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\title[A curious identity proved by Cauchy's integral formula]
{A curious identity proved by Cauchy's integral formula}


\author[H.~Prodinger]{Helmut Prodinger$^\dag{}$}
\thanks{$^\dag{}$Supported by NRF Grant 2053748}
\address{H.~Prodinger\\
The John Knopfmacher Centre for Applicable Analysis and Number Theory\\
 School of Mathematics\\
University of the Witwatersrand, P.~O. Wits\\ 
2050 Johannesburg, South Africa}
\email{helmut@maths.wits.ac.za}


\date{May 8, 2003}



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\maketitle

Simons \cite{Simons01} has proved the identity
\begin{equation}\label{simons_identity}
\sum_{r=0}^{q}\frac{(-1)^{q+r}(q+r)!(1+x)^r}{(q-r)!r!^2}=
\sum_{r=0}^q\frac{(q+r)!x^r}{(q-r)!r!^2};
\end{equation}
Chapman \cite{Chapman03} gave a nice and short proof of it.
In this note, I want to give another attractive proof. It uses 
Cauchy's integral formula to pull out coefficients of  generating functions.

We divide \eqref{simons_identity}  by $q!$ and prove the equivalent version
\begin{equation*}
S=\sum_{r=0}^{q}\binom{q}{r}\binom{q+r}{r}(-1)^{q+r}(1+x)^r=
\sum_{r=0}^{q}\binom{q}{r}\binom{q+r}{r}x^r.
\end{equation*}
We start with the righthand-side:
\begin{align*}
S&=[t^q]\sum_{i\ge0}\binom{q}{i}t^i\cdot\sum_{i\ge0}\binom{q+i}{i}(tx)^i\\
&=[t^q](1+t)^q\cdot(1-tx)^{-q-1}\\
&=\frac{1}{2\pi i}\oint \frac{dt}{t^{q+1}}(1+t)^q\cdot(1-tx)^{-q-1}.
\end{align*}
Now we substitute $t=u/(1-u)$, so that $dt=du/(1-u)^2$ and obtain
\begin{align*}
S
&=\frac{1}{2\pi i}\oint \frac{du}{(1-u)^2}
\frac{(1-u)^{q+1}}{u^{q+1}}(1-u)^{-q}\cdot\bigg(\frac{1-u(1+x)}{1-u}\bigg)^{-q-1}\\
&=[u^q]\big(1-u)^q(1-u(1+x)\big)^{-q-1}\\
&=\sum_{r=0}^q\binom{-q-1}{r}(-1)^r(1+x)^r\binom{q}{q-r}(-1)^{q-r}\\
&=\sum_{r=0}^q\binom{q+r}{r}\binom{q}{r}(1+x)^r(-1)^{q-r},
\end{align*}
which is the lefthand-side.
















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