Answer
The series is convergent.
Work Step by Step
$$\sum_{n=1}^\infty\left(\frac{1}{n^3}+\frac{1}{3^n}\right)=\sum_{n=1}^\infty\left(\frac{1}{n^3}\right)+\sum_{n=1}^\infty\left(\frac{1}{3^n}\right)$$Split the series and solve one at a time.
The series $\sum_{n=1}^\infty\left(\frac{1}{n^3}\right)$ is a $p$-series with $p=3>1$
Thus, the series $\sum_{n=1}^\infty\left(\frac{1}{n^3}\right)$ is convergent.
The series $\sum_{n=1}^\infty\left(\frac{1}{3^n}\right)=\sum_{n=1}^\infty\left(\frac{1}{3}\right)^n$
$\sum_{n=1}^\infty\left(\frac{1}{3}\right)^n$ is a Geometric Series with $[|r|=\frac{1}{3}<1]$
Thus the series $\sum_{n=1}^\infty\left(\frac{1}{3}\right)^n$ is convergent.
The sum of two convergent series is also convergent, so the series $$\sum_{n=1}^\infty\left(\frac{1}{n^3}+\frac{1}{3^n}\right)$$
is convergent.
