I was wondering how to determine whether this series converges:
$$ \sum_{k=1}^\infty \frac{(k+1)!}{k^k} \; ? $$
Here's how I started to tackle this problem:
$$ \begin{aligned} \lim_{k \to \infty}{\left|\frac{\cfrac{(k+2)!}{(k+1)^{k+1}}}{\cfrac{(k+1)!}{k^k}}\right|} &= \frac{(k+2)!}{(k+1)^{k+1}} \times \frac{k^k}{(k+1)!} \\[0.6em] &= \frac{k^k(k+2)!}{(k+1)^{k+1}(k+1)!} \\[0.6em] &= \frac{k^k(k+2)(k+1)!}{(k+1)^{k+1}(k+1)!} \\[0.6em] &= \frac{k^k(k+2)}{(k+1)^{k+1}} \\[0.6em] &= \frac{k^k(k+2)}{(k+1)^k(k+1)} \\[0.6em] &= \frac{k^k(k+2)}{(k+1)^k(k+1)} \\[0.6em] \end{aligned} $$
Any advice is greatly appreciated!
Noting $$ \lim_{k\to\infty}\frac{k^k}{(k+1)^k}=\lim_{k\to\infty}{(1+\frac1k)^{-k}}=e^{-1}$$ you have $$ \lim_{k\to\infty}\bigg|\frac{{\frac{(k+2)!}{(k+1)^{k+1}}}}{\frac{(k+1)!}{k^k}}\bigg|=e^{-1}<1 $$ which implies that your series converges.