Answer
Conditionally convergent
Work Step by Step
Let $b_n=\frac{1}{\ln n}$.
Notice that
$n+1>n$ (Take the natural logarithm)
$\ln (n+1)>\ln n$
$\frac{1}{\ln(n+1)}<\frac{1}{\ln n}$
The last expression tells us that $b_{n+1}\leq b_n$ for all $n>1$.
In addition, we can see that $\lim b_n =0$.
Using the Alternating Series Test, $\sum_{n=2}^{\infty }\frac{(-1)^n}{\ln n}$ is convergent.
However, we know $\frac{1}{\ln n}>\frac{1}{n}$ and the harmonic series $\sum_{n=2}^\infty \frac{1}{n}$ is divergent.
So, using the Comparison Test the series $\sum_{n=2}^\infty \frac{1}{\ln n}$ is divergent.
Thus, the series $\sum_{n=2}^\infty \frac{1}{\ln n}$ is conditionally convergent.
