Answer
Divergent
Work Step by Step
Let $a_k=\frac{2^kk!}{(k+2)!}$.
Simplify $\frac{a_{k+1}}{a_k}$:
$\frac{a_{k+1}}{a_k}=\frac{\frac{2^{k+1}(k+1)!}{(k+3)!}}{\frac{2^kk!}{(k+2)!}}=\frac{\frac{2\cdot 2^{k}(k+1)k!}{(k+3)(k+2)!}}{\frac{2^kk!}{(k+2)!}}=\frac{2k+2}{k+3}$
Find $\lim \frac{a_{k+1}}{a_k}$:
$\lim \frac{a_{k+1}}{a_k}=\lim \frac{2k+2}{k+3}=2>1$
Using the Ratio Test, the series $\sum_{k=1}^\infty \frac{2^kk!}{(k+2)!}$ is divergent.
