Answer
Convergent
Sum: $\dfrac{20}{3}$
Work Step by Step
We are given the geometric series:
$\sum_{k=1}^{\infty} 5\left(\dfrac{1}{4}\right)^{k-1}$
Determine the elements of the geometric series:
$a_1=5\left(\dfrac{1}{4}\right)^{1-1}=5$
$r=\dfrac{1}{4}$
We compute $|r|$:
$|r|=\left|\dfrac{1}{4}\right|=\dfrac{1}{4}$
Because $|r|=\dfrac{1}{4}<1$, the series converges. Determine its sum:
$S=\dfrac{a_1}{1-r}==\dfrac{5}{1-\dfrac{1}{4}}=\dfrac{5}{\dfrac{3}{4}}=\dfrac{20}{3}$