$\begin{align}  & {{S}_{\infty }}=\frac{a}{1-r} \\ & r=\frac{\tfrac{1}{12}}{\tfrac{1}{4}}=\frac{1}{12}\times \frac{4}{1}=\frac{1}{3} \\ & {{s}_{\infty }}=\frac{\tfrac{1}{4}}{1-\tfrac{1}{3}}=\frac{\tfrac{1}{4}}{\tfrac{2}{3}}=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8} \\\end{align}$