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  "body": "\n## Definition\n\nLet $F(n, k)$ and $G(n, k)$ be functions defined for integers $n \\ge 0$ and $k$.\nThe pair $(F, G)$ is a *WZ pair* if, for all $n$ and $k$,\n$$F(n + 1, k) - F(n, k) = G(n, k + 1) - G(n, k).$$\nThe *WZ method* proves an identity $\\sum_k f(n, k) = r(n)$, with $r(n) \\ne 0$, in three steps:\n\n1. set $F(n, k) = f(n, k) / r(n)$, so the identity reads $\\sum_k F(n, k) = 1$;\n2. give a rational function $R(n, k)$, the *WZ certificate*, such that $(F, G)$ with $G = R F$ is a WZ pair;\n3. check that $\\sum_k F(0, k) = 1$ and that $G(n, k) \\to 0$ as $k \\to \\pm\\infty$ for each $n$.\n\n## Remarks\n\nWhy it works: summing the WZ equation over all $k$ makes the right side telescope to $0$.\nSo $\\sum_k F(n + 1, k) = \\sum_k F(n, k)$ for all $n$, and the sum stays equal to its value $1$ at $n = 0$.\nWhen $F$ is a hypergeometric term, the WZ equation divided by $F(n, k)$ becomes an identity of rational functions, which is checked by algebra.\n\nExample: the identity $\\sum_k \\binom{n}{k} = 2^n$ gives $F(n, k) = \\binom{n}{k} / 2^n$.\nThe certificate is $R(n, k) = -\\frac{k}{2(n - k + 1)}$, so $G(n, k) = -\\binom{n}{k-1} / 2^{n+1}$.\nBoth sides of the WZ equation equal $\\big(\\binom{n}{k-1} - \\binom{n}{k}\\big) / 2^{n+1}$, by Pascal's rule.\nAlso $G(n, k) = 0$ outside $1 \\le k \\le n + 1$, and $\\sum_k F(0, k) = 1$.\n\nNon-example: for $F(n, k) = \\binom{n}{k} / 2^n$, the choice $G = 0$ does not give a WZ pair.\nThe equation would require $F(n + 1, k) = F(n, k)$, which fails for $n = 0$ and $k = 1$, since $1/2 \\ne 0$.\n\nThe Wilf–Zeilberger pair node treats the pairs themselves in more detail.\n",
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