Answer
Convergent
Work Step by Step
We can write the series as
$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n!}{1\times 3\times5\cdots(2n-1)}$
Apply Ratio Test, we have
$\lim_{n\to\infty}\biggl|\frac{(-1)^n(n+1)!}{1\times3\times5\cdots\times(2(n+1)-1)}\frac{1\times3\times5\cdots\times (2n-1)}{(-1)^{n-1}n!}\biggr|=\lim_{n\to\infty}\biggl|\frac{(-1)^n(n+1)!}{1\times3\times5\cdots\times(2n+1)}\frac{1\times3\times5\cdots\times (2n-1)}{(-1)^{n-1}n!}\biggr|$
$=\lim_{n\to\infty}\biggl|\frac{n+1}{(2n+1)}\biggr|=\frac{1}{2}$
Sinice the limit $ < 1$, by Ratio Test the series is absolutely convergent therefore convergent.
