What test can be used to show $\sum \dfrac{n!}{n^n}$ converges?

Solution 1:

$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!\cdot n^n}{(n+1)^{n+1}\cdot n!} =\frac{n^n}{(n+1)^n}=\frac1{\left(1+\frac1n\right)^n}\to\frac1e$$

Remark: While $\lim_{n\to \infty} \left(1+\frac1n\right)^n$ it is often used as a definition of $e$, we don't even need convergence here. It suffices to have $\left(1+\frac1n\right)^n\ge 2$ (for $n\ge 1$), which follows from the Bernoulli inequality or simply from the first two summands of the binomial expansion.

Solution 2:

If you don't know stirlings formula, by the inequality of arithmetic and geometric means we have, $$(\prod_{k=1}^n k)^{\frac{1}{n}}\leq\frac{1}{n}\sum_{k=1}^n k=\frac{n+1}{2}$$ $$\text{ So we have}$$ $$(n!)^\frac{1}{n}\leq\frac{n+1}{2}$$ $$\frac{n!}{(n+1)^n}\leq\frac{1}{2^n}$$ $$\frac{(n+1)!}{(n+1)^n}\leq\frac{(n+1)}{2^n}$$ $$\sum_{n=0}^\infty\frac{(n+1)!}{(n+1)^n}\leq\sum_{n=0}^\infty\frac{(n+1)}{2^n}=4$$ So, $\sum_{n=0}^\infty\frac{(n+1)!}{(n+1)^n}$, converges