Convergence of $\sum_{n=1}^{\infty}\frac{2^nn!}{n^n}$ $$\sum_{n=1}^{\infty}\frac{2^nn!}{n^n}$$
My attempt using root test:
$$\lim_{n \to{\infty}}\sqrt[n]{\frac{2^nn!}{n^n}}=\lim_{n \to{\infty}}2\sqrt[n]{\frac{n!}{n^n}}$$
It can easily be proved that for positive $n$ we have $n^n>n!$. Now what should I do with this limit?
Note: I want to solve the problem with root test not other tests.
 A: You can use Stirling's approximation, namely: $$n! \sim \sqrt{2\pi n} \left(\frac{n}{e} \right)^n, \text{ i.e.} \lim_{n \to \infty} \left(\frac{n!}{\sqrt{2\pi n} \left( \frac{n}{e}\right)^n} \right) = 1. $$ You can compute the limit using this approximation.
A: Apply ratio test. You will get $\lim \frac {a_{n+1}} {a_n}=\frac 2 e$ [using the fact that $(1+\frac 1 n )^{n} \to e$].   If you must use root test you will need Stirling's Formula. You will see from that formula that $a_n^{1/n}$ also has limit $\frac 2 e$. 
A: In general,
$$
\sum_{n=1}^\infty\frac{a^nn!}{n^n}, \quad a>0,
$$
converges if and only if $a<\mathrm{e}$.
Ratio test provides that
$$
\frac{a_{n+1}}{a_n}=\frac{\frac{a^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{a^nn!}{n^n}}=\frac{a}{(1+\frac{1}{n})^{n}}\to \frac{a}{\mathrm{e}}
$$
and hence, we have convergence for $a<\mathrm{e}$ and divergence for $a>\mathrm{e}$.
For $a=\mathrm{e}$, the test is inconclusive. However, in such case
$$
\frac{a_{n+1}}{a_n}=\frac{\mathrm{e}}{(1+\frac{1}{n})^{n}}>1
$$
and hence $\{a_n\}$ is increasing and thus it does not tend to zero, and therefore the sum does not converge.
A: Use the relation $\frac{nⁿ}{e^{n-1}}\leq n! \leq \frac{n^{n+1}}{e^{ⁿ-1}}$.
From this you can easily show by root test that $\sum_{n=1}^{\infty}\frac{2^{n}n^n}{n^{n}e^{n-1}}$ and $\sum_{n=1}^{\infty}\frac{2^{n}n^{n+1}}{n^{n}e^{n-1}}$ both converge.
The convergence of $\sum_{n=1}^{\infty}\frac{2^{n}n!}{n^{n}}$ follows from this. 
