<s>[INST] 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.
 [/INST] 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.
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