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

Question

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.

Answer 1

\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.

Answer 2

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

Answer 3

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