Answer
Convergent
Work Step by Step
Let $f(x)=\frac{x^2}{e^x}$. This function is continuous and decreasing on $[3,\infty)$.
We will evaluate $\int_3^\infty f(x) dx$.
Notice that:
$\int \frac{x^2}e^{x}dx=\int x^2e^{-x}dx$ (Use the IBP for $u=x^2\to du=2x$ and $dv=e^{-x}dx\to v=-e^{-x}$)
$=-x^2e^{-x}+\int 2xe^{-x} dx$ (Use the IBP for $u=2x\to du=2dx$ and $dv=e^{-x}dx\to v=-e^{-x}$)
$=-x^2e^{-x}-2xe^{-x}+\int 2e^{-x}dx$
$=-x^2e^{-x}-2xe^{-x}-2e^{-x}$
$=-(x^2+2x+2)e^{-x}$
Then,
$\int_3^\infty \frac{x^2}{e^{x}}dx=-(x^2+2x+2)e^{-x}]_3^\infty=17e^{-3}$
So, the integral $\int_3^\infty f(x)dx$ is convergent.
Using the Integral Test for Series, $\sum_{n=3}^\infty \frac{n^2}{e^n}$ is convergent.
