Binomial expansion?

1. Dec 17, 2007

gnome

The next-to-last step in the proof on pg 1 of this article

http://links.jstor.org/sici?sici=0006-3444(194511)33:3<222:OAMOEF>2.0.CO;2-Whttp://links.jstor.org/sici?sici=0006-3444%28194511%2933%3A3%3C222%3AOAMOEF%3E2.0.CO%3B2-W

makes this substitution

$$\sum_{r=0}^\infty \binom {m+r-2}{r} q^r = (1-q)^{1-m}$$

I don't see it. How does
$$(1-q)^{1-m} = \sum_{r=0}^\infty \binom{1-m}{r} (-q)^r$$
or
$$(1-q)^{1-m} = \sum_{r=0}^\infty \binom{1-m}{r} (-q)^{1-m-r}$$
transform to the expression given by Haldane?

Last edited by a moderator: Apr 23, 2017
2. Dec 17, 2007

gnome

I just remembered that this m is always a positive integer so (1-m) is 0 or negative.

The key to this must be here...

http://mathworld.wolfram.com/NegativeBinomialSeries.html" [Broken]

Last edited by a moderator: May 3, 2017
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