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?