If we factor 25 from the sum, we have
[tex]\displaystyle 25\left(1+\dfrac{1}{5}+\dfrac{1}{25}+\ldots\right)=25\sum_{i=0}^\infty \left(\dfrac{1}{5}\right)^i = 25 \dfrac{1}{1-\frac{1}{5}} = 25\dfrac{1}{\frac{4}{5}}=25\cdot \dfrac{5}{4} = \dfrac{125}{4}[/tex]
