Answer
The series is convergent by the Ratio Test.
Work Step by Step
$\sum_{n+1}^\infty k^2e^{-k}=\sum_{n+1}^\infty \frac{k^2}{e^k}$
Use the Ratio Test: $\lim\limits_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|$
$\lim\limits_{k\to\infty}\left|\frac{a_{k+1}}
{a_k}\right|=\lim\limits_{k\to\infty}\left|\frac{(k+1)^2}{e^{k+1}}\cdot\frac{e^k}{k^2}\right|$
Cancel out terms and put like terms together.
$\lim\limits_{k\to\infty}\left|\frac{(k+1)^2}{k^2}\cdot \frac{1}{e}\right|=\lim\limits_{k\to\infty}\left|\left(\frac{k+1}{k}\right)^2\cdot \frac{1}{e}\right|$
Divide the term with $k$ by the highest degree of $k$.
$\lim\limits_{k\to\infty}\left|\left(\frac{1+\frac{1}{k}}{1}\right)^2\cdot \frac{1}{e}\right|$
As $n\to\infty$ the limit becomes $(1)(\frac{1}{e})=\frac{1}{e}<1$, thus the series converges by the Ratio Test.