN} = {1 \over x} + \sum_{m = 1}^{\infty}\bracks{{\pars{-1}^{m} \over x - mN} + {\pars{-1}^{-m} \over x + mN}} \\[5mm] = &\ {1 \over x} + \sum_{m = 2,\ m\ \mrm{even}}^{\infty}\bracks{{1 \over x - mN} + {1 \over x + mN}} - \sum_{m = 1,\ m\ \mrm{odd}}^{\infty}\bracks{{1 \over x - mN} + {1 \over x + mN}} \\[5mm] = &\ {1 \over x} + \color{#f00}{2}\sum_{m = 2,\ m\ \mrm{even}}^{\infty} \bracks{{1 \over x - mN} + {1 \over x + mN}} - \sum_{m = 1}^{\infty}\bracks{{1 \over x - mN} + {1 \over x + mN}} \label{1}\tag{1} \end{align}

because $\ds{\sum_{m = 1}^{\infty}\cdots = \sum_{m = 2\,,\ m\ \mrm{even}}^{\infty}\cdots + \sum_{m = 1\,,\ m\ \mrm{odd}}^{\infty}\cdots}$ which yields the prefactor $\ds{\color{#f00}{2}}$ in expression \eqref{1}.

Then,

\begin{align} &\sum_{m = -\infty}^{\infty}{\pars{-1}^{m} \over x - mN} = {1 \over x} + 2\sum_{m = 0}^{\infty}\bracks{% {1 \over x - \pars{2m + 2}N} + {1 \over x + \pars{2m + 2}N}} \\[5mm] - &\ \sum_{m = 1}^{\infty}\bracks{% {1 \over x - mN} + {1 \over x + mN}} \\[1cm] = &\ {1 \over x} + {1 \over N}\sum_{m = 0}^{\infty}\bracks{-\,{1 \over m + 1 - x/\pars{2N}} + {1 \over m + 1 + x/\pars{2N}}} \\[5mm] - &\ {1 \over N}\sum_{m = 0}^{\infty}\bracks{-\,{1 \over m + 1 - x/N} + {1 \over m + 1 + x/N}} \\[1cm] = &\ {1 \over x} + {1 \over N}\bracks{\Psi\pars{1 - {x \over 2N}} - \Psi\pars{1 + {x \over 2N}}} - {1 \over N}\bracks{\Psi\pars{1 - {x \over N}} - \Psi\pars{1 + {x \over N}}} \end{align}

where $\ds{\Psi}$ is the Digamma Function.

Then, \begin{align} &\sum_{m = -\infty}^{\infty}{\pars{-1}^{m} \over x - mN} \\[5mm] = &\ {1 \over x} + {1 \over N}\bracks{\Psi\pars{-\,{x \over 2N}} - {2N \over x} - \Psi\pars{1 + {x \over 2N}}} \\[5mm] - &\ {1 \over N}\bracks{\Psi\pars{-\,{x \over N}} - {N \over x} - \Psi\pars{1 + {x \over N}}}\qquad\pars{~Recurrence Property~} \\[1cm] = &\ {1 \over N}\bracks{\Psi\pars{-\,{x \over 2N}} - \Psi\pars{1 + {x \over 2N}}} - {1 \over N}\bracks{\Psi\pars{-\,{x \over N}} - \Psi\pars{1 + {x \over N}}} \\[5mm] = &\ {1 \over N}\bracks{-\pi\cot\pars{\pi\bracks{-\,{x \over 2N}}}} - {1 \over N}\bracks{-\pi\cot\pars{\pi\bracks{-\,{x \over N}}}} \quad\pars{~Euler\ Reflection\ Formula~} \\[5mm] = & {\pi \over N}\bracks{\cot\pars{\pi x \over 2N} - \cot\pars{\pi x \over N}} = \bbx{\ds{\pi/N \over \sin\pars{\pi x/N}}} \end{align}

The other one can be evaluated in the same fashion !!!.

• what'sa Digamma function? – robert bristow-johnson Mar 12 '17 at 4:52
• @robertbristow-johnson It's a 'special function' which has many interesting properties. Check this link. – Felix Marin Mar 12 '17 at 4:55
• please check some of your math, Felix. i think there were a couple errors early on. but fixing those errors will propagate down the derivation. – robert bristow-johnson Mar 12 '17 at 18:27
• @robertbristow-johnson Please,$\color{#f00}{ask\ me}$ to check the answer. The factor $\color{#f00}{two}$ is not a mistake because $\sum_{all} = \sum_{even} + \sum_{odd} \implies \sum_{even} - \sum_{odd} = \sum_{even} - \left(\sum_{all} - \sum_{even}\right) = \color{#f00}{2}\sum_{even} - \sum_{all}$. Now, I have to check your edit. – Felix Marin Mar 13 '17 at 1:06
• sorry, Felix. . – robert bristow-johnson Mar 13 '17 at 3:58