<s>[INST] Want to show $e_1 (x) := \lim_{N\rightarrow \infty} \sum_{\mu=-N}^{N} \frac{1}{x+\mu} = \frac{1}{x} - \sum_{m=1}^{\infty} \gamma_m x^{2m-1}$ I want to prove 
\begin{align}
e_1 (x) := \lim_{N\rightarrow \infty} \sum_{\mu=-N}^{N} \frac{1}{x+\mu}
 \end{align}
for $\mu\neq 0$, $|x|<1$, 
\begin{align}
e_1(x) = \frac{1}{x} - \sum_{m=1}^{\infty} \gamma_m x^{**2**m-1}
\end{align}
where $\gamma_m = 2 \sum_{\mu=1}^{\infty} \mu^{-2m}$. 
This comes from A. Weil, "Elliptic functions according to Eisenstate and Kronecker"

The factor $2$ is missing in the original textbook. From its general expression for $e_n(x) $ i guess it is a typo. 
 [/INST] My trial is as follows 
\begin{align}
e_1 (x) &:= \lim_{N\rightarrow \infty} \sum_{\mu=-N}^{N} \frac{1}{x+\mu}  \\
 &= \frac{1}{x} + \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \frac{2x}{x^2-\mu^2} \\
& = \frac{1}{x} - \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \frac{2x}{\mu^2-x^2}  \\
&= \frac{1}{x} - \lim_{N\rightarrow \infty} \sum_{\mu=1}^N  \frac{2x}{\mu^2} \frac{1}{1-\frac{x^2}{\mu^2}} \\
&= \frac{1}{x} - \lim_{N\rightarrow \infty} \sum_{\mu=1}^N  \frac{2x}{\mu^2}
\sum_{m=1}^{\infty} \left( \frac{x^2}{\mu^2} \right)^{m-1} \\ 
& = \frac{1}{x} - \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \sum_{m=1}^{\infty} \frac{2}{\mu^{2m}} x^{2m-1}   \\
& = \frac{1}{x} -  \sum_{m=1}^{\infty} 2\zeta(2m) x^{2m-1}   \\
\end{align}
Since $\mu$ is integer, thus $\mu\geq 1$, thus $\frac{x}{\mu} < 1$

</s>