# Convergence of $\sum_n \frac{n!}{n^n}$

I’m working on a problem sheet and it ask to discuss the convergence of
$$\sum \frac{n!}{{n}^{n}}$$
By D’Lembert’s ratio test,
$$\lim_{n->\infty}\frac{{a}_{n+1}}{{a}_{n}} = 1$$
and so, is inconclusive.

Using Cauchy’s root test,

$$\lim_{n->\infty}({\frac{n!}{{n}^{n}}})^\frac{1}{n}=1$$

What are my alternatives?

Should I take the integral of the term of the series above? Would integrating factorial works?

#### Solutions Collecting From Web of "Convergence of $\sum_n \frac{n!}{n^n}$"

Actually the ratio test turns out to be conclusive :

\begin{align}\lim_{n\to\infty}\frac{{a}_{n+1}}{{a}_{n}} &=\lim_{n\to\infty}\dfrac{(n+1)!}{(n+1)^{n+1}} \cdot \dfrac{n^n}{n!} \\~\\&=\lim_{n\to\infty}\dfrac{n+1}{(n+1)^{n+1}} \cdot \dfrac{n^n}{1} \\~\\&=\lim_{n\to\infty} \left(\dfrac{n}{n+1}\right)^n\\~\\&=\lim_{n\to\infty} \left(\dfrac{\color{blue}{n+1}-1}{n+1}\right)^n\\~\\&=\lim_{n\to\infty} \left(\color{blue}{1}+\dfrac{-1}{n+1}\right)^n\\~\\&=e^{-1}~~\color{Red}{\star} \\~\\&\lt 1\end{align}

$\color{red}{\star}$ : please see $e^x$ limit definition

$$\frac{n!}{n^n}=\frac{1}{n}\frac{2}{n}\cdots\frac{n-1}{n}\frac{n}{n}<\frac2{n^2}$$

Hint: $\dfrac{n!}{n^n} < \dfrac{2}{n^2}$.