Series convergent or divergent

Question

trying to determine if the series is conditionally convergent or divergent. $$\sum_{n = 1}^\infty \frac{2^{n^{2}}}{n!}$$ with n! i tried the ratio test on the series $$\frac{2^{(n+1)^{2}}}{(n+1)!} * \frac{n!}{2^{n^{2}}} = \frac{2^{2n+1}}{(n+1)} $$ which is > 1 as $n\to \infty$ and is overall divergent ? not sure if I am on the right track.

Answer 1

Well, applying the ratio test:

$$\lim_{\text{n}\to\infty}\space\left|\frac{\frac{2^{\left(\text{n}+1\right)^2}}{\left(\text{n}+1\right)!}}{\frac{2^{\text{n}^2}}{\text{n}!}}\right|=\lim_{\text{n}\to\infty}\space\left|\frac{2^{2\text{n}+1}}{\text{n}+1}\right|\space\to\space\infty\tag1$$

Answer 2

Note that ${2^{n^2} \over n!} = { (2^n)^n \over n!} \ge { (2^n)^n \over n^n}= ({2^n \over n})^n \ge 1$. Hence $\sum_n {2^{n^2} \over n!} \ge N$ for all $N$.