Geometric Power Series Representation of ln(1+2x) at c=0

[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

 

[courtrigrad]

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

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

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


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

 

[fsm]

Thanks! I see what I did.