Answer
Convergent
Work Step by Step
$\sum_{k =1}^{\infty}\frac{2^{k-1}3^{k+1}}{k^k}=\frac{3}{2}\sum_{k =1}^{\infty}(\frac{6}{k})^{k}$
$\lim\limits_{k \to \infty}\frac{3}{2}\sqrt[k] {(\frac{6}{k})^{k}}=\frac{3}{2}\lim\limits_{k \to \infty}\frac{6}{k}$
$=0\lt 1$
Convergent.
