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

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.

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.

• Which tests have you applied? What was the result? May 20 '19 at 22:16
• Context. What sort of class is this for? May 20 '19 at 22:16
• How far did you get without any tricks? What are the numbers involved in (n!)^2 * 4^n / (2n)! ? May 20 '19 at 22:18
• Looking at the comments I feel like I am missing something, but any convergence test I have tried to use has failed. May 20 '19 at 22:19
• 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. May 20 '19 at 22:24

\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.

• +1 clean and elegant. May 20 '19 at 23:12

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

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