I was given the following series to find out if it absolutely converges or conditionally converges, or if diverges.
$$\sum_\limits{n=1}^\infty \dfrac{2^{n^2}}{n!}$$
I decided to venture on this one with the Root Test to find out if it converges, or diverges. I got this point:
$$\displaystyle \lim_\limits{n \to \infty} (\vert \dfrac{2^{n^2}}{n!} \vert)^{\frac{1}n}$$
Which lead me to the following:
$$\lim_\limits{n \to \infty} \dfrac{2^n}{(n!)^\frac{1}n}$$
I went to wolfram alpha and it says the limit is $\infty$, meaning the series diverges which it does. How do I solve this limit?
Best Answer
You can do it using Stirling's formula. But it is more natural to prove that the series diverges using the ratio test:$$\lim_{n\to\infty}\frac{\frac{2^{(n+1)^2}}{(n+1)!}}{\frac{2^{n^2}}{n!}}=\lim_{n\to\infty}\frac{2^{2n+1}}{n+1}=\infty.$$