Answer
$=\dfrac {5}{2}=2.5$
Work Step by Step
$\sum ^{\infty }_{m=2}\dfrac {5}{2^{m}}=\dfrac {a_{1}}{1-r};a_{1}=\dfrac {5}{2^{2}}=\dfrac {5}{4};r=\dfrac {1}{2}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {5}{4}}{1-\dfrac {1}{2}}=\dfrac {5}{2}=2.5$
