Answer
Convergent
Work Step by Step
The Comparison Test states that the geometric series $\sum_{n=1}^{\infty} r^n$ is convergent if $|r|\lt1$ and divergent if $|r|\geq1$.
$\sum_{n=1}^{\infty}\frac{1+cosn}{e^n}\lt\sum_{n=1}^{\infty}\frac{1}{e^n}$ given $-1\leq cosn \leq1$ as $n$ approaches $\infty$
Note that
$\sum_{n=1}^{\infty}\frac{1}{e^n}=\sum_{n=1}^{\infty}(\frac{1}{e})^n$
$|r|=|\frac{1}{e}|\lt1$
Since $\sum_{n=1}^{\infty}\frac{1}{e^n}$ is a convergent geometric series with $|r|=\lt1$, and $\sum_{n=1}^{\infty}\frac{1+cosn}{e^n}\lt\sum_{n=1}^{\infty}\frac{1}{e^n}$, $\sum_{n=1}^{\infty}\frac{1+cosn}{e^n}$ is convergent by the direct comparison test.