Convergence/Divergence of Series for $e^{-1}$

Question

How do I show that the series $\sum \frac{n^n}{n!} x^n$ diverges at $x =e^{-1}$?

I understand that $a(n) = (\frac{n}{n+1})^n$ is a strictly decreasing function and therefore $(\frac{n}{n+1})^n > \frac{1}{e}$.

Hope it helps.

Answer 1

If you don't mind a bit of an overkill, Stirling's formula says that for large $n$, $n!$ looks like $$n! \sim \sqrt{2\pi} n^{n+\frac{1}{2}} e^{-n},$$ so the terms in your series for large $n$ look like $$ \frac{n^n e^{-n}}{\sqrt{2\pi}n^{n+\frac{1}{2}} e^{-n}} = \frac{1}{\sqrt{2\pi n}} \sim n^{-\frac{1}{2}}.$$ You know that $$\sum n^{-\alpha}$$ converges precisely when $\alpha > 1$. Since $\frac{1}{2} < 1$, the series must diverge.