<s>[INST] Which expression about power series is correct? I found something weird about power series.
In my textbook,
$$ \frac{1}{1-x} = 1+x+x^2+x^3+\cdots=\sum_{n=0}^\infty x^n \quad |x|<1  $$
and $ \frac{1}{2+x} $ could be expressed using above equation.
$$ \frac{1}{2+x} = \frac{1}{2 \left(1 + \frac{x}{2} \right)} = \frac{1}{2 \left[ 1- \left( - \frac{x}{2} \right) \right]} \\ = \frac{1}{2} \sum_{n=0}^\infty \left( - \frac{x}{2} \right)^n = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^n $$
which has interval of convergence $ (-2,2) $.
And below is my solution.
$$ \frac{1}{2+x} = \frac{1}{1-(-1-x)} = 1+(-1-x)+(-1-x)^2+(-1-x)^3+\cdots \\ =\sum_{n=0}^\infty (-1-x)^n \quad |-1-x|<1 $$
which has interval of convergence $ (-2,0) $
Here are my questions.


*

*Why above two power series have different interval of convergence from one equation $ \frac{1}{2+x} $? Why does it appear?

*How do I find out right answer?

 [/INST] Notice that $\frac{1}{x+2}$ is undefined at $x = -2$.  Any series you produce will not converge at this point.
We can write the series $\sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^n$ as $\sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}(x-0)^n$.  We say this series is centered at $0$.  We are obstructed at $x = -2$ and $|(-2)-0| = 2$ is the radius of convergence.
We can write the series $\sum_{n=0}^\infty (-1-x)^n$ as $\sum_{n=0}^\infty (-1)^n(x--1)^n$.  We say this series is centered at $-1$.  We are obstructed at $x = -2$ and $|(-2)- -1| = 1$ is the radius of convergence.
Note that, in general, the obstructions can be anywhere in the complex plane, not just on the real line, so the bound on the radius of convergence may be set by features of the function that are not as easy to see as in this example.
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