Calculus 8th Edition

$$\sum_{n=1}^\infty\left(\frac{1}{n^3}+\frac{1}{3^n}\right)=\sum_{n=1}^\infty\left(\frac{1}{n^3}\right)+\sum_{n=1}^\infty\left(\frac{1}{3^n}\right)$$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.