<s>[INST] failed application of magicry in Taylor expansion of $1/x^2$ near $x=2$ It's straightforward to find the Taylor expansion for $\frac{1}{x^2}$ near $x=2$ using the the Taylor series definition.
This is turns out to be $\frac{1}{4} - \frac{1}{4} (x-2) + \frac{3}{16}(x-2)^2 + \cdots$
I was trying to be cute by finding the expansion using the geometric series:
$$
\begin{align}
\frac{1}{x^2} &= \frac{1}{1 -(1 - x^2)} = \frac{1}{1-z} &\text{ where } z  = 1-x^2\\
\end{align}
$$
This is where I run into trouble. My initial guess was to expand around $x=0$ and then shift over by 2.
$$
\begin{align}
&1 + z + z^2+ \cdots\\
&1 + (1-x^2) + (1-x^2) ^2 + \cdots \\
&1 + (\ 1-(x-2)^2\ ) + (\ 1-(x-2)^2\ )^2+\cdots
\end{align}
$$
I looked at the Wikipedia page on the geometric series and see that the formula is different when the common ratio is not 1. 
$$a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r}$$
This helped me to see that I had missed the $a$ term. I'm taking it on faith that the common ratio is less than one. The article points out that when $|r|<1$, the series will be $\frac{a}{1-r}$, which is the familiar geometric series.
Somehow, my substitution seems to be wrong. Any help will be appreciated.
 [/INST] Your idea is sound... iff
$$|1-x^2|<1\implies 0<x^2<2\implies |x|<\sqrt2$$
Now, you want to develop around $\,x=2\,$...not good, uh?
My idea: 
$$\frac1x=\frac1{2+(x-2)}=\frac12\frac1{1+\left(\frac{x-2}2\right)}=\frac12\left(1-\frac{x-2}2+\frac{(x-2)^2}4-\ldots\right)\implies$$
$$\frac1{x^2}=\frac1x\frac1x=\frac14\left(1-\frac{x-2}2+\frac{(x-2)^2}4-\ldots\right)^2=$$
$$=\frac14\left(1-(x-2)+\frac34(x-2)^2-\frac38(x-2)^3+\ldots\right)$$
The only thing you need now to justify the above is to prove that
$$\left|\frac{x-2}2\right|<1\iff |x-2|<2\iff\;\ldots$$
</s>