Convergence of binomial distribution with successes given 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!
 A: 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
$$\binom kk p^k=1\times1^k=1=\frac1p\ .$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
