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?
 A: 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.
A: Hint: If the general term of a series does not converge to $0$, then the series does not converge.
A: 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.
