Answer
The Taylor series for $f(x)=\ln(1+\frac{1}{3}x)$ for $r$ in $(-3,3]$ is
$x-\frac{1}{18}x^2+\frac{1}{81}x^3-\frac{1}{324}x^4+...+\frac{(-1)^n(\frac{1}{3}x)^{n+1}}{n+1}+...$
Work Step by Step
We are given $f(x)=\ln(1+\frac{1}{3}x)$
for $r$ in $(-3,3]$
The Taylor series for $\ln(1+x)$ is
$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...+\frac{(-1)^nx^{n+1}}{n+1}+...$
The Taylor series for $f(x)=\ln(1+\frac{1}{3}x)$ is
$f(x)=x-\frac{1}{2}(\frac{1}{3}x)^2+\frac{1}{3}(\frac{1}{3}x)^3-\frac{1}{4}(\frac{1}{3}x)^4+....+\frac{(-1)^n(\frac{1}{3}x)^{n+1}}{n+1}+...$
$=x-\frac{1}{18}x^2+\frac{1}{81}x^3-\frac{1}{324}x^4+...+\frac{(-1)^n(\frac{1}{3}x)^{n+1}}{n+1}+...$
