Given:

$\displaystyle \sum^{\infty}_{n=1}\frac{n}{2^{n-1}}x^{n-1}$

Radius of convergence is 2.

Defined as a function:

$\displaystyle f(x)=\sum^{\infty}_{n=1}\frac{n}{2^{n-1}}x^{n-1}$

$\displaystyle x\in]-2,2[$

Show that for $\displaystyle |x|<2$ the following is true:

$\displaystyle \int_{0}^{x}f(t)dt=\frac{2x}{2-x}$

My attempt:

I first start by changing the definition under the sum from $\displaystyle n=1$ to $\displaystyle n=0$:

$\displaystyle f(x)=\sum^{\infty}_{n=1}\frac{n}{2^{n-1}}x^{n-1}=\sum^{\infty}_{n=0}\frac{n+1}{2^{(n+1)-1}}x^{(n+1)-1}=\sum^{\infty}_{n=0}\frac{n+1}{2^n}x^n$

To integrate a series, I use the following formula:

$\displaystyle \int_{0}^{b}\sum^{\infty}_{n=0}c_nx^ndx=\sum^{\inf ty}_{n=0}\frac{c_n}{n+1}b^{n+1}$

Using that formula I get:

$\displaystyle \int_{0}^{x}\sum^{\infty}_{n=0}\frac{n+1}{2^n}t^nd t=\sum^{\infty}_{n=0}\frac{\frac{n+1}{2^n}}{n+1}x^ {n+1}=\sum^{\infty}_{n=0}\frac{n+1}{(n+1)2^n}x^{n+ 1}=\sum^{\infty}_{n=0}2^{-n}x^{n+1}$

But now I have no idea what to do next.

I know that the sum for a power series is:

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

But the problem with this formula is that it only counts for $\displaystyle |x|<1$ whereas I have $\displaystyle |x|<2$

I also don't know how to express my series as $\displaystyle x^n$.