# 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.

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 $$c_a$$. $$c_a=\frac1 {(1-p)^{a+1}}$$

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.

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:

$$S = \frac{1}{(1-p)\cdot(1-p)^a} = \frac{1}{(1-p)^{a+1}}$$