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

Jun 27, 2018

$\ln \left(1 + x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{n + 1} / \left(n + 1\right)$ for $x \in \left(- 1 , 1\right)$

#### Explanation:

Start from the geometric series:

${\sum}_{n = 0}^{\infty} {q}^{n} = \frac{1}{1 - q}$

converging for $\left\mid q \right\mid < 1$.

Let $q = - t$, then:

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

converging for $\left\mid t \right\mid < 1$

Inside the interval of convergence, that is for $t \in \left(- 1 , 1\right)$, we can integrate term by term:

${\int}_{0}^{x} \frac{\mathrm{dt}}{1 + t} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\int}_{0}^{x} {t}^{n} \mathrm{dt}$

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

obtaining a series with at least the same radius of convergence $R = 1$.