Limit of $n$th root of $n$! I am asked to determine if a series converges or not:

$$\displaystyle\sum\limits_{n=1}^{\infty} \frac{(2^n)n!}{(n^n)}$$

So I'm using the $n$th root test and came up with $\lim_{n \to {\infty}}\frac{2}{n}\times(\sqrt[n]{n!})$
I know that the limit of $\frac{2}{n}$ goes to $0$ when $n$ goes to infinity but what about the $(\sqrt[n]{n!})$? 
 A: Note that we do not need to actually evaluate the limit, we just need to find an upper bound.
Consider that, for $n>m$,
$$
\frac{n!}{m!}\leq n^{n-m}
$$
As such, if we let $n=6k-a$, where $0\leq a\leq5$, we can observe that
$$
n!\leq (6k)!\leq \prod_{i=1}^6 (ik)^k=(720k^6)^k<\left
(\frac{20}{6^4}\right)^k(6k)^{6k}
$$
Therefore,
$$
\sqrt[n]{n!}<\left(\frac{20}{6^4}\right)^{(n+a)/6n}(n+a)^{1+a/n}
$$
and thus
$$
\frac{2\sqrt[n]{n!}}{n}<2\left(\frac{20}{6^4}\right)^{1/6}\left(\frac{20}{6^4}\right)^{a/6n}(1+a/n)(n+a)^{a/n}
$$
Now, as $a$ cannot be larger than 5, we can easily take the limit of each term as $n\to\infty$, to give
$$
\lim_{n\to\infty}\frac{2\sqrt[n]{n!}}{n}<2\left(\frac{20}{6^4}\right)^{1/6}\approx 0.997932
$$
Therefore, as the limit is less than 1, it converges.
Note that the $\lim$ in the final line isn't strictly correct notation, as we have not proven that the limit exists. That said, it captures the intent, that for sufficiently large $n$, the expression will be less than $0.997932$.
A: 
I thought it might be instructive to present an approach that relies on elementary tools only.  To that end, we now proceed.

Note that we can write
$$\begin{align}
\log\left(\frac{\sqrt[n]{n!}}{n}\right)&=\frac1n\log(n!)-\log(n)\\\\
&=\frac1n\sum_{k=1}^n\log(k)-\log(n)\\\\
&=\underbrace{\frac1n\sum_{k=1}^n\log(k/n)}_{\text{Riemann Sum for}\,\,\int_0^1 \log(x)\,dx=-1}\\\\
\end{align}$$
Hence, we have
$$\lim_{n\to \infty}\left(\frac{2\sqrt[n]{n!}}{n}\right)=2e^{-1}$$
And we are done!

Tools Used.  Straightforward arithmetic and Riemann sums.

A: Since OP is taking Calculus III, perhaps the ratio test from calculus II is a more suitable way.
Let $a_n = (2^n)n!/n^n$.
\begin{align}
\frac{a_{n+1}}{a_n} &= \frac{(2^{n+1})(n+1)!/(n+1)^{n+1}}{(2^n)n!/n^n} \\
&= 2(n+1) \frac{n^n}{(n+1)^n} \frac1{n+1} \\
&= 2 \frac1{\left(1+\frac{1}{n}\right)^n} \\
\end{align}
$$L = \lim_{n \to +\infty} \frac{a_{n+1}}{a_n} = \lim_{n \to +\infty} 2 \frac1{\left(1+\frac{1}{n}\right)^n} = \frac2e < 1 $$
So the series converges.

Alternative method by Stirling's approximation
I type this for fun and to show the power of this formula for $\sum\limits_{n=1}^{\infty} \frac{2^n n!}{n^n}$
Use the root test on $a_n = (2^n)n!/n^n$.
\begin{align}
a_n =& \frac{2^n n!}{n^n} \\
\sim& \frac{2^n\sqrt{2\pi n} e^{-n} n^n}{n^n} \\
=& \sqrt{2\pi} \cdot \frac{2^n}{e^n} \cdot \sqrt{n}
\end{align}
The limit $1 \le \sqrt{n}^{1/n} \le n^{1/n} \to 1$ as $n \to +\infty$ allows us to recover the ratio $2/e$ in the previous section.
$$L = \lim\limits_{n\to+\infty} a_n^{1/n} = \lim\limits_{n\to+\infty} \frac2e \sqrt{n}^{1/n} = \frac2e$$
