Is $\sum_{n=0}^\infty (-2)^n\frac{n!}{(2n)!}$ abs. convergent, conditionally convergent or divergent?

Question

My task is this:

Decide whether $$\sum_{n=0}^\infty (-2)^n\frac{n!}{(2n)!}$$ is abs. convergent, conditionally convergent or divergent.

My work so far:

We observe that $(-2)^n\frac{n!}{(2n)!} = (-1)^n\frac{2^nn!}{(2n)!}$. Now trying $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|= \frac{2^{n+1}(n+1)!}{[2(n+1)]!}\frac{(2n)!}{2^nn!}= \frac{2*2^n(n+1)n!(2n)!}{2(n+1)2(n+1/2)(2n)!2^nn!}= \frac{1}{2(n+1/2)} = 0$, which shows it's absolutely covergent. Another approach would be to notice that $\frac{n!}{(2n)!} = \frac{n(n-1)(n-2)\ldots 3*2*1}{2n(2n-1)(2n-2)\ldots n(n-1)(n-2)\ldots 3*2*1}= \frac{1}{2n(2n-1)(2n-2)\ldots (n+2)(n+1)}$, and then find an expression less than this.

Now if i could write the factorial funciton for $I = [n+1, 2n]$ s.t $x! = 2n(2n-1)(2n-2)\ldots (n+2)(n+1)$ it probably could be easier to decide the convergance of this series. Any other hints and tips would be more than welcome, but note that in addition to a better approach to this problem, I also want a solution to the factorial of $I$.

Thanks in advance!

Edit: Thanks to user62498, and Wojowu who made me take a closer look at $\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$ which allowed me to correct my statement. And to Cm7F7Bb for showing a greater convergent series.

Answer 1

It converges absolutely since $$ \sum_{n=0}^\infty\biggl|(-2)^n\frac{n!}{(2n)!}\biggr| =\sum_{n=0}^\infty\biggl|(-1)^n\frac{2^n n!}{(2n)!}\biggr| =\sum_{n=0}^\infty\frac{2^n}{2n\cdot(2n-1)\cdot\ldots\cdot (n+1)} \le\sum_{n=0}^\infty\frac{2^n}{n!}=e^2. $$

Answer 2

We have

$\lim\limits_{n\to\infty}|\frac{a_{n+1}}{a_n}|=\lim\limits_{n\to\infty}\frac{2(n+1)}{(2n+1)(2n+2)}=0$

this result shows that the series abs. convergent and therefore convergent.