TITLE: Trouble understanding how this identity is derived: $\sum_{j=0}^{\infty}\binom{a+j}{j}x^j=(1-x)^{-a-1}$
QUESTION [5 upvotes]: $$\sum_{j=0}^{\infty}\binom{a+j}{j}x^j=(1-x)^{-a-1}$$
The $-a-1$ is throwing me off. Can anyone help me understand this identity.
I have tried letting $m=-a-1$ and then applying the binomial theorem, and letting the sum run up to $\infty$ since anything past $m$ will be $0$. But I didn't get anywhere because we'll still have $(-1)^i$ in the sum.
Since there is the $\infty$ in the sum, I have also tried thinking about it in terms of generating functions but can't get anywhere.

REPLY [5 votes]: First, in the binomal identity
$$(1+u)^m=\sum_{j=0}^\infty\binom mju^j$$
set $u=-x$ and $m=-a-1$ to obtain
$$(1-x)^{-a-1}=\sum_{j=0}^\infty\binom{-a-1}j(-x)^j=\sum_{j=0}^\infty(-1)^j\binom{-a-1}jx^j.\tag{1}$$
Next, use the identity
$$\binom{-t}j=(-1)^j\binom{t+j-1}j\tag{2}$$
with $t=a+1$ to obtain
$$(-1)^j\binom{-a-1}j=(-1)^j(-1)^j\binom{a+j}j=\binom{a+j}j.\tag{3}$$
From (1) and (3) we get
$$\boxed{(1-x)^{-a-1}=\sum_{j=0}^\infty\binom{a+j}jx^j.}$$
P.S. The identity (2) follows directly from the definition of the binomial coefficient:
$$\binom{-t}j=\frac{(-t)(-t-1)(-t-2)\cdots(-t-j+1)}{j!}=(-1)^j\cdot\frac{t(t+1)(t+2)\cdots(t+j-1)}{j!}=(-1)^j\binom{t+j-1}j.$$