$\sum \limits_{x=k}^{\infty}~ {x-1 \choose k-1}~ q^{x-k}~z^x = z^k(1-qz)^{-k} $

Question

x and k are integers, and q is a real number between, $0 \le q \le 1$, and z is a complex number.

How to prove that f(z):

$$f(z) = \sum \limits_{x=k}^{\infty}~ {x-1 \choose k-1}~ q^{x-k}~z^x$$

equals:

$$f(z) = z^k(1-qz)^{-k}$$

I'm thinking it has something to do with binomial theorem, but i could be wrong.

$$(a+b)^n = \sum_{j=0}^{n}~ {n \choose j}~a^j~ b^{n-j}$$

---

$$f(z) = \sum \limits_{x=k}^{\infty}~ {x-1 \choose k-1}~ q^{x-k}~z^x$$

$$f(z) = \sum \limits_{x=0}^{\infty}~ {x+k-1 \choose k-1}~ q^{x+k-k}~z^{x+k}$$

$$f(z) = \sum \limits_{x=0}^{\infty}~ {x+k-1 \choose k-1}~ q^{x}~z^{x+k}$$

no idea how to get it to equal... $f(z) = z^k(1-qz)^{-k}$

seems that bionomial theorem has an upper limit of n and this summation has an upper limit of $\infty$...

Answer 1

Start with $\frac1{1-qz}=\sum_{n\ge 0}q^nz^n$ and differentiate $k-1$ times with respect to $z$ to get

$$\frac{q^{k-1}(k-1)!}{(1-qz)^k}=\sum_{n\ge 0}n^{\underline{k-1}}q^nz^{n-k+1}\;,$$

where

$$n^{\underline{m}}=\underbrace{n(n-1)(n-2)\ldots(n-m+1)}_{m\text{ factors}}$$

is the falling factorial. Multiply both sides by $z^k$ and do a bit of algebraic manipulation to get the desired result. If you’ve not seen falling factorials before, it may be helpful to note that $\binom{n}m=\frac{n^{\underline{m}}}{m!}$.

Answer 2

First, you may want to expand $\frac{1}{1-qz} = \sum_{n=0}^\infty (qz)^n$ for $|qz|<1$. We obtain $$\left(\frac{z}{1-qz}\right)^k = z^k\left(\sum_{n=0}^\infty (qz)^n \right)^k.$$ By using $$\left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{n=0}^\infty b_nx^n\right) = \sum_{n=0}^\infty \left(\sum_{k=0}^n a_{k}b_{n-k}\right) x^n,$$ we obtain $$ z^2\left(\sum_{n=0}^\infty (qz)^n \right)^2 = z^2\sum_{n=0}^\infty n(qz)^n$$ and then $$\left(\sum_{n=0}^\infty n(qz)^n\right)\left(\sum_{n=0}^\infty (qz)^n\right) = \sum_{n=0}^\infty \left(\sum_{k=0}^n k\right)(qz)^n = \sum_{n=0}^\infty \binom{n+1}{2}(qz)^n.$$ Using induction on $k$ and the identity $$\sum_{k=0}^n \binom{k}{r} = \binom{n+1}{r+1},$$ we get exactly what was initally claimed.