Answer
Diverges.
Work Step by Step
The kth term formula of the series is given by: $a_k=(\dfrac{k}{k+1})^{k+1}$
Thus, $\lim\limits_{k \to \infty}a_k=\lim\limits_{k \to \infty}(\dfrac{k+1-1}{k+1})^{k+1}\\=\lim\limits_{k \to \infty}(1-\dfrac{1}{k+1})^{k+1}\\=\dfrac{1}{e}$
Therefore, we can see that the limit of kth term is not zero, this implies that the series diverges.
