# How do you find the Maclaurin series of f(x)=ln(1+x) ?

Aug 29, 2014

The Maclaurin series of $f \left(x\right) = \ln \left(1 + x\right)$ is:
$f \left(x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{{x}^{n + 1}}{n + 1}$,
where $| x | < 1$.

First, let us find the Maclaurin series for
$f ' \left(x\right) = \frac{1}{1 + x} = \frac{1}{1 - \left(- x\right)}$.

Remember that
$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$ if $| x | < 1$.
(Note: This can be justified by viewing it as a geometric series.)

By replacing $x$ by $- x$,
$f ' \left(x\right) = \frac{1}{1 - \left(- x\right)} = {\sum}_{n = 0}^{\infty} {\left(- x\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{n}$

By integrating using Power Rule,
$f \left(x\right) = \int {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{n} \mathrm{dx} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{{x}^{n + 1}}{n + 1} + C$
(Note: integration can be done term by term.)

Since $f \left(0\right) = \ln \left[1 + \left(0\right)\right] = 0$,
$f \left(0\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{{\left(0\right)}^{n + 1}}{n + 1} + C = C = 0$.

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