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:


*

*Does this summation converge to a simple function of $\displaystyle p$ and $\displaystyle r$ ?.

*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 $\displaystyle r$ ) ?.

 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\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}
A: 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.
