Evaluating $\sum_{y=a}^{\infty}{y \choose a} \cdot p^{y-a}$ for $p \in [0,1]$ I am trying to evaluate the following sum $$\sum_{y=a}^{\infty}{y \choose a} \cdot p^{y-a}$$ for $p \in [0,1]$. This looks somewhat like the binomial theorem, but I don't know how I would go about applying it, as the index of summation is $y$ and it's at the top of the binomial coefficient.
I evaluated the sum using Mathematica, and I got $-\frac{(1-p)^{-a}}{p-1}$ which does make it seem like it's been obtained using the binomial theorem, but I am unable to find a way to use it.
Any help would be appreciated.
 A: Let $$c_a=\sum_{y=a}^{+\infty}{y \choose a}p^{y-a}$$
Define the generating function $$\begin{split}
 f(z)&=\sum_{a=0}^{+\infty}c_a z^a\\
&=\sum_{a=0}^{+\infty}\sum_{y=a}^{+\infty}{y \choose a}p^{y-a}z^a\\
&=\sum_{y=0}^{+\infty}\sum_{a=0}^{y}{y \choose a}p^{y-a}z^a\\
&=\sum_{y=0}^{+\infty}p^y\sum_{a=0}^{y}{y \choose a}\left(\frac z p\right)^a\\
&=\sum_{y=0}^{+\infty}p^y\left(1+\frac z p\right)^y\\
&= \frac 1 {1-p-z}\\
&= \frac 1 {1-p} \cdot \frac 1 {1-\frac z { 1-p}}\\
&= \frac 1 {1-p} \sum_{a=0}^{+\infty}\frac 1 {(1-p)^a}z^a
\end{split}$$
This gives the desired formula for $c_a$.
$$c_a=\frac1 {(1-p)^{a+1}}$$
A: 
We obtain
  \begin{align*}
\color{blue}{\sum_{y=a}^\infty\binom{y}{a}p^{y-a}}&=\sum_{y=0}^\infty\binom{y+a}{y}p^y\tag{1}\\
&=\sum_{y=0}^\infty\binom{-a-1}{y}(-p)^y\tag{2}\\
&\,\,\color{blue}{=\frac{1}{(1-p)^{a+1}}}\tag{3}
\end{align*}

Comment:


*

*In (1) we shift the index to start with $y=0$ and use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (3) we apply the binomial series expansion.
A: So we have
$$S  = \binom{a}{a} + \binom{a+1}{a}p + \binom{a+2}{a}p^2 + \space ...$$
$$pS = \binom{a}{a}p + \binom{a+1}{a}p^2 + \binom{a+2}{a}p^3 + \space ...$$
subtracting, we obtain
$$(1-p)S = \binom{a}{a} + \binom{a}{a-1}p + \binom{a+1}{a-1}p^2 + \space ...$$
(I've used the identity $\binom{a}{a-1} + \binom{a}{a} = \binom{a+1}{a}$)
If we do the same with the above expression by taking $p(1-p)S$ and subtracting it, we obtain
$$(1-p)^2S = \binom{a}{a} + (\binom{a}{a-1} -\binom{a}{a})p + \binom{a}{a-2}p^2 + \space ...$$
notice that for $(1-p)^n$, the $(n+1)th$ binomial coefficient is reducing into $a$ at the top and the $nth$ term is reducing by the previous one's coefficient. If we extrapolate and do this a times, we get:
$$(1-p)^aS = \binom{a}{a} + (\binom{a}{a-1} - (a-1)\binom{a}{a})p + ( \binom{a}{a-2} - (a-2)\binom{a}{a-1} + \frac{(a-2)(a-1)}{2}\binom{a}{a})p^2 + \space ...$$
if we end up expanding the coefficients, they reduce to:
$$(1-p)^aS = 1 + p + p^2 + p^3 + \space ...$$
and vóila! The right hand side is now an infinite GP, which converges to $\frac{1}{1-p}$. Rearranging the terms, we obtain:
$$ S = \frac{1}{(1-p)\cdot(1-p)^a} = \frac{1}{(1-p)^{a+1}}$$
which is the final answer.
