Calculate the sum of $\lim_{N\rightarrow \infty} \sum_{-N}^N 1/(n+e)$ Problem

Calculate the sum of $S = \lim_{N\rightarrow \infty} \sum_{-N}^N 1/(n+e)$

My attempt
We quickly realize that $\frac{1}{e+n} + \frac{1}{e-n} = \frac{2e}{e^2-n^2} = \frac{-2e}{n^2+(ie)^2}$. Hence, the sum can be rewritten as:
$$ S = 1/e + \sum_{n\geq 1} \frac{-2e}{n^2+(ie)^2} = 1/e - 2e \sum_{n\geq 1} \frac{1}{n^2+(ie)^2}$$
We also know that the Fourier series of $e^{bx}$ on $(-\pi, \pi)$ is given by:
$$ e^{bx} = \frac{\sinh(b\pi)}{\pi} \sum_{n\in\mathbb{Z}} \frac{(-1)^n}{b-in}e^{inx}$$
for $x \in(-\pi, \pi)$. By applying Parsevals formula we get:
$$ 2 \pi = \frac{\sinh^2(b\pi)}{\pi^2}\sum_{n\in\mathbb{Z}} \frac{1}{b^2+n^2} \cdot 2\pi $$
Which after rearranging gives us:
$$ \frac{\pi^2}{\sinh^2(b\pi)} = \sum_{n\in\mathbb{Z}} \frac{1}{b^2+n^2} = 1/b^2 + 2\sum_{n\geq1} \frac{1}{b^2+n^2} $$
Plugging in $b = ie$ and utilizing the fact that $\sinh(ix) = i\sin(x)$ then gives us that:
$$2\sum_{n\geq1} \frac{1}{(ie)^2+n^2} =  \frac{\pi^2}{-\sin^2(\pi e)} + 1/e^2 $$
This can be substituted back into our sought sum $S$:
$$ S = 1/e - e \left( \frac{\pi^2}{-\sin^2(\pi e)} + 1/e^2 \right) = 1/e + \frac{e\pi^2}{\sin^2(\pi e)} - 1/e = \frac{e\pi^2}{\sin^2(\pi e)} $$
Which after calculating the original sum numerically is nowhere near the correct answer. What's going wrong here exactly?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
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\begin{align}
\color{#44f}{S} & \equiv\color{#44f}{\lim_{N \to \infty}\
\sum_{n\ =\ -N}^{N}\ {1 \over n + \expo{}}} =
\lim_{N \to \infty}\pars{%
\sum_{n\ =\ -N}^{-1}\ {1 \over n + \expo{}} + {1 \over \expo{}} + 
\sum_{n\ =\ 1}^{N}\ {1 \over n + \expo{}}}
\\[5mm] & = {1 \over \expo{}} + \sum_{n = 1}^{\infty}
\pars{{1 \over n + \expo{}} + {1 \over -n + \expo{}}} =
{1 \over \expo{}} + \sum_{n = 0}^{\infty}
\pars{{1 \over n + 1 + \expo{}} - {1 \over n + 1 - \expo{}}}
\\[5mm] & =
{1 \over \expo{}} +
\bracks{\Psi\pars{1 - \expo{}} - \Psi\pars{1 + \expo{}}}
\\[5mm] & =
{1 \over \expo{}} +
\braces{\Psi\pars{1 - \expo{}} -
\overbrace{\bracks{\Psi\pars{\expo{}} + {1 \over \expo{}}}}^{\ds{Recursive\ Property}}}
\\[5mm] & = \Psi\pars{1 - \expo{}} - \Psi\pars{\expo{}} = \bbx{\color{#44f}{\pi\cot\pars{\pi\expo{}}}} \approx -2.5705\\ &
\end{align}
$\ds{\Psi:\ Digamma\ Function}$.
