Answer
$\dfrac {5}{6}$
Work Step by Step
$\sum ^{\infty }_{n=0}\left( -\dfrac {1}{5}\right) ^{n}=\lim _{n\rightarrow \infty }\left( \dfrac {1-r^{n}}{1-r}\right) =\dfrac {1-\left( -\dfrac {1}{5}\right) ^{n}}{1-\left( -\dfrac {1}{5}\right) }=\dfrac {1-0}{1-\left( -\dfrac {1}{5}\right) }=\dfrac {5}{6}$
