Answer
Converges
Work Step by Step
Given $$ \sum_{n=1}^{\infty} \frac{e^{n}+n}{e^{2 n}-n^{2}}$$
Compare with the convergent series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{e^{n}}$ (geometric series , |r|<1) and by using the Limit Comparison Test, we get:
\begin{align*}
\lim_{n\to \infty} \frac{a_n}{b_n}&=\lim_{n\to \infty}\frac{e^n(e^{n}+n)}{e^{2 n}-n^{2}}\\
&=\lim_{n\to \infty}\frac{e^n(e^{n}+n)}{(e^{n}+n)(e^{n}-n)}\\
&=\lim_{n\to \infty}\frac{e^n }{ (e^{n}-n)}\\
&=1
\end{align*}
Then $\displaystyle \sum_{n=1}^{\infty}\frac{e^{n}+n}{e^{2 n}-n^{2}}$ also converges
