## Calculus (3rd Edition)

The series $\sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{ (1.1)^n}$ converges absolutely.
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.