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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?

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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.

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

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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.

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