Chapter 11 - Section 11.5 - Alternating Series - 11.5 Exercises - Page 772: 19

Convergent

Let $b_n=\frac{n^2}{5^n}$. For all $n\geq 1$, we have $2n+1\leq 3n< 4n^2$ $2n+1< 4n^2$ $n^2+2n+1< 5n^2$ $\frac{n^2+2n+1}{5^n5}< \frac{n^2}{5^n}$ $\frac{(n+1)^2}{5^{n+1}}< \frac{n^2}{5^n}$ and we know that as $n$ is large enough the value of $\frac{n^2}{5^n}$ is close to zero. It means $b_{n+1}0$. By the Alternating Series Test, the series converges.