Answer
The series $\sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{ (1.1)^n}$ converges absolutely.
Work Step by Step
We have the absolute series
$$\sum_{n=0}^{\infty} |\frac{(-1)^{n-1}}{ (1.1)^n}|=\sum_{n=0}^{\infty} \frac{1}{ (1.1)^n}$$
which is a convergent geometric series as $\frac{1}{1.1}\lt 1$.
Thus, the alternating series $\sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{ (1.1)^n}$ converges absolutely.
