I was proofing a formula when I meet a summation that I culdn't solve. After some efforts and investigations I've successfully recognized it in its generalized formula: $$\sum_{a=b}^{\infty}{\binom{a-1}{b-1}x^{a-b}}=(1-x)^{-b}$$ that I saw online in a list of knowed series.
I've searched for a long time now, but I can't find information about it, even it's name, can you help me please?
I would like to prove it.
It's a form of the generalized binomial theorem.
It can be rewritten like this:
$\begin{array}\\ \sum_{a=b}^{\infty}{\binom{a-1}{b-1}x^{a-b}} &=\sum_{a=0}^{\infty}{\binom{a+b-1}{b-1}x^{a}}\\ \end{array} $
From the generalized binomial theorem:
$\begin{array}\\ (1-x)^{-b} &=\sum_{n=0}^{\infty} \binom{-b}{n} (-x)^n\\ &=\sum_{n=0}^{\infty} \dfrac{\prod_{k=0}^{n-1}(-b-k)}{n!} (-1)^nx^n\\ &=\sum_{n=0}^{\infty} \dfrac{(-1)^n\prod_{k=0}^{n-1}(b+k)}{n!} (-1)^nx^n\\ &=\sum_{n=0}^{\infty} \dfrac{\prod_{k=b}^{n+b-1}(k)}{n!} x^n\\ &=\sum_{n=0}^{\infty} \dfrac{(n+b-1)!}{(b-1)!n!} x^n\\ &=\sum_{n=0}^{\infty} \binom{n+b-1}{n} x^n\\ \end{array} $
and this is the same with $n$ instead of $a$.