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While studying a problem about orthogonal polynomials I encountered the following expressions \begin{equation} f(n)=\sum_{k=0}^{n}(-1)^k\binom{n+k}{2k} \frac{1}{k+1}\binom{2k}{k} \end{equation} and \begin{equation} g(n)=\sum_{k=0}^{n-1}(-1)^k\binom{n+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1} \end{equation}

I can prove that $f(n)=0$ for all integers $n\geq 1$ and $g(n)=0$ for all integers $n>1$ using properties of orthogonal polynomials. However, I would like to find an elementary proof, which might also be more illuminating. While doing some experiments with calculations using Mathematica, I defined the functions $f(n)$ and $g(n)$ with the above formulas but forgot to specify that $n$ is an integer value. When I typed $f(n)$ and $g(n)$ for a generic variable $n$, I guess that Mathematica "assumed" that the variables involved were real and provided the following simple formulas:

\begin{equation} f(x)=\frac{\sin \pi x}{x(x+1)\pi} \end{equation} and \begin{equation} g(x)=-\frac{2\sin \pi x}{(x+1)(x-1)\pi} \end{equation} which makes it evident that these functions are zero for all positive (and negative) integers with the possible exceptions of $0,1,-1$.

Why are these equalities true?

I realize it might have to do, perhaps, with properties of the Euler Beta and Gamma functions, but I know too little about these functions to figure out a proof along these lines. Can anybody help? Thank you!

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When non-integer $n$ (or $x$) occurs as the upper bound on a summation, I'd interpret the sum as ranging just over integers below that bound. And I'd interpret binomial coefficients $\binom{y}{m}$, with integer $m$ but non-integer $y$, as polynomials in $y$. So I'd get piecewise polynomial results for the sum. Apparently, Mathematica interprets things differently than I do --- but how? Maybe the sums become infinite series, because it keeps adding terms until the summation variable equals the upper bound? –  Andreas Blass Oct 11 '12 at 12:40
For $k>n$ the first binomial in $f(n)$ vanishes, for $k>n-1$ the first binomial in $g(n)$ vanishes. So you can ignore the upper limit, i.e. replace it with infinity. –  Martin Rubey Oct 11 '12 at 13:03
It seems to me that the vanishing of $\binom{n+k}{2k}$ for $n>k$ depends on $n$'s being an integer. –  Andreas Blass Oct 11 '12 at 13:16
Yes it does. All I'm saying is that $f$ and $g$ can be first understood as infinite series without changing the value. In a second step we replace $n$ in the summands by a continues variable $x$, which again doesn't change the value for integer $x$. This explains how mathematica arrives at the nice closed forms. –  Martin Rubey Oct 11 '12 at 13:30
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2 Answers

up vote 7 down vote accepted

Let's consider more generally the sums (from $0$ to $n$, resp. from $0$ to $n-1$) with a real or complex number $x$ in place of the $n$ inside the sums:

\begin{equation} \sum_{k=0}^{n}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k} \end{equation} and \begin{equation} \sum_{k=0}^{n-1}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}\, . \end{equation}

These sums can be evaluated by induction as:

\begin{equation} (-1)^n\frac{n+1}{x(x+1)} \binom{x+n+1}{2n+2}\binom{2n+2}{n+1} \end{equation} and \begin{equation} (-1)^n\frac{n+1}{x^2-1} \binom{x+n+1}{2n+3}\binom{2n+4}{n+2}\, . \end{equation} We can write these expression respectively as

\begin{equation} \frac{1}{x+1}\left(1+\frac{x}{n+1} \right)\prod_{k=1}^n\left(1-\frac{x^2}{k^2}\right) \end{equation} and \begin{equation} -\frac{2x}{x^2-1}\frac{n+1}{n+2}\prod_{k=1}^{n+1}\left(1-\frac{x^2}{k^2}\right)\, . \end{equation}

The limit of these expression can be evaluated by means of the Euler's infinite product for $\sin(\pi x)$, obtaining respectively the values of the series :

\begin{equation} \sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}=\frac{\sin(\pi x)}{\pi x(x+1)} \end{equation} and \begin{equation} \sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}=-\frac{2\sin(\pi x)}{\pi(x^2-1)}\, . \end{equation} As observed in the comments, for a positive integer $x=n$, these series coincide respectively with the initially considered finite sums $f$ and $g$, of which they may be considered therefore as a natural extension to complex values of $x$.

