Why does changing the center of a geometric power series change the interval of convergence? I know that the interval of convergence of the geometric power series
$$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ is $(-1,1)$. 
Why is it that if I do the following manipulation $$\frac{1}{1-x}=\frac{1}{2-(x+1)}=\frac{1/2}{1-(x+1)/2},$$
which causes the center to change to $x=-1$ change the interval of convergence to $(-3,1)$? I understand what is going on algebraically, but do not understand what is actually going on. I don't understand why not even the radius remains the same! 
Could someone clearly explain this to me? Thank you!
Correction: Let me be more specific. I do not understand why I am getting two different intervals of convergence for the same geometric power series.
 A: You are not finding different intervals of convergence for the same power series. You are finding two different power series, centered at different points and converging on different intervals, for the same function. That function is of course $f(x) = 1/(1-x).$ Note that $f$ blows up at $1.$ So if you center the power series of $f$ at $0,$ you can't expect it to converge at $x=1$! But if you center the power series of $f$ at $-1,$ the distance to the blow up point is now $2.$ That is why you found a radius of convergence of $2$ for the other power series.
A: Hint. One may write
$$
\frac{1}{\large 1-\frac{x+1}2}=\frac1{1-z}
$$
then, as trying to expand the latter in power series of $z$, one needs that
$$
|z|\lt1
$$ that is $$\left|\frac{x+1}2\right|\lt1$$ giving
$$
-1\lt\frac{x+1}2\lt1
$$ which one solves to get $-3\lt x\lt1$.
A: Complex Analysis
I think the subject becomes clear when viewed using complex analysis. In this Wikipedia article it is stated that

The radius of convergence of a power series $f$ centered on a point $a$ is equal to the distance from $a$ to the nearest point where $f$ cannot be defined in a way that makes it holomorphic.

This can be proven using Cauchy's Integral Formula.
The function
$$
f(x)=\frac1{1-x}
$$
can be extended to an analytic function on $\mathbb{C}$, and as such, has only one singularity, at $x=1$.  The radius of convergence of the power series for $f$ is the distance from the center of the expansion to that singularity.
For example,
$$
\frac1{1-x}=1+x+x^2+x^3+x^4+\dots
$$
is centered at $x=0$.  The radius of convergence is $\left|\,1-0\,\right|=1$. Therefore, its real interval of convergence is $(-1,1)$.
Another way to write $f$ is
$$
\frac1{2-(x+1)}=\frac12+\frac{x+1}4+\frac{(x+1)^2}8+\frac{(x+1)^3}{16}+\dots
$$
is centered at $x=-1$ and its radius of convergence is $\left|\,1-(-1)\,\right|=2$. Therefore, its real interval of convergence is $(-3,1)$.
We could also write $f$ as
$$
\frac1{3-(x+2)}=\frac13+\frac{x+2}9+\frac{(x+2)^2}{27}+\frac{(x+2)^3}{81}+\dots
$$
which is centered at $-2$. Its radius of convergence is $\left|\,1-(-2)\,\right|=3$. Therefore, its real interval of convergence is $(-5,1)$.
