Calculus 8th Edition

In the given problem: $a_{n}=\frac{(-5)^{n}}{n^{2}9^{n}}=\frac{(25)^{n}}{n^{2}9^{n}}$ and $a_{n+1}=\frac{(25)^{n+1}}{(n+1)^{2}9^{n+1}}$ Re-write $a_{n+1}$ as $a_{n+1}=\frac{(25)^{n+1}}{(n+1)^{2}9^{n+1}}$ Therefore, $\frac{a_{n+1}}{a_{n}}= \frac{\frac{(25)^{n+1}}{(n+1)^{2}9^{n+1}}}{\frac{(25)^{n}}{(n)^{2}9^{n}}}$ $=\frac{25}{9}(\frac{n}{n+1})^{2}$ and $L=\lim\limits_{n \to \infty}|\frac{25}{9}(\frac{n}{n+1})^{2}|$ $=\frac{25}{9} \gt 1$ Thus, $L\gt 1$ Hence, the series is divergent by ratio test.