Simple series convergence/divergence: $\sum_{k=1}^{\infty}\frac{2^{k}k!}{k^{k}}$
$$ \frac{2k^{k}(k+1)}{(k+1)^{k+1}} = \frac {2}{ \left(1+1/k\right)^k} \to \frac 2e <1 $$
You can also find the result using the Stirling formula.
$$\lim\frac{2^{k+1}\cdot(k+1)!\cdot k^k}{(k+1)^{k+1} \cdot 2^k\cdot k!}=\lim\frac{2\cdot (k+1)\cdot k^k}{(k+1)^{k+1}}=2 \lim \frac{k^k}{(k+1)^k}=$$
$$=2\lim \left(\frac{k}{k+1}\right)^k=2\lim \left( 1-\frac{1}{k+1}\right)^k$$
Now let $t=k+1$, so
$$2 \lim\left(1-\frac{1}{t} \right)^{t-1}= 2 \lim \left(1-\frac{1}{t} \right)^t\cdot \left(1-\frac{1}{t} \right)^{-1}=\frac{2}{e}<1$$