As per the title, I am trying to find if the series $$ \sum\limits_{n = 1}^{+\infty} \frac{(-1)^n ((n - 1)!!)^2}{n!} $$ converges absolutely, conditionally, or diverges. The $“!!”$ is the double factorial function. I tried to use the fact that $n! = n!! \cdot (n - 1)!!$ in order to simplify the expression to $\sum\nolimits_{n = 1}^{+\infty} \frac{(-1)^n (n - 1)!!}{n!!}$, but it didn't help a lot.
Any help would be greatly appreciated.
Let $a_n=\frac{(n-1)!!}{n!!}$. Using the identities $(2n-1)!!\cdot(2n)!!=(2n)!$ and $(2n)!!=2^n\cdot n!$, one gets $a_{2n}=\frac{(2n-1)!!}{(2n)!!}=\frac{(2n)!}{4^n\cdot(n!)^2}$ hence Stirling formula yields $\color{red}{a_{2n}\sim\frac1{\sqrt{\pi n}}}$. On the other hand, $a_{2n+1}=\frac{(2n)!!}{(2n+1)!!}=\frac1{2n+1}\cdot\frac1{a_{2n}}$ hence $\color{red}{a_{2n+1}\sim\frac{\sqrt\pi}{2\sqrt{n}}}$. Thus, $a_{2n}-a_{2n-1}\sim-\frac{c}{\sqrt{n}}$ with $c=\frac{\pi-2}{2\sqrt\pi}\gt0$ hence the series $\sum\limits_n(-1)^na_n$ diverges and $\lim\limits_{N\to\infty}\sum\limits_{n=1}^N(-1)^na_n=-\infty$.