In order to study the convergence of the serie of general term $u_n=(-1)^n \ln (1-\frac{1}{n^{\alpha}})$, I remark that for $\alpha \leq 0$, the sequence $u_n$ does not tend towards zero. Suppose that $\alpha > 0$; then $|u_n|\sim n^{-\alpha}$, which means that the serie $\sum u_n$ is absolutly convergent, if and only if, $\alpha > 1$.
Remain the case where $0 < \alpha \leq 1$.
If $0 < \alpha \leq 1$, I have seen in a similar examples that we must rewrite $u_n$ as
$$
u_n= X_n + v_n
$$
where $\sum X_n$ convergent and consequently the series $\sum u_n$ and $\sum v_n$ are in the same nature.
My question is to find the value of $X_n$ and $v_n$ and conclude.
any helps are welcome.
hint
near $x=0$,
$$\ln(1-x)=-x-\frac{x^2}{2}(1+\epsilon(x))$$ with $$\lim_{x\to 0}\epsilon(x)=0$$
Your method will work if we had
$$u_n=\ln(1-\frac{(-1)^n}{n^\alpha})$$
As pointed by Steven, the alternating series test give the answer. $(|u_n|) $ is decreasing and goes to zero, so the series $\sum (-1)^nu_n$ converges.