#### Answer

$\dfrac{5}{7}$

#### Work Step by Step

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, $S_{\infty}=\dfrac{5}{7}$