Answer
(a) $\sum_{n=1}^{\infty}-\frac{x^{n}}{n}$, $R=1$
(b) $-\sum_{n=0}^{\infty}\frac{x^{n+2}}{n+2}$
(c) $ln2=\sum_{n=1}^{\infty}-\frac{1}{n2^{n}}$
Work Step by Step
(a) $f(x)=ln(1-x)=\sum_{n=0}^{\infty}-\frac{x^{n+1}}{n+1}=\sum_{n=1}^{\infty}-\frac{x^{n}}{n}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{x^{n+1}}{n+1}}{\frac{x^{n}}{n}}|$
$=|x|\lt 1$
The given series converges with $R=1$
(b) $f(x)=xln(1-x)=-\sum_{n=0}^{\infty}\frac{x^{n+2}}{n+2}$
(c) From part (a)
$f(x)=ln(1-x)=\sum_{n=1}^{\infty}-\frac{x^{n}}{n}$
if we let $x=\frac{1}{2}$
$ln2=\sum_{n=1}^{\infty}-\frac{1}{n2^{n}}$
