## Calculus 8th Edition

In the given problem: $a_{n}=\frac{1.3.5....(2n-1)}{5^{n}n!}$ and $a_{n+1}=\frac{1.3.5....(2n-1)(2n+1)}{5^{n+1}(n+1)!}$ Re-write $a_{n+1}$ as $a_{n+1}=\frac{1.3.5....(2n-1)}{5^{n}n!}\times \frac{(2n+1)}{5(n+1)}$ Therefore, $\frac{a_{n+1}}{a_{n}}= \frac{(2n+1)}{5(n+1)}$ and $L=\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}\frac{(2n+1)}{5(n+1)}$ Divide numerator and denominator by $n$ $=\lim\limits_{n \to \infty}\frac{(2+\frac{1}{n})}{5(1+\frac{1}{n})}$ $=\frac{2+0}{5(1+0)}$ $=\frac{2}{5}\lt 1$ Hence, the series converges by ratio test.