Answer
Converges
Work Step by Step
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,
$\frac{a_{n}}{a_{n+1}}= \frac{(2n+1)}{5(n+1)}$
and
$L=\lim\limits_{n \to \infty}|\frac{a_{n}}{a_{n+1}}|=\lim\limits_{n \to \infty}\frac{(2n+1)}{5(n+1)}$
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.