# What is the power series representation of ln((1+x)/(1-x))?

Dec 4, 2016

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

#### Explanation:

I would use the following

1. The log rule; $\log \left(\frac{A}{B}\right) = \log A - \log B$
2. The known power series :
$\ln \left(1 + x\right) = 1 - {x}^{2} / 2 + {x}^{3} / 3 - {x}^{4} \ldots = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} {x}^{n} / n$

So:
$\setminus \setminus \setminus \setminus \setminus \ln \left(\frac{1 + x}{1 - x}\right) = \ln \left(1 + x\right) - \ln \left(1 - x\right)$
$\therefore \ln \left(\frac{1 + x}{1 - x}\right) = \left\{1 - {x}^{2} / 2 + {x}^{3} / 3 - {x}^{4} + \ldots\right\} - \left\{1 - {\left(- x\right)}^{2} / 2 + {\left(- x\right)}^{3} / 3 - {\left(- x\right)}^{4} + \ldots\right\}$
$\therefore \ln \left(\frac{1 + x}{1 - x}\right) = \left\{1 - {x}^{2} / 2 + {x}^{3} / 3 - {x}^{4} + \ldots\right\} - \left\{1 - {x}^{2} / 2 - {x}^{3} / 3 - {x}^{4} + \ldots\right\}$
$\therefore \ln \left(\frac{1 + x}{1 - x}\right) = 1 - {x}^{2} / 2 + {x}^{3} / 3 - {x}^{4} + \ldots - 1 + {x}^{2} / 2 + {x}^{3} / 3 + {x}^{4} + \ldots$
$\therefore \ln \left(\frac{1 + x}{1 - x}\right) = 2 {x}^{3} / 3 + 2 {x}^{5} / 5 + 2 {x}^{7} / 7 \ldots = 2 {\sum}_{n = 1}^{\infty} {x}^{2 n + 1} / \left(2 n + 1\right)$