# Infinite binomial sum

Let $$\displaystyle\pi_{lr}\left(p\right) := {l \choose r}p^{r}\left(1 - p\right)^{l - r}\quad$$ ( i.e., the binomial probability with parameters $$\displaystyle l$$ and $$\displaystyle r$$ ).

I'm interested in computing the following sum: $$\displaystyle\sum_{l = r}^{\infty}\pi_{lr}\left(p\right)$$

I've two questions:

1. Does this summation converge to a simple function of $$\displaystyle p$$ and $$\displaystyle r$$ ?.
2. If I were to settle on approximating it with $$\displaystyle\sum_{l = r}^{N}\pi_{lr}(p)$$ for some $$\displaystyle N$$. How large should I choose $$\displaystyle N$$ ( as a function of $$\displaystyle p$$ and $$\displaystyle r$$ ) ?.
• Could you clarify the second part of the second question ? Feb 10, 2019 at 3:39
• sure, if the summation didn't admit a clean expression (such as $1/p$), I was planning to study it computationally. In which case, the obvious thing to do is to try to approximate the infinite sum with a finite one. However, one needs to choose a large enough $N$ so that this approximation is a good one. My second question was how to choose $N$. But it is now irrelevant, given the answer below.
– Ozzy
Feb 10, 2019 at 3:54
• Thank you for the answer ! It was just not clear to me; now, I understand. Cheers. Feb 10, 2019 at 3:57

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$

$$\ds{\pi_{\ell r}\pars{p} \equiv {\ell \choose r}p^{r}\pars{1 - p}^{\ell - r}.\qquad \sum_{\ell = r}^{\infty}\pi_{\ell r}\pars{p}:\ {\LARGE ?}}$$.

\begin{align} &\bbox[10px,#ffd]{\sum_{\ell = r}^{\infty}\pi_{\ell r}\pars{p}} = \sum_{\ell = r}^{\infty}{\ell \choose r}p^{r}\pars{1 - p}^{\ell - r} = p^{r}\sum_{\ell = r}^{\infty}\overbrace{\ell \choose \ell - r} ^{\ds{=\ {\ell \choose r}}}\ \pars{1 - p}^{\ell - r} \\[5mm] = &\ p^{r}\sum_{\ell = r}^{\infty} \overbrace{{-r - 1 \choose \ell - r}\pars{-1}^{\ell - r}} ^{\ds{=\ {\ell \choose \ell - r}}}\ \pars{1 - p}^{\ell - r} \\[5mm] \stackrel{\ell\ -\ r\ \mapsto\ \ell}{=}\,\,\,& p^{r}\sum_{\ell = 0}^{\infty} {-r - 1 \choose \ell}\pars{-1}^{\ell}\pars{1 - p}^{\ell} \\[5mm] = &\ p^{r}\sum_{\ell = 0}^{\infty}{-r - 1 \choose \ell}\pars{p - 1}^{\ell} = p^{r}\,\bracks{1 + \pars{p - 1}}^{-r - 1} = \bbx{1 \over p} \end{align}

• very nice, thank you!
– Ozzy
Feb 10, 2019 at 0:32
• @Ozzy You're welcome. Feb 10, 2019 at 0:35

As far as approximations goes, we have $$\sum_{l=r}^\infty{l\choose r}p^r(1-p)^{l-r}=\left({p\over 1-p}\right)^r\sum_{l=r}^\infty{l\choose r}(1-p)^l$$ Write $$a_l = {l\choose r}(1-p)^l.$$ Then $${a_{l+1}\over a_l} = {l+1\over l+1-r}(1-p)\to 1-p$$ so that we essentially have a geometric series. Observing that $${l+1\over l+1-r}$$ decreases to $$1$$ as $$l\to\infty$$ makes it easy to bound the omitted terms.

As to the first part of the question, I don't know, but I doubt it, except perhaps when $$r$$ is very small. Since the ratio of successive terms is a rational function of $$l$$ for fixed $$p$$ and $$r$$, this is a hypergeometric series. There is a large theory of these, which I unfortunately know nothing about. I've added the hypergeometric-function tag in hopes of attracting someone knowledgeable.