# Radius of Convergence for Power Expansion?

## Expanded as a power series in terms of $x - \frac{2}{3}$, what is the radius of convergence of $f \left(x\right) = \frac{1}{x \left(1 - x\right)}$?

Feb 28, 2017

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

with radius of convergence $R = \frac{1}{3}$

#### Explanation:

We have:

$f \left(x\right) = \frac{1}{x \left(1 - x\right)} = \frac{1}{x} + \frac{1}{1 - x}$

Substitute: $t = \left(x - \frac{2}{3}\right)$

$f \left(t\right) = \frac{1}{t + \frac{2}{3}} + \frac{1}{1 - \left(t + \frac{2}{3}\right)} = \frac{1}{\frac{2}{3} + t} + \frac{1}{\frac{1}{3} - t} = \frac{3}{2} \frac{1}{1 + \frac{3}{2} t} - \frac{3}{1 - 3 t}$

Both terms can be expressed as sum of a geometric series with ratio: $\left(- \frac{3 t}{2}\right)$ and $\left(3 t\right)$ respectively:

$\frac{1}{1 + \frac{3 t}{2}} = {\sum}_{n = 0}^{\infty} {\left(\frac{- 3 t}{2}\right)}^{n}$ for $\left\mid \frac{3 t}{2} \right\mid < 1 \implies t \in \left(- \frac{2}{3} , \frac{2}{3}\right)$

$\frac{1}{1 - 3 t} = {\sum}_{n = 0}^{\infty} {\left(3 t\right)}^{n}$ for $\left\mid 3 t \right\mid < 1 \implies t \in \left(- \frac{1}{3} , \frac{1}{3}\right)$

So we have that in the intersection of the two intervals, that is for $t \in \left(- \frac{1}{3} , \frac{1}{3}\right)$

$f \left(t\right) = \frac{3}{2} {\sum}_{n = 0}^{\infty} {\left(\frac{- 3 t}{2}\right)}^{n} - 3 {\sum}_{n = 0}^{\infty} {\left(3 t\right)}^{n}$

$f \left(t\right) = - {\sum}_{n = 0}^{\infty} {\left(- \frac{3}{2}\right)}^{n + 1} {t}^{n} - {\sum}_{n = 0}^{\infty} {3}^{n + 1} {t}^{n}$

$f \left(t\right) = - {\sum}_{n = 0}^{\infty} {3}^{n + 1} {t}^{n} \left(1 + {\left(- \frac{1}{2}\right)}^{n + 1}\right)$

and substituting back $x$:

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

with radius of convergence $R = \frac{1}{3}$