Answer
The series converges.
Work Step by Step
We have the given series
$\sum_{n=1}^{\infty} \frac{2^n}{3^n-n}$
We apply the limit comparison test with $b_n=(\frac{2}{3})^n$ (a convergent geometric series with $r=2/3\lt 1$):
$L=\lim_{n\rightarrow\infty} \frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{2^n}{3^n-n}\times(\frac{3}{2})^n=\lim_{n\rightarrow\infty}\frac{1}{1-\frac{n}{3^n}}$
The $\frac{n}{3^n}$ term is indeterminate, so we apply L'Hopital's rule and derive it:
$\lim_{n\rightarrow\infty}\frac{n}{3^n}=\lim_{n\rightarrow\infty}\frac{1}{3^n\ln 3}=0$
Thus the overall limit is:
$L=\lim_{n\rightarrow\infty}\frac{1}{1-\frac{n}{3^n}}=1$
Since $L=1$, our starting series converges.
