Trouble understanding how this identity is derived: $\sum_{j=0}^{\infty}\binom{a+j}{j}x^j=(1-x)^{-a-1}$ $$\sum_{j=0}^{\infty}\binom{a+j}{j}x^j=(1-x)^{-a-1}$$
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.
 A: There are several ways to prove it. Here’s one. I’ll take advantage of the fact that $\binom{a+j}j=\binom{a+j}a$ to prove instead that
$$\frac1{(1-x)^{a+1}}=\sum_{j\ge 0}\binom{a+j}ax^j\;.$$
Start with the case $a=0$:
$$\frac1{1-x}=\sum_{j\ge 0}x^j\;.$$
Now differentiate with respect to $x$:
$$\frac1{(1-x)^2}=\sum_{j\ge 0}jx^{j-1}=\sum_{j\ge 0}(j+1)x^j=\sum_{j\ge 0}\binom{1+j}1x^j\;.$$
Do it again:
$$\frac2{(1-x)^3}=\sum_{j\ge 0}j\binom{1+j}1x^{j-1}\;,$$
so
$$\begin{align*}
\frac1{(1-x)^3}&=\frac12\sum_{j\ge 0}(j+1)\binom{2+j}1x^j\\
&=\frac12\sum_{j\ge 0}(j+1)\binom{2+j}{j+1}\\
&=\frac12\sum_{j\ge 0}(2+j)\binom{1+j}jx^j\\
&=\sum_{j\ge 0}\frac{2+j}2\binom{1+j}1x^j\\
&=\sum_{j\ge 0}\binom{2+j}2x^j\;.
\end{align*}$$
Now use that as a model to prove the general result by induction on $a$.
Another approach, more directly using generating functions, is to note that when you convolve a sequence $\sigma=\langle a_n:n\in\Bbb N\rangle$ with the constant sequence $\langle 1,1,1,\ldots\rangle$, you get the sequence of partial sums of $\sigma$: $\langle a_0,a_0+a_1,a_0+a_1+a_2,\ldots\rangle$. In particular, if 
$$\frac1{(1-x)^{a+1}}=\sum_{j\ge 0}\binom{a+j}ax^j\;,$$
then 
$$\frac1{(1-x)^{a+2}}$$
is the generating function of the convolution of $\langle 1,1,1,\ldots\rangle$ with
$$\left\langle\binom{a+j}a:j\in\Bbb N\right\rangle\;,$$
so the coefficient of $x^n$ must be
$$\sum_{j=0}^n\binom{a+j}a=\binom{a+1+n}{a+1}$$
by what is sometimes called the hockey stick identity. That is,
$$\frac1{(1-x)^{a+2}}=\sum_{j\ge 0}\binom{a+1+j}{a+1}x^j\;,$$
as desired.
A: 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:
$$\binom{-t}j=\frac{(-t)(-t-1)(-t-2)\cdots(-t-j+1)}{j!}=(-1)^j\cdot\frac{t(t+1)(t+2)\cdots(t+j-1)}{j!}=(-1)^j\binom{t+j-1}j.$$
