Absolute convergence $\frac{(-1)^n}{n \ln n}$ $$\sum_2^\infty \frac{(-1)^n}{n \ln n}$$
So I claim that it is absolutely convergent because it only coverges when it is an absolute value. This is wrong but I don't see how.

What is written below is really the Cauchy Condensation Test, without mention of Cauchy, and with calculations concretely carried out.  We could work with partial sums, but we will do things somewhat more informally. Our series is 
$$\left( \frac{1}{2\log 2}\right)+\left( \frac{1}{3\log 3}+\frac{1}{4\log 4}\right)+ 
\left( \frac{1}{5\log 5}+\frac{1}{6\log 6}+\frac{1}{7\log 7}+\frac{1}{8\log 8}\right)+\cdots$$
We have $1$ term of $\frac{1}{2\log 2}$, $2$ terms each $\ge \frac{1}{4\log 4}$, and $4$ terms each $\ge \frac{1}{8\log 8}$, and so on. So our sum is greater than 
$$\frac{1}{2}\left(\frac{1}{\log 2}+\frac{1}{\log 4}+\frac{1}{\log 8}+\frac{1}{\log 16}+\cdots\right).$$
Note that $\log 4=2\log 2$, $\log 6=3\log 2$, $\log 16=4\log 2$, and so on. So our sum is greater than 
$$\frac{1}{2\log 2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right).$$
But $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ (undramatically) blows up. It follows that our original series diverges.
