Convergence of series of $\frac{n!}{n^n}$ and its reciprocal $\frac{n^n}{n!}$ I read somewhere, correct me if I am wrong, that series given below is convergent. How do I prove that with comparison and ratio test? 
$<u_n> = 1+\frac{2!}{2^2}+\frac{3!}{3^3}+.....\frac{n!}{n^n}$
It was written that its reciprocal where $T_n=\frac{n^n}{n!}$ is also convergent. Will deeply appreciate your help on this.
 A: Applying the ratio test for $\sum_{n=0}^{\infty}\frac{n!}{n^n}$, you have
$$\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac{n^n}{(n+1)^n}=\left(1+\frac{1}{n}\right)^{-n}\to e^{-1}<1 $$
Hence the series converges. Note in the second equality, we took the reciprocal and wrote $-1$ in the exponent. The limit follows from the definition of $e$. The other series diverges because the convergence of this one implies $\lim_{n\to\infty}\frac{n!}{n^n}=0\Rightarrow \lim_{n\to\infty}\frac{n^n}{n!}=+\infty$.
If you're interested in a comparison, you may show that $$n!\leq \left(\frac n2\right)^n\quad\text{for }n\geq 6$$
using induction, so the series is less than a (convergent) geometric series with ratio $\frac 12$.
A: Observe that for $n\ge 2,$
$$\frac{n!}{n^n} = \frac{n}{n}\frac{n-1}{n}\cdots \frac{2}{n}\frac{1}{n} \le \frac{2}{n^2}.$$
Since $\sum \dfrac{2}{n^2}<\infty,$ the given series converges by the comparison test.
A: We can also use Stirling's approximation to to solve it.
$$\lim_{n\to\infty}\dfrac{n!}{\sqrt{2\pi n}(\frac{n}{e})^n}=1$$
Another form:
$$\lim_{n\to\infty}\dfrac{e^nn!}{n^n\sqrt{n}}=1$$
A: We know that $\frac{(n+1)^n}{(n+1)!}=\frac{n^n}{n!}\cdot\frac{(n+1)^n}{n^n}\cdot\frac{1}{n+1}=\frac{n^n}{n!}\cdot\left(\frac{n+1}{n}\right)^n\cdot\frac{1}{n+1}=\frac{n^n}{n!}\cdot\left(1+\frac{1}{n}\right)^n\cdot\frac{1}{n+1}$ but I am not fully sure about how this can be rigorously applied to your problem. I would appreciate it if someone could take over from here or show that this may not be useful for solving the problem.
