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

Sep 29, 2015

Write out a power series that when multiplied by $1 + {x}^{2}$ gives $x$.

Find ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1}$ works and has radius of convergence $1$.

#### Explanation:

Consider ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} = x - {x}^{3} + {x}^{5} - {x}^{7} + \ldots$

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

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

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

$= {\left(- 1\right)}^{0} {x}^{1} = x$

So:

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

...if the sums converge.

The sum ${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1}$ is a geometric series with common ratio $- {x}^{2}$.

To converge, the absolute value of the common ratio must be less than $1$.

That is $\left\mid - {x}^{2} \right\mid < 1$, so $\left\mid x \right\mid < 1$

That is: the radius of convergence is $1$.