Answer
converges absolutely
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1.1} \ln (n+1)}$$
We compare $\sum_{n=1}^{\infty} \dfrac{1}{n^{1.1} \ln (n+1)}$ with $ \dfrac{1}{n^{1.1}}$, a convergent series:
\begin{align*}
\lim_{n\to \infty } \dfrac{n^{1.1}}{n^{1.1} \ln (n+1)}=0
\end{align*}
Then $ \sum_{n=1}^{\infty} \dfrac{1}{n^{1.1} \ln (n+1)} $ converges.
Since $a_n = \dfrac{1}{n^{1.1} \ln (n+1)}$ is decreasing and $\lim_{n\to \infty } a_n =0$, then $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1.1} \ln (n+1)}$ converges. Hence, $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1.1} \ln (n+1)}$ converges absolutely.