I was doing the partial fraction of $\frac{1}{x^6}(\frac{x-2}{x^2+1})$. By using long division I get: $$\frac{1}{x^6}(\frac{x-2}{x^2+1}) = \frac{1}{x^6}\left [ -2 + x + 2x^2 -x^3 -2x^4 + x^5+\frac{2x^6-x^7}{1+x^2} \right ]$$
$$\\\frac{1}{x^6}(\frac{x-2}{x^2+1}) = \left [ \frac{-2}{x^6} + \frac{1}{x^5} + \frac{2x}{x^4} -\frac{1}{x^3} -\frac{2}{x^2} + \frac{1}{x}+\frac{2-x}{1+x^2} \right ] $$
I noticed that $\frac{x-2}{x^2+1}$ looks like a geometrical sum to infinity $\frac{a}{1-r}$ with $r = -x^2$ and $a = x - 2$. So:
$$\frac{1}{x^6}\frac{x-2}{x^2+1} = \frac{1}{x^6}\left [(x-2)-x^2(x-2)+x^4(x-2)-\frac{x^6(x-2)}{x^2+1}\right ]$$
$$= \left [ \frac{-2}{x^6} + \frac{1}{x^5} + \frac{2x}{x^4} -\frac{1}{x^3} -\frac{2}{x^2} + \frac{1}{x}+\frac{2-x}{1+x^2} \right ]$$
Which yield the same result, I know that $|r| = |-x^2| < 1$ has to be true for the convergence. But the question obviously didn't state that. So my question is can I use this way to do replace the process of long division which is slower. Is it even legal to do this? And will this always be true?