TITLE: Convergent or divergent: $\sum_{k=1}^{\infty}\frac{2^{k}\cdot k!}{k^{k}}$
QUESTION [6 upvotes]: Convergent, absolutely convergent or divergent:
  $$\sum_{k=1}^{\infty}\frac{2^{k}\cdot k!}{k^{k}}$$

I have used ratio test because we got a fraction here and I think I did alright till the end:
$$\lim_{k\rightarrow \infty}\frac{\frac{2^{k+1}\cdot (k+1)!}{(k+1)^{k+1}}}{\frac{2^{k}\cdot k!}{k^{k}}} = \lim_{k\rightarrow \infty}\frac{2^{k+1}\cdot (k+1)! \cdot k^{k}}{(k+1)^{k+1}\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2^{k}\cdot 2^1\cdot k! \cdot (k+1)\cdot k^{k}}{(k+1)^{k}\cdot (k+1)\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2k^{k}}{(k+1)^{k}}$$ $$=2\lim_{k\rightarrow \infty} \left( \frac{k}{k+1}\right)^{k}$$
Now I don't know (without a calculator) to what this would converge / diverge to... In the exam we are not allowed to use a calculator... So what to do?
The denominator will be greater than the enumerator by 1, so dividing each other we got something $<1$. We take exponent $k$ which is $\geq 1$, so we will end up with $<1$ again. Multiply this with 2 we get something $< 1$ but $> 0$ and thus the series is convergent...?
I hope I have described it well? Would you give me full points to this task? :D
Edit: I haven't described it well in the end. See the accepted answer and its comments! Thanks a lot to everyone - from every question I ask here, I always learn new things :-)

REPLY [2 votes]: Observe
\begin{align}
\lim_{k\rightarrow \infty}\left(\frac{k}{k+1} \right)^k = \lim_{k \rightarrow \infty}\left(1-\frac{1}{k+1} \right)^{k+1}\left( 1-\frac{1}{k+1}\right)^{-1} = e^{-1}.
\end{align}