Convergence of binomial distribution with successes given

Question

Looking for the sum of the infinite series:

$\sum_{n=k}^{\infty}\binom{n}{k}p^{k}(1-p)^{n-k}$

You can pull a few terms out to get:

$\frac{p^k}{k!}\sum_{n=k}^{\infty}\frac{n!}{(n-k)!}(1-p)^{n-k}$

Strangely enough, I've actually calculated it for a bunch of different values of $p$ in R and the sum always converges to $\frac{1}{p}$ but I can't seem to derive it on paper. Thanks!

Answer 1

Use the binomial theorem with negative exponent: $$(1-x)^{-(k+1)}=\sum_{m=0}^\infty \binom{-(k+1)}m(-x)^m\ ,\quad |x|<1\ .$$ Note that $$\binom{-(k+1)}m=(-1)^m\binom{k+m}m=(-1)^m\binom{k+m}k$$ so this can be written $$(1-x)^{-(k+1)}=\sum_{m=0}^\infty \binom{k+m}kx^m\ .$$ Substituting $n=k+m$ gives $$(1-x)^{-(k+1)}=\sum_{n=k}^\infty \binom{n}kx^{n-k}\ .$$ As long as $q\ne1$ we have $$\sum_{n=k}^\infty \binom nk p^kq^{n-k}=p^k(1-q)^{-(k+1)}=\frac1p\ .$$

Comment: $q=0$ is not actually a probem. The series is $$\binom kk p^k+\binom{k+1}k p^kq+\binom{k+2}kp^kq^2+\cdots\ .$$ So if $q=0$ the first term is not zero and the whole sum is $$\binom kk p^k=1\times1^k=1=\frac1p\ .$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n\ =\ k}^{\infty}{n \choose k}p^{k}\,\pars{1 - p}^{n - k} & \,\,\,\stackrel{n\ \mapsto\ n + k}{=}\,\,\, p^{k}\,\sum_{n\ =\ 0}^{\infty}{n + k \choose k}\pars{1 - p}^{n} \\[5mm] & = p^{k}\,\sum_{n\ =\ 0}^{\infty}{n + k \choose n}\pars{1 - p}^{n} \qquad\pars{~Binomial\ Symmetry~} \\[5mm] & = p^{k}\,\sum_{n\ =\ 0}^{\infty}{-k - 1\choose n}\pars{-1}^{n}\pars{1 - p}^{n} \qquad\pars{~Binomial\ Negation~} \\[5mm] & = p^{k}\,\sum_{n\ =\ 0}^{\infty}{-k - 1\choose n}\pars{p - 1}^{n} \\[5mm] & = p^{k}\,\bracks{1 + \pars{p - 1}}^{-k - 1} \qquad\pars{~Newton\ Binomial\ Formula~} \\[5mm] & = p^{k}\, p^{-k - 1} = \bbx{\ds{1 \over p}} \end{align}