I need to expand $1/z$ about $z_0=-1$. I decided to do it using both methods, which don't agree.
Using Taylor: Finding coefficients: $$f^{(n)}(z)=(-1)^n n!/z^{n+1} \Rightarrow f^{(n)}(-1)=-n!$$ Pluggin into taylor series formula:
$$\frac{1}{z}=\sum \frac{f^{(n)}(-1)}{n!}(z+1)^n=\sum -(z+1)^n$$
On the other hand using Power Series I get:
$$\frac{1}{z}=\frac{1}{z+1-1}=\frac{1}{z+1}\frac{1}{1-1/(z+1)}=\frac{1}{z+1}\sum \frac{1}{(z+1)}^n=\sum \frac{1}{(z+1)^{n+1}}$$
Why didn't you do the following?:
$$\frac1z=\frac1{z+1-1}=-\frac1{1-(z+1)}=-\sum_{n=1}^\infty(z+1)^n$$
and you get the same as with the Taylor series, derivatives and etc.
This is valid whenever $\;|z+1|<1\;$ , which is what you want (I presume)...and in your power series method you can't use this geometric series development since
$$\frac1{|z+1|}<1\iff|z+1|>1\;$$
which is not (I presume) what you want.