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Does the series $\sum_{n=1}^{\infty} \frac{(-1)^n*(n!)^2*4^n}{(2n)!} $ converge?

I have no idea how to do this. I have tried to use any trick I am aware of but can't figure this out.

Can anyone help please?

EDIT:

I have already found out that the series $\sum_{n=1}^{\infty} \frac{(n!)^2*4^n}{(2n)!} $ diverges using Raabe's test.

The ratio test is inconclusive for this series.

EDIT 2:

Using Stirling's approximation for $n!$ is not allowed.

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  • $\begingroup$ Which tests have you applied? What was the result? $\endgroup$ May 20 '19 at 22:16
  • $\begingroup$ Context. What sort of class is this for? $\endgroup$
    – The Count
    May 20 '19 at 22:16
  • $\begingroup$ How far did you get without any tricks? What are the numbers involved in (n!)^2 * 4^n / (2n)! ? $\endgroup$
    – gnasher729
    May 20 '19 at 22:18
  • $\begingroup$ Looking at the comments I feel like I am missing something, but any convergence test I have tried to use has failed. $\endgroup$ May 20 '19 at 22:19
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    $\begingroup$ Hint: in the ratio test, it turns out $|a_{n+1} / a_n| > 1$ for all $n$. Therefore, $|a_n|$ is strictly increasing so it can't converge to 0. $\endgroup$ May 20 '19 at 22:24
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\begin{align*} &&4^n&=(1+1)^{2n}\\[4pt] &&&=\sum_{k=0}^{2n}\binom{2n}{k}\\[4pt] &&& > \binom{2n}{n}\\[4pt] \end{align*} hence $$|a_n|=\frac{(n!)^2 4^n}{(2n)!}=\frac{4^n}{\binom{2n}{n}} > 1$$ so the series diverges.

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    $\begingroup$ +1 clean and elegant. $\endgroup$ May 20 '19 at 23:12
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Note $(n!)*2^n = (2n)!!$, so your sum is $$ \sum_1^\infty \frac{(-1)^n * (n!)^2 * 4^n}{(2n)!} = \sum_1^\infty \frac{(-1)^n (2n)!!}{(2n-1)!!} $$

Obviously $$ \left| \frac{(-1)^2(2n)!!}{(2n-1)!!} \right| > 1 $$

So it does not converge

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Given

$$ a_n=\frac{4^n(n!)^2}{(2n)!} $$

it is a simple exercise to show that

$$ a_{n+1}>a_n $$

so

$$ \sum_{n=1}^\infty (-1)^na_n $$

fails to converge since $\lim_{n\to\infty}(-1)^na_n\ne0$.

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