Finding a Power Series representation for the function $f(x) = \frac{2}{3-x}$ Let's say I want to find a Power Series representation of the function $f(x) = \frac{2}{3-x}$
Now I know we can write this as a geometric series
$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}$$
But I see two possible ways two write it as a geometric series. $(1)$ and $(2)$ below

Method $(1)$:
We let $a=2$, and $r = x-2$, thus we transform $\frac{2}{3-x}$ into the form $\frac{a}{1-r}$ and we write the Power Series for $f$ as follows:
$$f(x) = \sum_{n=0}^{\infty}\ 2(x-2)^n$$

Method $(2)$:
We divide both numerator and denominator by a factor of $3$ to go from $\frac{2}{3-x}$ to $$\frac{\frac{2}{3}}{1-\frac{x}{3}}$$
and we can then write $f$ as follows:
$$\begin{aligned}f(x) &= \sum_{n=0}^{\infty}\ \frac{2}{3}\left(\frac{x}{3}\right)^n
\\
&= \sum_{n=0}^{\infty}\left(\frac{2}{3^{n+1}}\right)x^n
\end{aligned}$$

But only $(2)$ is correct, and $(1)$ is incorrect, but I can't seem to see why. What I did in $(1)$ seemed like perfectly valid algebraic manipulations, so why does $(1)$ result in an erroneous answer?

Both methods are quite ok and valid and there are even many more possibilities to expand the function into a power series.

In order to better see what's going on, we should at first look somewhat closer at the function $f$. We consider the full specification of $f$ and choose as domain and codomain 
\begin{align*}
&f:\mathbb{R}\setminus\{3\}\longrightarrow\mathbb{R}\\
&f(x)=\frac{2}{3-x}
\end{align*}
  Note that we have a singularity, a simple pole at $x=3$. This is crucial when expanding a function as power series.

As important as  domain and codomain is the radius of convergence when expanding a function as power series. 
\begin{align*}
\sum_{n=0}^\infty a x^n=\frac{a}{1-x}\qquad\qquad |x|<1
\end{align*}
It is the radius of convergence which determines where the representation of $f$ as power series is valid.

If we look at the series expansion of the first method
  \begin{align*}
f(x)=\sum_{n=0}^\infty2(x-2)^n\qquad\qquad\qquad |x-2|<1\tag{1}
\end{align*}
  and the second method gives
  \begin{align*}
f(x)=\sum_{n=0}^\infty\left(\frac{2}{3^{n+1}}\right)x^{n}=\frac{\frac{2}{3}}{1-\frac{x}{3}}\qquad\qquad \left|\frac{x}{3}\right|<1\tag{2}
\end{align*}
Attention: We have to explicitely state in (1) and (2) the range of validity since the power series is not defined outside this range. On the other hand $f$ can be defined on a much larger domain $\mathbb{R}\setminus\{3\}$.

Here's a graphic which illustrates both methods. We  see the graph of $f$ with the asymptote at $x=3$.
                  


*

*Method 1: The point $A=(2,2)$ is the center of an interval with length $2$ showing the validity of the power expansion. We clearly see the interval  is bounded by the asymptote.

*Method 2: The point $B=\left(0,\frac{2}{3}\right)$ is the center of an interval with length $6$ showing the validity of the power expansion.

More expansions:
The graphic indicates what's going on, when we expand $f$ in a power series at a point $x=x_0$. The point $x_0$ is the center of an interval and the length is determined by the distance from $x_0$ to the asymptote $x=3$.
We can now expand $f$ at any point $x_0\in \mathbb{R}\setminus \{3\}$ as follows:
\begin{align*}
\frac{2}{3-x}&=\frac{2}{(3-x_0)-(x-x_0)}\\
&=\frac{2}{3-x_0}\cdot\frac{1}{1-\frac{x-x_0}{3-x_0}}\\
&=\frac{2}{3-x_0}\sum_{n=0}^\infty\left(\frac{x-x_0}{3-x_0}\right)^n\\
&=\sum_{n=0}^\infty\frac{2}{(3-x_0)^{n+1}}(x-x_0)^n\qquad\qquad \left|\frac{x-x_0}{3-x_0}\right|<1
\end{align*}
Of course, setting $x_0=2$ we obtain $\sum_{n=0}2(x-2)^n$ and setting $x_0=0$ we obtain $\sum_{n=0}^\infty\frac{2}{3^{n+1}}x^n$ and get back both methods as special cases.

