Determine whether the series $\sum_{n=0}^\infty\frac{2^{n^2}}{n!}$ is convergent or divergent.

Question

Determine whether the series $$\sum_{n=0}^\infty\frac{2^{n^2}}{n!},$$ is convergent or divergent.

I know I have to use the ratio test.

Answer 1

Yes it diverges, and the simplest test to use is indeed the ratio test: $$ \frac{a_{n+1}}{a_n}=\frac{2^{(n+1)^2} n!}{2^{n^2}(n+1)!}=\frac{2^{2n+1}}{n+1}\to \infty, $$ as $n\to\infty$.

Hence the series diverges.

Answer 2

As an alternative

$$\frac{2^{n^2}}{n!}\ge \frac{2^{n^2}}{n^n}=\left(\frac{2^n}{n}\right)^n\ge \left(\frac{2n}{n}\right)^n=2^n\to \infty$$