Need help following my textbook (series)

[isukatphysics69]

Homework Statement

Homework Equations

The Attempt at a Solution

I am not following what is going on here, how are they getting that part that is circled. i am just completely lost here

 

[FactChecker]

They want to expand in powers of $(x-(-1)) = (x+1)$ so that the series will converge in a region centered at $x=-1$. And they want it in the form $\frac{a}{1-r}$. That line gives the answer.

 

[isukatphysics69]

this isn't making sense to me. i don't think I'm cut out for this stuff idk why I'm here doing this too much math and physics can't keep up

 

[FactChecker]

The Macclaurin series, $\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$ converges in the interval from $x=-1$ to $x=1$ (centered at $x=0$). They say that they want a series centered at $x=-1$. So transform $\frac{1}{1-x}$ with $x-(-1)=x+1$ in the x position of the denominator $\frac{1}{2-\textbf{(x+1)}}$. But that puts a 2 in the denominator $\frac{1}{\textbf{2}-(x+1)}$. So divide both numerator and denominator by 2 to get $\frac{1/2}{1-[(x+1)/2]}$. This is exactly the standard form of $\frac{a}{1-r}$, where $a=1/2$ and $r=(x+1)/2$.
The series $\frac{a}{1-r} = a\sum_{n=0}^{\infty}r^n$ converges in the interval from $r=-1$ to $r=1$.
That is from $-1=r=\frac{(x+1)}{2}$ to $1=r=\frac{(x+1)}{2}$. Or from $x=-3$ to $x=1$. This is centered at $x=-1$, as desired.

 

[isukatphysics69]

@FactChecker Thank you! i will take a look at this when i get back to calculus i have been spending the past few days on other things