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I got this answer from WolframAlpha. Does anyone know how even to approach it to obtain the solution using digamma function. Please don't solve it, just show me in the right direction!

$$ \sum_{k=1}^{n}{1 \over n^{2} - 4\left(k - 1\right)^{2}} ={\pi n\cot\left(\pi n/2\right) - n\Psi\left(n/2\right) + n\Psi\left(3 n/2\right) + 2 \over 4 n^{2}} $$

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Here is a series for digamma. Look for a similar series for cot. (I hope I didn't reveal too much :)

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{k = 1}^{n}{1 \over n^{2} - 4\pars{k - 1}^{2}} ={\pi n\cot\pars{\pi n/2} - n\Psi\pars{n/2} + n\Psi\pars{3 n/2} + 2 \over 4 n^{2}}:\ {\large ?}}$

\begin{align} &\color{#c00000}{\sum_{k = 1}^{n}{1 \over n^{2} - 4\pars{k - 1}^{2}}} =-\,{1 \over 4}\sum_{k = 0}^{n - 1}{1 \over k^{2} - n^{2}/4} =-\,{1 \over 4}\sum_{k = 0}^{n - 1}{1 \over \pars{k + n/2}\pars{k - n/2}} \\[3mm]&=-\,{1 \over 4}\sum_{k = 0}^{\infty}\bracks{% {1 \over \pars{k + n/2}\pars{k - n/2}} - {1 \over \pars{k + 3n/2}\pars{k + n/2}}} \\[3mm]&=-\,{1 \over 4}\, {\Psi\pars{n/2} - \Psi\pars{-n/2} \over \pars{n/2} - \pars{-n/2}} + {1 \over 4}\, {\Psi\pars{3n/2} - \Psi\pars{n/2} \over \pars{3n/2} - \pars{n/2}} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function. See ${\bf\mbox{6.3.1}}$ in that link.

$$ \color{#c00000}{\sum_{k = 1}^{n}{1 \over n^{2} - 4\pars{k - 1}^{2}}} ={\Psi\pars{3n/2} - 2\Psi\pars{n/2} + \Psi\pars{-n/2} \over 4n}\tag{1} $$

With formula ${\bf\mbox{6.3.7}}$ we'll get $$ \Psi\pars{-\,{n \over 2}} = \Psi\pars{1 + {n \over 2}} -\pi\cot\pars{\pi\bracks{-\,{n \over 2}}} $$

Moreover, $\ds{\Psi\pars{1 + n/2} = \Psi\pars{n/2} + 1/\pars{n/2}}$ where we used the identity ${\bf\mbox{6.3.5}}$ such that: $$ \Psi\pars{-\,{n \over 2}} =\Psi\pars{n \over 2} + {2 \over n} + \pi\cot\pars{\pi n \over 2} $$

The final answer is found by replacing this result in expression $\pars{1}$.

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