## Precalculus (6th Edition) Blitzer

$\dfrac{5}{7}$
As we are given that $\sum_{i=1}^{\infty}(\dfrac{-2}{5})^{i-1}$ we have $r=\dfrac{-2}{5}$ and $a_1=1$ This shows an infinite geometric sequence whose sum will be $S_{\infty}=\dfrac{a_1}{1-r}=\dfrac{1}{1-(\dfrac{-2}{5})}$ and $=\dfrac{1}{7/5}$ Thus, $S_{\infty}=\dfrac{5}{7}$