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Is this series convergent $$sum_{n=0}^inftyfrac{(-2)^{n^2}}{n!},?$$

I am supposed to find out if this series is convergent, absolutely convergent or divergent.
No test has given me information.
By this point i am aware that I'm supposed to simplify something but i don't know how. How do I get rid of $2^{n^2}$? Below is my attempt.

From $n!le n^n$ and $2^ngt n$ for all $nge1$, we have

$${2^{n^2}over n!}ge{2^{n^2}over n^n}=left(2^nover nright)^ngt1$$

so $(-2)^{n^2}/n!notto0$, and therefore the series diverges.

HINT

Let consider

$$lim_{nto infty}left|frac{(-2)^{n^2}}{n!}right|=lim_{nto infty}frac{2^{n^2}}{n!}$$

$|(-2)^{(n+1)^2}| = |2^{(n+1)^2}| = |2^{n^2 + 2n+ 1}| = (2^{n^2})(2^{2n})(2)$

Now apply the ratio test.

You seem to have an algebra error, using the Ratio Test, you have

$$
lim_{n to infty} left| dfrac{dfrac{(-2)^{(n+1)^2}}{(n+1)!}}{dfrac{(-2)^{n^2}}{n!}} right|= lim_{n to infty} dfrac{2^{n^2+2n+1}}{2^{n^2}} cdot dfrac{n!}{(n+1)!}= lim_{n to infty} dfrac{2^{n^2} cdot 2^{2n} cdot 2}{2^{n^2}} cdot dfrac{1}{n+1}= lim_{n to infty} dfrac{2^{2n+1}}{n+1}stackrel{L.H.}{=} lim_{n to infty} dfrac{2^{2n+1} ln 2 cdot 2}{1}= infty
$$