Answer
Converges for all exponents $k$
Work Step by Step
Given
$$\sum_{n=1}^{\infty} n^k3^{-n}$$
By using the Ratio Test, we get:
\begin{align*}
\rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\
&=\lim _{n \rightarrow \infty}\frac{ (n+1)^{k}}{3^{n+1} }\frac{3^n}{n^k} \\
&= \lim _{n \rightarrow \infty} \frac{ (n+1)^{k}}{3n^k}\\
&= \frac{1}{3}<1
\end{align*}
Thus the series converges for all exponents $k$.
