# $\sum_{m=0}^\infty {{n+m-1}\choose{m}}(s(1-p))^m p^n = ({\frac{p}{1-s(1-p)}})^n$ if $|s|<(1-p)^{-1}$

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 $$(p \sum_{m=0}^\infty(s(1-p))^m)^n$$

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

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. $$p^n\sum_{m=0}^\infty {{n+m-1}\choose{m}}(s(1-p))^m=\frac{p^n}{(1-s(1-p))^n}$$
• 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.