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I am looking for tips how to solve this sum: $$ \sum_{m=0}^\infty {{n+m-1}\choose{m}}(s(1-p))^m p^n = ({\frac{p}{1-s(1-p)}})^n$$ $$|s|<(1-p)^{-1}$$ I suspect it must be somehow linked to geometric series but can´t see how. I tried working backward but don`t really know what to do with sum like $$(p \sum_{m=0}^\infty(s(1-p))^m)^n$$

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  • $\begingroup$ For $n=1$ the sum is easy, have you tried induction for higher $n$? $\endgroup$ May 16, 2017 at 9:35
  • $\begingroup$ If I understand you correctlly:result is not given, just wrote it here so people know solution. $\endgroup$
    – econ
    May 16, 2017 at 9:37

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Since $|s|<(1-p)^{-1} \Rightarrow |s(1-p)|<1$, the sum converges and following identity can be used: $$\sum_{m=0}^\infty {{n+m-1}\choose{m}}x^m=\frac{1}{(1-x)^n},$$ from which the result instantly follows. $$p^n\sum_{m=0}^\infty {{n+m-1}\choose{m}}(s(1-p))^m=\frac{p^n}{(1-s(1-p))^n}$$

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  • $\begingroup$ Thanks! Can you pleas explain identity: $$\sum_{m=0}^\infty {{n+m-1}\choose{m}}x^m=\frac{1}{(1-x)^n}$$ . I was looking for something like this but could not find it. $\endgroup$
    – econ
    May 16, 2017 at 9:54
  • $\begingroup$ Don't really know the name for identity,but theres a proof for general case: [link]gotohaggstrom.com/… . I only know combinatorial proof which is kinda hard to explain here. $\endgroup$
    – M.P
    May 16, 2017 at 10:14

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