# Series

• September 25th 2010, 04:47 PM
rondo09
Series
Show by integrating the series for 1/(1 + x), that if |x| < 1, then

$\ln (1 + x)$ = ${\Sigma_{n=1}^{\infty}}\frac {-1^{(n+1)}}{n} x^n$
• September 25th 2010, 05:45 PM
bobby76
isn't that use Taylor expansion series?
• September 25th 2010, 06:56 PM
TheCoffeeMachine
Consider the suggested series and integrate both sides (the RHS term-by-term and let u = 1+x for the LHS):

$\displaystyle \frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^nx^n \Rightarrow \int\frac{1}{1+x}\;{dx} = \int\sum_{n=0}^{\infty}(-1)^nx^n \;{dx} \Rightarrow \ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}.$