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?
Best Answer
I don't see anything wrong with $(1)$, but just remember one little thing:
$$\sum_{n=0}^\infty ar^n=\frac a{1-r}$$
if $|r|<1$.
For yours, this translates to $|x-2|<1$. As far as I see, method $(1)$ is perfectly valid with those restrictions. For example, $x=2.5$:
$$\sum_{n=0}^\infty2(1/2)^n=\frac2{1-1/2}=\frac2{1/2}=\frac2{3-(2.5)}=f(2.5)$$
There is nothing wrong here.