TITLE: taylor of $\frac{1}{z}$ at $a=-2$
QUESTION [5 upvotes]: I want to find the taylor series representation of $f(z)=\frac{1}{z}$ at $a=-2$.
The point of this exercise is not to find some pattern in the derivatives, infact we are not meant to find any derivatives at all.
The idea is to somehow use the known formula $\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$ or other manipulations.
This is what the teacher did, and i find it severely lacking, I can't understand how he reached his answer and would appreciate an explanation
$\frac{1}{z}=\frac{1}{z+2-2}=-\sum_{n=0}^{\infty}2^{-n-1}(z+2)^n$, without any explanation.
I understand why $\frac{1}{z}=\frac{1}{z+2-2}$ obviously, but why is that equal to that sum?

REPLY [2 votes]: you understand that the radius of convergence of the geometric series $\dfrac{1}{1- u} = 1 + u + u^2 + \cdots $ is $|u| < 1.$ if you want to expand $\dfrac{1}{z}$  about $z = -2,$ then the radius convergence can at most be $2$ because $\dfrac{1}{z}$ is undefined at $z = 0$  that is why you write 
$$\dfrac{1}{z} = \dfrac{1}{-2 + (z+2)} = -\dfrac{1}{2}\dfrac{1}{1 - (z+2)/2} = -\dfrac{1}{2}\dfrac{1}{1 - u}=-\dfrac{1}{2}\{ 1 + u + u^2 + \cdots\} \text{ for } |u| < 1$$ where we have used $u = \dfrac{z+2}{2}$ and $|u| < 1$ translates to $|z=2| < 2.$