Answer
$\dfrac{5}{7}$
Work Step by Step
since,we have $\sum_{i=1}^{\infty}(\dfrac{-2}{5})^{i-1}$
Here, $r=\dfrac{-2}{5}$ and $a_1=1$; which is an infinite geometric sequence.
Their sum is $S_{\infty}=\dfrac{a_1}{1-r}$
Thus,
$S_{\infty}=\dfrac{1}{1-(\dfrac{-2}{5})}$
or, $=\dfrac{1}{\dfrac{7}{5}}$
Hence,
$=\dfrac{5}{7}$
