Convergent or divergent: $\sum_{k=1}^{\infty}\frac{2^{k}\cdot k!}{k^{k}}$

Hint:

Use that $$\lim_{k\to\infty}\left(1+\frac{1}{k}\right)^k=e$$

So, $$2\lim_{k\to\infty}\left(\frac{k}{k+1}\right)^k=\frac{2}{e}\lt1$$

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}