# How do you find the taylor series for ln(1+x)?

Mar 11, 2017

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

Replacing $x$ with $- x$:

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

Note that integrating $\frac{1}{1 + x}$ gives $\ln \left(1 + x\right) + C$:

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

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

Letting $x = 0$ shows that $C = 0$:

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