# How to figure out whether $\sum_{n = 1}^\infty\frac{(-1)^nn!2^n}{1\times3\times\cdots\times(2n-1)}$ is convergent or divergent

How to figure out $$\sum_{n = 1}^{\infty}\frac{(-1)^nn!2^n}{1\times3\times\cdots\times(2n-1)}$$ is convergent or divergent? I tried to use ratio test, but the limit is 1 which means it is inconclusive. What else test should I use?

I reckon that $n!2^n = 2\times4\times6\times\cdots\times 2n$ is the product of the first $n$ even numbers, and that is bigger than the product of the first $n$ odd numbers.
Hint: If the general term of a series does not converge to $0$, then the series does not converge.
Let be $$a_n = \frac{{n!2^n }} {{1 \cdot 3 \cdot \ldots \left( {2n - 1} \right)}}$$ Then $$\begin{gathered} \frac{{a_{n + 1} }} {{a_n }} = \frac{{\left( {n + 1} \right)!2^{n + 1} }} {{1 \cdot 3 \cdot \ldots \left( {2n - 1} \right)\left( {2n + 1} \right)}} \cdot \frac{{1 \cdot 3 \cdot \ldots \left( {2n - 1} \right)}} {{n!2^n }} = \hfill \\ \hfill \\ = \frac{{2\left( {n + 1} \right)}} {{2n + 1}} = 1 + \frac{1} {{2n + 1}} > 1,\,\,\,\,\forall n \in \mathbb{N} \hfill \\ \end{gathered}$$ Therefore $$a_{n+1}>a_n$$ for each $$n \in \mathbb N$$. In particular $$a_n>a_1>0$$. This means that it can not be true that $$\mathop {\lim }\limits_{n \to + \infty } a_n = 0$$ Since this one is a necessary condition for the convergence, it follows that the series is not convergent.