I have tried the integral test, and I don't know what to do with the numerator that is not defined when n approaches infinity. The thing is that I know that I could try to "take"/find out if the series is absolute convergence:
$$\sum^{∞}_{n=1} \; \left|{\frac{(-1)^{n}}{\ln(n)^{2}}}\right| \implies \frac{\left|{(-1)^{n}}\right|}{\left|{\ln(n)^{2}}\right|} = {\frac{1}{\ln(n)^{2}}}$$
$$ \lim_{n\to∞} {\frac{1}{\ln(n)^{2}}} = 0 \;\;?$$
...and because the sequence is absolute convergent, it is also convergent? The thing that is throwing me off is the $(-1)^{n}$ that is not defined when n approaches ∞.
Thanks for all help!
As @JMoravitz linked, there is a handy test for this kind of series:
Given a series $Σ_{n \in ℕ} (-1)^n a_n$ with $(a_n)_n$ monotonically decreasing, $a_n \geq 0$ for all $n$ and $a_n → 0$ for $n→∞$, the series converges. See the wiki article for a proof.
In your case $a_n = \frac1{\ln (n)^2}$. Since $\ln$ is monotonically increasing (even if slow) and squaring as well, we indeed have that $a_n$ is monotonically decreasing. Furthermore $\ln (n)^2$ is unbounded (goes to $∝$ for $n→∞$ and therefore $a_n → 0$.
Therefore the alternating series test shows that your series converges. It does not tell us directly what the limit is and at least I have no idea what it might be.
About absolute convergence
Absolute convergence is a different concept. A series $Σ_{n ∈ ℕ} a_n$ converges absolutely if $Σ_{n∈ℕ} |a_n|$ converges. The convergence of $a_n$ -- what you showed -- is necessary but not sufficiant for that.
You mention that absolute convergence implies converges, which is correct -- but in the definition just mentioned. In particular your series is not absolutely convergent since for sufficiantly large $n$, $\frac1{\ln (n)^2} > \frac1n$ and the series $Σ_{n ∈ ℕ} \frac1n$ already diverges.