Simple series convergence/divergence: $\sum_{k=1}^{\infty}\frac{2^{k}k!}{k^{k}}$ I have the following problem:
$$\sum_{k=1}^{\infty}\frac{2^{k}k!}{k^{k}}$$
I only need to find whether the series converges or diverges. My initial thinking was to use the ratio test. I hit a stump after simplifying my limit, though:
$$\lim_{k}\frac{2k^{k}(k+1)}{(k+1)^{k+1}}$$
Normally I'd just derive by L'Hopital's Rule, but the derivative is pretty long and complicated. I doubt that's the correct route.
 A: $$
\frac{2k^{k}(k+1)}{(k+1)^{k+1}}
= \frac {2}{ \left(1+1/k\right)^k} \to \frac 2e <1
$$

You can also find the result using the Stirling formula.
A: $$\lim\frac{2^{k+1}\cdot(k+1)!\cdot k^k}{(k+1)^{k+1} \cdot 2^k\cdot k!}=\lim\frac{2\cdot (k+1)\cdot k^k}{(k+1)^{k+1}}=2 \lim \frac{k^k}{(k+1)^k}=$$
$$=2\lim \left(\frac{k}{k+1}\right)^k=2\lim \left( 1-\frac{1}{k+1}\right)^k$$
Now let $t=k+1$, so
$$2 \lim\left(1-\frac{1}{t} \right)^{t-1}= 2 \lim \left(1-\frac{1}{t} \right)^t\cdot \left(1-\frac{1}{t} \right)^{-1}=\frac{2}{e}<1$$
A: As suggested by mookid, if you use Stirling approximation $$k! \simeq \sqrt{2 \pi } e^{-k} k^{k+\frac{1}{2}}$$ defining $$u_k=\frac{2^{k}k!}{k^{k}}$$ and replacing $k!$ by the above expression you have $$\frac{u_{k+1}}{u_k}=\frac{2 \sqrt{k+1}}{e \sqrt{k}}$$ the limit of which being $\frac {2}{e}$ as already given to you in other answers. 
What is nice it that using the above, the summation can be approximated and $$\sum_{k=1}^{\infty}\frac{2^{k}k!}{k^{k}}\simeq \sqrt{2 \pi } \text{Li}_{-\frac{1}{2}}\left(\frac{2}{e}\right)=12.5684$$ while the exact value is equal to $12.9490$
A: 
I thought it might be instructive to present an approach that relies on the root test and elementary tools only.  To that end, we now proceed.

Rather than use the ratio test, we use the root test.  Hence we need to evaluate the limit 
$$\lim_{k\to \infty}\sqrt[k]{\frac{2^k\,k!}{k^k}}=\lim_{k\to \infty}\frac{2\sqrt[k]{k!}}{k}$$
Note that we can write
$$\begin{align}
\log\left(\frac{\sqrt[k]{k!}}{k}\right)&=\frac1k\log(k!)-\log(k)\\\\
&=\frac1k\sum_{j=1}^k\log(j)-\log(k)\\\\
&=\underbrace{\frac1k\sum_{j=1}^k\log(j/k)}_{\text{Riemann Sum for}\,\,\int_0^1 \log(x)\,dx=1}\\\\
\end{align}$$
Hence, we have
$$\lim_{k\to \infty}\sqrt[k]{\frac{2^k\,k!}{k^k}}=2e^{-1}$$
And we are done!

Tools Used.  Straightforward arithmetic and Riemann sums.

