## Calculus (3rd Edition)

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.