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.
 A: Hint Look at $$\int_2^a\frac{dx}{x\log  x}\;\;;\;\;a\to\infty$$
Note that $$\left(\log\log x\right)^\prime=\frac{1}{x\log x}$$
Thus $$\int_2^a\frac{dx}{x\log x}=\log\log a-\log\log 2$$
And this diverges as $a\to\infty$. The integral test says the series of absolute values diverges.
ADD Since $$\frac{1}{n\log n}\to 0$$ and the sequence is decreasing, Leibniz' test says the series of alternate terms converges.
A: Another cool way of proving that 
$$
\sum_{n=2}^{\infty}\frac{1}{n\ln n}
$$
diverges is to use Cauchy Condensation Test. If you let $f(n)=\frac{1}{n\ln n}$ (which is non-increasing for $n\ge 2$) we get that
$$
\sum_{n=2}^{\infty}2^{n}f(2^n)=\sum_{n=2}^{\infty}\frac{2^n}{2^n\ln(2^{n})}=\sum_{n=2}^{\infty}\frac{1}{n\ln2}=\frac{1}{\ln2}\sum_{n=2}^{\infty}\frac{1}{n}
$$
Since harmonic series diverges, we conclude that the above sum also diverges.
A: Let 
$$u_n=\frac{1}{n\log n}$$
so the harmonic series
$$2^nu_{2^n}=\frac{1}{n\log 2}$$
is divergent hence by the Cauchy condensation test  the given series isn't absolute convergent.
A: 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.
A: The absolute value of the general term $a_n$ of the series is $\frac{1}{n\ln n}$. To see why it diverges, you can do a comparison series/integral — hint: $$\frac{d}{dx}\ln\ln x = \frac{1}{x\ln x}$$
