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!

integersbelow 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 variableequalsthe upper bound? $\endgroup$ – Andreas Blass Oct 11 '12 at 12:40