Thread: Integrating a series and finding its sum

1. Integrating a series and finding its sum

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

Defined as a function:
$f(x)=\sum^{\infty}_{n=1}\frac{n}{2^{n-1}}x^{n-1}$
$x\in]-2,2[$

Show that for $|x|<2$ the following is true:
$\int_{0}^{x}f(t)dt=\frac{2x}{2-x}$

My attempt:
I first start by changing the definition under the sum from $n=1$ to $n=0$:
$f(x)=\sum^{\infty}_{n=1}\frac{n}{2^{n-1}}x^{n-1}=\sum^{\infty}_{n=0}\frac{n+1}{2^{(n+1)-1}}x^{(n+1)-1}=\sum^{\infty}_{n=0}\frac{n+1}{2^n}x^n$

To integrate a series, I use the following formula:
$\int_{0}^{b}\sum^{\infty}_{n=0}c_nx^ndx=\sum^{\inf ty}_{n=0}\frac{c_n}{n+1}b^{n+1}$

Using that formula I get:
$\int_{0}^{x}\sum^{\infty}_{n=0}\frac{n+1}{2^n}t^nd t=\sum^{\infty}_{n=0}\frac{\frac{n+1}{2^n}}{n+1}x^ {n+1}=\sum^{\infty}_{n=0}\frac{n+1}{(n+1)2^n}x^{n+ 1}=\sum^{\infty}_{n=0}2^{-n}x^{n+1}$

But now I have no idea what to do next.
I know that the sum for a power series is:
$\sum^{\infty}_{n=0}x^n=\frac{1}{1-x}$
But the problem with this formula is that it only counts for $|x|<1$ whereas I have $|x|<2$

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

2. Re: Integrating a series and finding its sum

Note that
$\sum_{n=0}^{\infty} 2^{-n}x^{n+1} = \sum_{n=0}^{\infty} \frac{x^n x}{2^n} = x \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$
This geometric serie converges if $\left|\frac{x}{2}\right|<1 \Rightarrow |x|<2$ with sum
$\frac{1}{1-\frac{x}{2}} = \frac{2}{2-x}$
therefore (with the factor $x$) the sum is $\frac{2x}{2-x}$