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

May 6, 2017

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

The radius of convergence is equal to $1$ by the ratio test.

#### Explanation:

Remember the MacLaurin series representation for $\ln \left(1 + u\right)$:
$\textcolor{b l u e}{\ln \left(1 + u\right) = {\sum}_{n = 0}^{\infty} \frac{{\left(- 1\right)}^{n} {\left(u\right)}^{n + 1}}{n + 1}}$

Substitute $u = - {x}^{2}$:

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

$\textcolor{w h i t e}{\ln \left(1 - {x}^{2}\right)} = {\sum}_{n = 0}^{\infty} \frac{{\left(- 1\right)}^{n} {\left(- 1\right)}^{n + 1} {\left(x\right)}^{2 n + 2}}{n + 1}$

$\textcolor{w h i t e}{\ln \left(1 - {x}^{2}\right)} = {\sum}_{n = 0}^{\infty} \frac{- {\left(x\right)}^{2 n + 2}}{n + 1}$

To find radius of convergence, use ratio test, which states that if ${\lim}_{x \to \infty} | \frac{{a}_{n + 1}}{{a}_{n}} | < 1$, then ${\sum}_{n = 0}^{\infty} {a}_{n}$ converges

${\lim}_{n \to \infty} | \frac{\textcolor{red}{-} {\left(x\right)}^{2 n + 4}}{n + 2} \cdot \frac{n + 1}{\textcolor{red}{-} {\left(x\right)}^{2 n + 2}} |$

$= {\lim}_{x \to \infty} | \frac{\textcolor{red}{{\left(x\right)}^{2 n}} {\left(x\right)}^{4} \left(n + 1\right)}{\left(n + 2\right) \textcolor{red}{{\left(x\right)}^{2 n}} {\left(x\right)}^{2}} |$

$= | {x}^{2} |$

In order for the series to converge, set this less than 1:
${x}^{2} < 1 , \textcolor{red}{{x}^{2} > - 1}$

The radius of convergence is equal to $1$ by the ratio test.