Answer
Diverges by the n-th term test
Work Step by Step
$\Sigma^{\infty}_{n=1} = \frac{(-1)^{n+1} (n+1)}{\ln(n+1)} $
Apply the n-th term test
$\lim\limits_{n \to \infty} \frac{n+1}{\ln(n+1)}= \frac{\infty}{\infty}$
Use L'hopital to evaluate the limit
$\lim\limits_{n \to \infty} n+1= \infty$
Diverges by the n-th term test
