How do you prove that this infinite sum is just a power of a geometric series?

Question

How do you prove that the following series converges to $1/p^s$? $$\sum_{k=0}^\infty \binom{k+s-1}{s-1}(1-p)^k = (1-(1-p))^{-s} = p^{-s}$$ I got this from Iqbal Shahid's derivation of the expected value of the negative binomial distribution but he did not prove it.

Youtube: Negative Binomial Distribution: Expected Value

Answer 1

I'm sure there's another trick P Vanchinathan is getting at, but note that the Taylor coefficients of $(1-t)^{-s}$ are $a_k=\frac{f^{(k)}(0)}{k!}=\frac{s\dots(s-(k-1))}{n!}$. Once you see where to go with this, you should then verify that the radius is exactly what you'd want it to be using the ratio test.

Answer 2

Observe that $$ {k+s-1\choose s-1}=\frac{(k+s-1)!}{(s-1)!k!}=\frac{(k+s-1)\cdots (s+1)s}{k!}$$ $$=(-1)^k\frac{(-s)(-s-1)\cdots (-s-k+1)}{k!}=(-1)^k{-s\choose k}$$ (hence the name negative binomial distribution). Then by Newton's binomial theorem, $$ \sum_{k=0}^{\infty}{k+s-1\choose s-1}(1-p)^k=\sum_{k=0}^{\infty}{-s\choose k}(-1)^k(1-p)^k=(1-(1-p))^{-s}=p^{-s}$$