Answer
The power series for $f(x)=\frac{x}{9+x^2}$ is $\sum_{n=0}^\infty \frac{x^{2n+1}}{9^{n+1}}$ with the interval of convergence is $(-3,3)$.
Work Step by Step
We know the following power series:
$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$ where $|x|<1$
Replacing $x$ by $-x^2/9$,
$\frac{1}{1+x^2/9}=\sum_{n=0}^\infty (-x^2/9)^n$ where $|-x^2/9|<1$
$\frac{9}{9+x^2}=\sum_{n=0}^\infty \frac{x^{2n}}{9^n}$ where $|x^2|<9$ (Multiply by $x/9$)
$\frac{x}{9+x^2}=\sum_{n=0}^\infty \frac{x^{2n+1}}{9^{n+1}}$ where $|x|<3$
Thus, the power series for $f(x)=\frac{x}{9+x^2}$ is $\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{9^{n+1}}$ with the interval of convergence is $(-3,3)$.
