infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem in the Bayesian treatment of the German tank problem in Wikipedia here, they use:
$\displaystyle \sum_{n=m}^\infty \dfrac{1}{\binom{n}{k}}=\dfrac{k}{k-1}\dfrac{1}{\binom{m-1}{k-1}}$
how can I prove this in a clever combinatorics fashion?
I found this paper, see eqn. (9), which uses Gauss' hypergeometric function-- a bit beyond me.
there must be some way via a recursion relation, like I found in this old paper. theorem 1 in that reference has a similar infinite sum of an inverse binomial coefficient.
 A: Let $$S=\sum_{n=m}^{\infty} {n \choose k}^{-1}~~~~(1)$$
Use $${n \choose k}^{-1}=(n+1)\int_{0}^{1} x^k (1-x)^{n-k} dx~~~~(2)$$
Then $$S=\int_{0}^{1} \sum_{n=m}^{\infty} x^k(1-x)^{-k} \sum_{n=m}^{\infty} [(n+1) (1-x)^n]~~~~(3)$$
Use sum of infinite GP: $$\sum_{j=m}^{\infty} (j+1)z^j= x^{-2} (1-x)^m(1+mx)~~~~(4)$$
Then $$S=\int_{0}^{1} x^{k-2} (1-x)^{m-k}(1+mx) dx~~~~(5)$$
Use $\beta$ integral:
$$\int_{0}^{1} z^u (1-z)^v dz=\frac{\Gamma(1+u) \Gamma(1+v)}{\Gamma(2+u+v)}~~~~(6).$$
Then $$S=\frac{\Gamma(k-1) \Gamma(1+m-k)}{\Gamma(m)}+\frac{m\Gamma(k) \Gamma(1+m-k)}{\Gamma(1+m)}~~~~(7)$$
$$\implies S=\frac{k (k-2)! (m-k)!}{(m-1)!}= \frac{k}{k-1} {m-1 \choose k-1}^{-1}~~~~(8)$$
A: we can write this as a telescoping sum using the identity:
$\dfrac{1}{\binom{n}{k}}-\dfrac{1}{\binom{n+1}{k}}=\dfrac{k}{k+1}\dfrac{1}{\binom{n+1}{k+1}}$
write the left side as factorials and get a common denominator to see :)
then we can write the sum as:
\begin{equation}
\lim_{\Omega\rightarrow\infty}\dfrac{k}{k-1}\displaystyle\sum_{n=m}^\Omega [a_{n-1}-a_n]=\dfrac{k}{k-1} [a_{m-1}-a_\Omega]
\end{equation}
with:
\begin{equation}
   a_n:=\frac{1}{\binom{n}{k-1}}
\end{equation}
the right-most equality follows from the telescoping property of the sum.
as $\Omega \rightarrow \infty$, $a_\Omega \rightarrow 0$. thus:
$\displaystyle\sum_{n=m}^\infty \dfrac{1}{\binom{n}{k}}=\dfrac{k}{k-1}\dfrac{1}{\binom{m-1}{k-1}}$
