Q:

Not sure how to evaluate this series $\sum_{n=2}^\infty \frac{(-1)^n}{\ln(n)}$

Stumped on this absolute convergence problem!  (Converge conditionally, absolutely, or diverges)
$$\sum_{n=2}^\infty \frac{(-1)^n}{\ln(n)} $$
First, I try to do the absolute convergence test:
$$\sum_{n=2}^\infty \left\lvert \frac{-1^n}{\ln(n)}\right\rvert = \sum_{n=2}^\infty \frac{1}{\ln(n)} $$
Not sure how to evaluate this series.  Which test to use?

A:

The series fulfills the conditions of Leibniz alternating series and then it converges, but it doesn't converge absolutely since by comparison
$$\frac1{\log n}\ge\frac1n$$
and the harmonic series diverges.

A:

You know that for $n\ge 1$, $\ln(n) \le n$ which in turn says that $\frac{1}{n}\le \frac{1}{\ln(n)}$. What does this tell you about the absolute convergence?
As for conditional, think about the alternating series test.

