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I am trying to expand into a series (sorry, I'm not sure of proper terminology here but hopefully it is clear) the ratio of a polynomial in $x^2$ at two consecutive values:

$$\frac{a_0 (x+1)^n + a_2 (x+1)^{n-2} + a_3 (x+1)^{n-4} + ...}{a_0 x^n + a_2 x^{n-2} + a_3 x^{n-4} + ...}.$$

Fairly easily this can be expanded to

$$1+ \frac{n}{x} + \frac{n(n-1)}{2x^2} + ...$$

My question is: what do I do after that? I want to stay under the true value, and continuing with

$$\frac{n(n-1)(n-2)}{6}$$

goes over the true value often. Is there a way of finding the next coefficients, or do I have to stop at $x^{-2}$?

Ziggy
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    I think your question is not clear! What is $a$? How did you get the second equation? It is weird that the second equation does not depend on $x$! – Hosein Rahnama Jun 21 '16 at 17:19
  • Oh jeez, you're right. I confused myself with notation! Those a's are meant to be x's - I'll correct it. Thanks! – Ziggy Jun 21 '16 at 17:29
  • That seems reasonable now! :) How did you get the second equation easily? :) – Hosein Rahnama Jun 21 '16 at 17:46
  • Quite simply, by expanding the $(x+1)^i$'s. So each expanded bracket will start with $x^i$ - that's why you have $1$ - and then the only $x^{n-1}$ and $x^{n-2}$ terms appear from the expansion of $(x+1)^n$ (because the $x^{n-2}$ term that comes from $(x+1)^{n-2}$ has been taken by the 1). And the $a_0$ cancels out with the leading term in the denominator. (Right?) – Ziggy Jun 21 '16 at 18:17
  • It will be better to write it down in your question. I am not still sure that this is possible. – Hosein Rahnama Jun 21 '16 at 20:21

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