#### Answer

$4$

#### Work Step by Step

$\sum ^{\infty }_{i=1}5\left( -\dfrac {1}{4}\right) ^{i-1}=\dfrac {a_{1}}{1-r};a_{1}=5\times \left( \dfrac {-1}{4}\right) ^{1-1}=5;r=\left( -\dfrac {1}{4}\right) \Rightarrow S_{\infty }=\dfrac {5}{1-\left( -\dfrac {1}{4}\right) }=4$