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:
$$ $$
$$ \sum\limits_{m=-\infty}^{\infty} \frac{(-1)^{mN}}{x-mN} \ = \ \frac{\pi }{2N\sin\left(\tfrac{\pi}{N} x\right)} \bigg( \cos\left(\tfrac{\pi}{N} x\right) + 1 + (-1)^N \Big( \cos\left(\tfrac{\pi}{N} x\right) - 1 \Big) \bigg) $$
$$ $$
with $ N \in \mathbb{Z} $ and $ x \in \mathbb{R} $.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\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 !!!.

