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Find a power series around $a=0$ for the function $f:\mathbb{R}\backslash\{2,3\} \to \mathbb{R}$ with $f(x) = \frac{1}{x^2-5x+6}$.

It is

$f(x) = \frac{1}{x^2-5x+6} = \frac{1}{2-x}\frac{1}{3-x} = \frac{1}{1-(x-1)}\frac{1}{1-(x-2)}$

Now I could use the geometric series but I need to take care of the range of $x$ then, right?

$= \sum_{n=0}^{\infty} (x-1)^n \cdot \sum_{n=0}^{\infty} (x-2)^n$

But anyway, I don't know how to proceed from there.. Any hints

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Hint: Before attempting to expand, express $\frac{1}{x^2-5x+6}$ as $\frac{A}{2-x}+\frac{B}{3-x}$, using the partial fractions idea. (The idea has many uses, it's not just for integration!)

After the partial fractions process, you will want to use a method other than the one you proposed. Perhaps (with different numbers) something like $$\frac{1}{7-3x}=\frac{1}{7}\frac{1}{1-\frac{3x}{7}}.$$

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Andre is correct--it is possible to write this as a product of power series, but by writing it as a sum it will be far easier to expand. – Chris K. Caldwell Jun 4 '12 at 14:36
I don't see how to expand it to partial fractions.. – menfredk Jun 4 '12 at 20:32
$\frac{A}{2-x}+\frac{B}{3-x}=\frac{A(3-x)+B(2-x)}{(2-x)(3-x)}$. Want top to be identically $1$. So $3A+2B=1$, $-A-B=0$. So $B=-A$. Substitute in $3A+2B=1$. We get $A=1$, so $B=-1$, and therefore $\frac{1}{x^2-5x+6}=\frac{1}{2-x}-\frac{1}{3-x}$. Now use the "After" part on each of these. – André Nicolas Jun 4 '12 at 20:49
Thanks. At the end I get $\sum_{n=0}^{\infty} \left(\frac{1}{2^{n+1}} - \frac{1}{3^{n+1}} \right) x$. To calculate the radius of convergence I then need $\operatorname{lim sup}_{n \to \infty} \sqrt[n]{\frac{1}{2^{n+1}} - \frac{1}{3^{n+1}}}$, richt? Any hints for this limit? – menfredk Jun 5 '12 at 4:30
@menfredk: The general first term is $\frac{1}{2}\left(\frac{x}{2}\right)^n$. This has radius of convergence $2$. The other guy is similar. It has radius of convergence $3$. The sum (difference) has radius of convergence the smaller of $2$ and $3$. So the whole thing has radius of convergence $2$. Where you have $x$ it sould be $x^n$. – André Nicolas Jun 5 '12 at 5:10

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