Chapter 9.10, Problem 38ES

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

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