<s>[INST] Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\\
&\leq \frac{1}{k}\cdot\frac{2}{k}\cdots\cdot\frac{k-2}{k} \\
&\qquad \vdots \\
&\leq \frac{1}{k}\cdot\frac{2}{k}
 = \frac{2}{k^2}
\end{align*}
That is
$$\sum_{k=1}^{\infty}\frac{k!}{k^k} \leq \sum_{k=1}^{\infty}\frac{2}{k^2} $$
And since right-hand side of the inequality is finite, so is left-hand side and therefore the series is convergent.
However, I dont find this way of solving the assignment elegant and I believe there is a cleaner way. Appreciates all help I can get.
 [/INST] You can also use the root test together with the inequality between geometric and arithmetic mean (GM-AM):
$$\sqrt[k]{\frac{k!}{k^k}}= \frac{\sqrt[k]{k!}}{k}\stackrel{GM-AM}{\leq}\frac{\frac{k(k+1)}{2k}}{k}=\frac{k+1}{2k}\stackrel{k\to\infty}{\longrightarrow}\frac 12 <1$$
</s>