Chapter 9.10, Problem 36ES

(a) Differentiate the Maclaurin series for $1/(1-x),$ and use the result to show that

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}k{x}^k}=\frac{x}{{(1-x)}^2}\text{for}-1<x<1$

(b) Integrate the Maclaurin series for $1/(1-x),$ and use the result to show that

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{x^k}{k}=-\ln (1-x)\text{for}-1<x<1}$

(c) Use the result in part (b) to show that

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{(-1)}^{k+1}\frac{x^k}{k}=\ln (1+x)\text{for}-1<x<1}$

(d) Show that the series in part (c) converges if $x=1.$

(e) Use the remark following Example 3 to show that

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{(-1)}^{k+1}\frac{x^k}{k}=\ln (1+x)\text{for}-1<x\le 1}$