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mmmh, wait, actually evaluating the sums I assumed x to be an integer! so I'm not sure of the result. –  Pietro Majer Oct 11 '12 at 15:17
The sums are polynomials in $x$ so if they agree on the integers then they are equal everywhere. I personally don't see how Maple is computing the sums, but that's another story. –  Matt Young Oct 11 '12 at 17:34
Yes, sorry, I was in a big hurry and I was taken by some sudden silly doubt. Checking now the details, there were no problem at all, so I roll back to the first version. The identities for the sums are easily proven by induction, and contain a partial product of the infinite product for $\sin(\pi x)/\pi x$. –  Pietro Majer Oct 12 '12 at 13:56
Thanks a lot. I really enjoyed your solution. Grazie mille. –  Stefano Capparelli Oct 16 '12 at 10:38
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The two identities are both special cases of Vandermonde's theorem (also called the Chu-Vandermonde theorem), which is the most well-known binomial coefficient identity after the binomial theorem. The first sum may be written $$f(n) = \frac 1n \sum_{k=0}^n (-1)^k\binom nk \binom {n+k}{k+1}.$$ Applying the identity $\binom{-c}m = (-1)^{m}\binom{c+m-1}{m}$, we get $$f(n)= -\frac 1n \sum_{k=0}^n \binom nk \binom {-n}{k+1},$$ and reversing the order of summation gives \begin{equation*} f(n) =-\frac 1n \sum_{k=0}^n \binom nk \binom {-n}{n-k+1}. \end{equation*}

Vandermonde's theorem is often written \begin{equation*} \sum_{k=0}^m \binom ak \binom b{m-k} = \binom{a+b}m. \end{equation*} There are many simple proofs of Vandermonde's theorem. For example, we can prove this formula by equating coefficients of $x^m$ in $(1+x)^a (1+x)^b = (1+x)^{a+b}$.

Setting $a=n$, $b=-n$, and $m=n+1$ in this formula (and observing that the $k=n+1$ term vanishes) and using the last formula for $f(n)$ shows that for $n>0$, $f(n) =- \frac 1n\binom 0{n+1}=0.$ Similarly, $g(n)$ can be evaluated by Vandermonde's theorem.

I believe that Pietro's indefinite summation approach is correct, but I don't think that this is what Mathematica is doing. There is a nonterminating generalization of Vandermonde's theorem, called Gauss's theorem (see, e.g., http://mathworld.wolfram.com/GausssHypergeometricTheorem.html) that evaluates the sum $\sum_{k=0}^\infty (-1)^k \binom ak \binom {b+k}{m+k}$ in terms of gamma functions when it converges, where $a$ and $b$ are arbitrary and $m$ is an integer. (Actually Gauss's theorem is a little more general than this.) In the particular case $a=n$, $b=n$, $m=1$, the reflection formula for the gamma function can be applied to give Stefano's formula for $f(x)$ in terms of $\sin \pi x$, and similarly for $g(x)$. Mathematica is probably applying Gauss's theorem, since it knows how to convert binomial coefficient sums to hypergeometric series and it knows how to evaluate them in cases like this one.

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Thank you! This is indeed a nice short proof. I am new here and if I have to choose a single answer as a preferred answer I am going to go for Pietro's as it also gives me a proof of the natural extension of the summation formulas. Of course, you might also be right that the actual Mathematica procedure is a different story. Thanks to all those that took interest in this. –  Stefano Capparelli Oct 16 '12 at 10:37
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