Chapter 9.10, Problem 37ES

Use the results in Exercise 36 to find the sum of the series.

$\begin{array}{l}\left(\text{a}\right){\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{k}{3^k}}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\mathrm{...}\\ \left(\text{b}\right){\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{1}{k(4^k)}=\frac{1}{4}+\frac{1}{2(4^2)}+\frac{1}{3(4^3)}+\frac{1}{4(4^4)}+\mathrm{...}}\end{array}$