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

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.

• I am afraid it is not comprehensible. – Yiorgos S. Smyrlis Nov 9 '14 at 21:21
• is it comprehensible now? im new to this website and do not know how to format the questions – John Nov 9 '14 at 21:23
• So, what does the ratio test tell you? – Hagen von Eitzen Nov 9 '14 at 21:25
• It is fine. Let me fix it with LaTeX. – Yiorgos S. Smyrlis Nov 9 '14 at 21:25
• it is actually 2 to the power of n squared in the numerator – John Nov 9 '14 at 21:29

## 1 Answer

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.

• Yes, indeed, corrected now. – Yiorgos S. Smyrlis Nov 18 '14 at 23:28
• It should be divergent. – Crostul Nov 18 '14 at 23:32
• Corrected! Thanks! – Yiorgos S. Smyrlis Nov 18 '14 at 23:35