Answer
The series converges by the Ratio Test.
Work Step by Step
Use the Ratio Test:$\lim\limits_{n \to \infty}\Big|\frac{a_{n+1}}{a_n}\Big|$.
Let $a_n=\frac{n^22^{n-1}}{(-5)^n}$. Then, $\lim\limits_{n\to \infty}\Bigg|\frac{(n+1)^22^{n}}{(-5)^{n+1}} \cdot \frac{(-5)^n}{n^22^{n-1}}\Bigg|$.
Because $2^n$ and $n^2$ are positive for all $n>0$ we can remove the absolute value lines and the negative signs on the alternating terms $(-5)^n$ and $(-5)^{n+1}$. Now, $\lim\limits_{n \to \infty}\Bigg|\frac{(n+1)^22^{n}}{(-5)^{n+1}} \cdot \frac{(-5)^n}{n^22^{n-1}}\Bigg|= \lim\limits_{n \to \infty}\Bigg[\frac{(n+1)^22^{n}}{(5)^{n+1}} \cdot \frac{(5)^n}{n^22^{n-1}}\Bigg]=\lim\limits_{n \to \infty}\Bigg[\frac{2(n+1)^2}{5n^2}\Bigg]$
Simplify the expression by dividing the top and bottom by $n^2$.
$\frac{2}{5} \lim\limits_{n \to \infty}\left[(1+\frac{1}{n})^2\right]$
As $n\to \infty, 1+\frac{1}{n}\to 1$. Therefore, the limit is $(\frac{2}{5})(1)=\frac{2}{5}<1$, and the series converges by the Ratio Test.