# How do you find a power series representation for f(x)= x/(9+x^2) and what is the radius of convergence?

Jun 27, 2018

$\frac{x}{9 + {x}^{2}} = {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {x}^{2 k + 1} / {3}^{2 k + 2}$

with radius of convergence $R = 3$.

#### Explanation:

Note that:

$\frac{x}{9 + {x}^{2}} = \frac{x}{9} \frac{1}{1 + {\left(\frac{x}{3}\right)}^{2}}$

Consider the sum of the geometric series:

${\sum}_{k = 0}^{\infty} {q}^{k} = \frac{1}{1 - q}$

converging for $\left\mid q \right\mid < 1$

Let $q = - {\left(\frac{x}{3}\right)}^{2}$ then:

$\frac{1}{1 + {\left(\frac{x}{3}\right)}^{2}} = {\sum}_{k = 0}^{\infty} {\left(- {\left(\frac{x}{3}\right)}^{2}\right)}^{k} = {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {x}^{2 k} / {3}^{2 k}$

converging for ${\left(\frac{x}{3}\right)}^{2} < 1$, that is for $x \in \left(- 3 , 3\right)$.

Now:

$\frac{x}{9 + {x}^{2}} = \frac{x}{9} {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {x}^{2 k} / {3}^{2 k} = {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {x}^{2 k + 1} / {3}^{2 k + 2}$

with radius of convergence $R = 3$.