By the Ratio Test we have $$L=\lim_{n\rightarrow \infty} \frac{\frac{3}{5^{n+1}-e^{n+1}}}{\frac{3}{5^n-e^n}}=\lim_{n\rightarrow \infty}\frac{5^n-e^n}{5^{n+1}-e^{n+1}}=\lim_{n\rightarrow \infty}\frac{1-\frac{e^n}{5}}{5-e\frac{e^n}{5^n}}$$
Since $\lim_{n\rightarrow \infty}e\frac{e^n}{5^n}=\lim_{n\rightarrow \infty}e(\frac{e}{5})^n=0$ it follows that
$L=\frac15\lt1$, thus the series is convergant.