<s>[INST] Power series $\ \frac{x}{(x-2)(x-3)}$, radius of convergence I got a little trouble with this function while expanding this to power series:
$$\ f(x) = \frac{x}{(x-2)(x-3)} $$
After partial fraction decomposition:
$$\ f(x) = \frac{3}{x-3} - \frac{2}{x-2} $$
I can see here geometric series and rewrite as the sums:
$$\ \frac{1}{1-t} = \sum^{\infty}_{n=0} t^n $$
$$\ f(x) = 3  \frac{1}{\frac{1}{3}x-1} - 2 \frac{1}{1-\frac{1}{2}x} $$
I can set $ t = \frac{1}{2}x $, but what can I do with $ \frac{1}{3}x $? The signs don't match in the denominator to the formula.
As a radius of convergence, I would choose $-2<x<2$ as the smaller range for the sum of the series
EDIT: I have tried this - with your suggestion:
$$\ f(x) = \frac{3}{3}  \frac{1}{\frac{1}{3}x-1} - \frac{2}{2} \frac{1}{1-\frac{1}{2}x} $$
$$\ f(x) = \frac{3}{3}  \frac{1}{-(1-\frac{1}{3}x)} - \frac{2}{2} \frac{1}{1-\frac{1}{2}x} $$
$$\ f(x) = -\frac{3}{3}  \frac{1}{-(1-\frac{1}{3}x)} - \frac{2}{2} \frac{1}{1-\frac{1}{2}x} $$
$$\ -(\sum^{\infty}_{n=0} (\frac{1}{2}x)^n + \sum^{\infty}_{n=0} (\frac{1}{3}x)^n) $$
$$\ \min\{2,3\} = 2 ~~\text{radius of convergence} $$
Is it right?
 [/INST] Hint:
We can write $$\dfrac3{x-3}=-\dfrac1{1-\dfrac x3}$$ right?
So, we need $|x|<$min$(2,3)$
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