On p. 49 in Gould's book Combinatorial Identities, the author states that the sum $$\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{2n}{2k}^{-1}$$ "... arises naturally in a statistical problem; it amounts to the evaluation of the moments of a certain distribution". Does someone know what exactly Gould is referring to? I am looking for references...
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$\begingroup$ this sum is just $n/(n+1)$ for even $n$, and 1 for odd $n$ $\endgroup$– Carlo BeenakkerMar 22 at 15:14
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$\renewcommand{\b}{\binom}\renewcommand{\B}{\text{B}}$As noted in Carlo Beenakker's comment,
\begin{equation*}
s_n:=\sum_{k=0}^{n-1}(-1)^k\b nk\b{2n}{2k}^{-1}=\frac{2 n+1-(-1)^n}{2 (n+1)}. \tag{1}\label{1}
\end{equation*}
The immediately related formula
\begin{equation*}
\sum_{k=0}^n(-1)^k\b nk\b{2n}{2k}^{-1}=\frac{[1+(-1)^n](2n+1)}{2 (n+1)}
\end{equation*}
(with the upper summation limit $n$)
is formula 4.2.8.5 in Vol.1 of the handbook by Prudnikov, Brychkov, and Marichev (in Russian).
It is immediate from \eqref{1} that $(s_n)$ is not the sequence of moments of any probability distribution. Indeed, otherwise the sequence $(s_{2n})$ would be convex, which it is not. So, the statement "it amounts to the evaluation of the moments of a certain distribution" cannot be true, unless "amounts to" is understood as something like "has something to do with".
However, it may be of help to present a (quite standard) derivation of \eqref{1}. It is based on the following representation of the reciprocals of the binomial coefficients in terms of the beta function (which of course has to do with the beta distribution): \begin{equation*} \b sr^{-1}=r\B(r,s-r+1) \tag{2}\label{2} \end{equation*} for integers $r$ and $s$ such that $0<r\le s$, where $\B(a,b):=\int_0^1 dx\, x^{a-1}(1-x)^{b-1}$.
By the definition of $s_n$ in \eqref{1} and identity \eqref{2}, we get \begin{equation*} \begin{aligned} s_n&=1+\sum_{k=1}^{n-1}(-1)^k\b nk\b{2n}{2k}^{-1} \\ &=1+\int_0^1 dx\, \sum_{k=1}^n(-1)^k\b nk 2k x^{2k-1}(1-x)^{2n-2k} \\ &=1-2n\int_0^1 dx\,[(-1)^n x^{2n-1}+x(1-2x)^{n-1}] \\ &=\frac{2 n+1-(-1)^n}{2 (n+1)}.\quad\Box \end{aligned} \end{equation*} One may note that the latter integral looks something like an expression for the $n$th moment of a probability distribution.
answered Mar 22 at 17:02