Geometric power series

1. Nov 27, 2006

fsm

I was wondering if someone could check my work:

Find the geometric power series representation of
f(x)=ln(1+2x), c=0

I get $$\\sum_{n=0}^ \\infty$$2(-2x)^n+1 on -1/2<x<1/2

2. Nov 27, 2006

$$f(x) = \ln(1+2x)$$

$$\ln(1+2x) = \frac{1}{2} \int \frac{1}{1+2x}$$

$$\int \frac{1}{1+2x} = \int \sum_{n=0}^{\infty} (-2x)^{n} = \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{n+1}$$

$$\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{n+1} = \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{2n+2}$$

3. Nov 27, 2006

fsm

Thanks!! I see what I did.