# How do you find a power series representation for x/(1-x^2) and what is the radius of convergence?

Oct 24, 2015

Use the Maclaurin series for $\frac{1}{1 - t}$ and substitution to find:

$\frac{x}{1 - {x}^{2}} = {\sum}_{n = 0}^{\infty} {x}^{2 n + 1}$

with radius of convergence $1$.

#### Explanation:

The Maclaurin series for $\frac{1}{1 - t}$ is ${\sum}_{n = 0}^{\infty} {t}^{n}$

since $\left(1 - t\right) {\sum}_{n = 0}^{\infty} {t}^{n} = {\sum}_{n = 0}^{\infty} {t}^{n} - t {\sum}_{n = 0}^{\infty} {t}^{n} = {\sum}_{n = 0}^{\infty} {t}^{n} - {\sum}_{n = 1}^{\infty} {t}^{n} = {t}^{0} = 1$

Substitute $t = {x}^{2}$ to get:

$\frac{1}{1 - {x}^{2}} = {\sum}_{n = 0}^{\infty} {\left({x}^{2}\right)}^{n} = {\sum}_{n = 0}^{\infty} {x}^{2 n}$

Multiply by $x$ to get:

$\frac{x}{1 - {x}^{2}} = x {\sum}_{n = 0}^{\infty} {x}^{2 n} = {\sum}_{n = 0}^{\infty} {x}^{2 n + 1}$

This is a geometric series with common ratio ${x}^{2}$, so it will converge if $\left\mid {x}^{2} \right\mid < 1$, which is when $\left\mid x \right\mid < 1$. That is, the radius of convergence is $1$.