Taylor series of a certain function I'm interested in how would one find Taylor series of a function 
$$f(z)=\left(1-\frac1z \right)^3$$ around $z_0=2$.
I have no clue where to start, the $3$ in the exponent baffles me. I tried to use the cube of the sum formula, but to no avail. I tried to find the series of a function $f(z)=1-1/z$ but I'm not sure where to go from there.
Any help would be appreciated.
 A: *

*First: it's easier around $0$ (as you can use "standard results" and  compose known Taylor expansions). So write $z=2+x$, from which
$$
f(z) = \left(1-\frac{1}{2+x}\right)^3  \tag{1}
$$
which you now want to expand around $x_0=0$.

*Once you have (1),
$$
\left(1-\frac{1}{2+x}\right)^3 =  \frac{1}{2^3}\left(2-\frac{1}{1+x/2}\right)^3
=  \frac{1}{2^3}\left(2-\left(1-\frac{x}{2}+o(x)\right)\right)^3\tag{2}
$$
using the series of $\frac{1}{1+u}$ for $u$ around $0$. So you get
$$
\left(1-\frac{1}{2+x}\right)^3
=  \frac{1}{2^3}\left(1+\frac{x}{2}+o(x)\right)^3
=  \frac{1}{2^3}\left(1+\frac{3x}{2}+o(x)\right)
= \frac{1}{8}+\frac{3x}{16}+o(x) \tag{3}
$$
using the series of $(1+u)^3$ for $u$ around $0$.

*Finally, you can "get back to $z$" from (3):
$$
f(z) = \frac{1}{8} + \frac{3}{16}(z-2)+ o(z-2)
$$
Of course, the above works as well if you want to expand to a higher order. Just do the same order for $x$ around $0$.
A: Note that$$\left(1-\frac1z\right)^3=1-\frac3z+\frac3{z^2}-\frac1{z^3}.\tag1$$Now,\begin{align}\frac1z&=\frac1{2+(z-2)}\\&=\frac12\cdot\frac1{1+\frac{z-2}2}\\&=\frac12-\frac{z-2}{2^2}+\frac{(z-2)^2}{2^3}-\cdots\end{align}if $|z-2|<2$. Therefore\begin{align}\frac1{z^2}&=-\left(\frac1z\right)'\\&=\frac1{2^2}-\frac{2(z-2)}{2^3}+\frac{3(z-2)^2}{2^4}-\cdots\end{align}and\begin{align}\frac1{z^3}&=-\frac12\left(\frac1{z^2}\right)'\\&=\frac{2}{2^4}-\frac{2\times3(z-2)}{2^5}+\frac{3\times4(z-2)^2}{2^6}-\cdots\end{align}Now, use $(1)$ to get everything as a single series.
