Wilf–Zeilberger pair

Remarks

Consequence: suppose that for each $n$ the sum $\sum_k F(n, k)$ converges and $G(n, k) \to 0$ as $k \to \pm\infty$. Then summing clause 1 over $k$ gives a telescoping sum on the right. So $\sum_k F(n + 1, k) = \sum_k F(n, k)$, and the sum is constant in $n$.

Example: let $F(n, k) = \binom{n}{k}/2^n$ and $G(n, k) = -\binom{n}{k - 1}/2^{n + 1}$, with binomial coefficients that vanish outside $0 \le k \le n$. Pascal's rule gives $F(n + 1, k) - F(n, k) = \big(\binom{n}{k - 1} - \binom{n}{k}\big)/2^{n + 1}$, which is $G(n, k + 1) - G(n, k)$. The certificate is $R(n, k) = -k/(2(n - k + 1))$. So $\sum_k \binom{n}{k}/2^n$ is constant, and its value $1$ at $n = 0$ proves $\sum_k \binom{n}{k} = 2^n$.

Non-example: the same $F$ with $G = 0$ is not a WZ pair, since clause 1 would need $F(n + 1, k) = F(n, k)$, which fails for $n = 0$, $k = 0$: $F(1, 0) = 1/2$ but $F(0, 0) = 1$.