1. ## Logarithm property

Can someone help me proof that for any $\displaystyle x \in (-1,1)$ we have $\displaystyle ln(1-x) = -\sum_{n=1}^{\infty}{\frac{x^n}{n}}$
thanks

2. Originally Posted by bram kierkels
Can someone help me proof that for any $\displaystyle x \in (-1,1)$ we have $\displaystyle ln(1-x) = -\sum_{n=1}^{\infty}{\frac{x^n}{n}}$
thanks
Hints:
$\displaystyle \sum\limits_{n = 0}^\infty {x^n } = \frac{1} {{1 - x}}\;\& \,\frac{d} {{dx}}\ln (1 - x) = ?$

3. or $\displaystyle -\ln (1-x)=\int_{0}^{x}{\frac{dt}{1-t}},$ and you can use geometric series there to change the sum by the integral.

note that it's actually for $\displaystyle x\in[-1,1).$

4. Originally Posted by bram kierkels
Can someone help me proof that for any $\displaystyle x \in (-1,1)$ we have $\displaystyle ln(1-x) = -\sum_{n=1}^{\infty}{\frac{x^n}{n}}$
thanks

From power series we know that $\displaystyle \frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n\,,\,\,\forall\,x\in\,(-1,1)\,\Longrightarrow -\ln(1-x)=\int\frac{1}{1-x}\,dx=\sum\limits_{n=0}^\infty\int x^n dx$ $\displaystyle =\sum\limits_{n=0}^\infty\frac{x^{n+1}}{n+1}\,,\,\ ,\forall\,x\in\,(-1,1)$