# An addendum to a Z talk

** Published:**

Doron Zeilberger recently gave a talk at RUMA, the Rutgers Undergraduate Math Association, about mathematical gambling. In the talk he mentioned a lovely half-page article he wrote confirming a centuries-old conjecture by Newton and Pepys. The proof is very short but relies on some hidden machinery, which I will explain here.

The basic idea is that we want to prove the identity

\[\begin{equation} \label{goal} \sum_{k = 0}^{n - 1} {6n \choose k} \left( \frac{1}{6} \right)^k \left( \frac{5}{6} \right)^{6n - k} = \sum_{k = 0}^{n - 1} R(k) {6k \choose k} \left( \frac{1}{6} \right)^k \left( \frac{5}{6} \right)^{5k} \end{equation}\]where $R(k)$ is some explicit rational function. The reason is that the sum on the right has no $n$ in it, and after seeing $R(k)$ written out, is obviously monotonic because the terms are positive. The sum on the left contains a bunch of $n$’s, so it isn’t obvious if it gets bigger or smaller as $n$ grows.

First, a quick lesson on WZ theory. A sequence $a(n, k)$ is called
*hypergeometric* provided that $a(n + 1, k) / a(n, k)$ and $a(n, k + 1) / a(n,
k)$ are both explicit rational functions in $n$ and $k$. Certain nice
hypergeometric sequences have a “partner” $b(n, k)$ which satisfies the
identity

In this case $(a, b)$ is called a *WZ pair*. It is simple to algorithmically
determine whether any explicit sequence is hypergeometric or part of a WZ pair.
Our summand happens to be both.

# The theory

Say that you have a WZ pair $(a, b)$. That is,

\[\begin{equation} \label{wz} a(n + 1, k) - a(n, k) = b(n, k + 1) - b(n, k). \end{equation}\]Let

\[S(n) = \sum_{k = 0}^{n - 1} a(n, k)\]be the sum that we are interested in. Summing over \eqref{wz} from $k = 0$ to $k = n$ gives

\[S(n + 1) - S(n) - a(n, n) = b(n, n + 1) - b(n, 0).\]If we move $a(n, n)$ to the right-hand side, then the left-hand side is telescoping, so summing again recovers $S(n)$ itself. The end-result will be

\[S(n) = \sum_k (a(k, k) + b(k, k + 1) - b(k, 0)),\]where the bounds are too tedious to determine. In our case it turns out that $b(k, 0) = 0$, and by some standard theory about hypergeometric sequences $b(k, k + 1)$ is a rational multiples of $a(k, k)$, so we can write

\[S(n) = \sum_k R(k) a(k, k),\]the kind of sum we desired.

# The proof

The proof now is merely a matter of executing the steps outlined in the
previous section. All of these have been implemented in *Maple* or
*Mathematica* for decades. If we download
EKHAD and read it
into *Maple*, then executing `zeillim(a(n, k), k, n, N, 0, 1)`

produces output
that implies \eqref{goal}. The rational function is