## Multivariable Calculus, 7th Edition

Use the Comparison Test. Let $a_n=\frac{1}{n+3^n}$ and $b_n=\frac{1}{3^n}$. The series $\frac{1}{3^n}>\frac{1}{n+3^n}$ for all $n>1$. $\frac{1}{3^n}=\left(\frac{1}{3}\right)^n$ This is a geometric series with $[|r|=\frac{1}{3}<1]$; therefore, the series is convergent. Because $\sum_{n=1}^\infty \frac{1}{3^n}$ is convergent and greater than $\sum_{n=1}^{\infty}\frac{1}{n+3^n}$ for all $n>1$ the series $\sum_{n=1}^{\infty}\frac{1}{n+3^n}$ is also convergent.