Let \(A\) be a set and let \(f_n\colon A\to\CC\) for \(n\geq 0\). Put \(\norm{f_n}_{\infty, A}\coloneqq\sup_{x\in A}\abs{f_n(x)}\). If \(\sum_{n\geq 0}\norm{f_n}_{\infty, A} < \infty\), then for every \(x\in A\) the series \(\sum_{n\geq 0} f_n(x)\) converges absolutely, and its partial sums converge uniformly on \(A\) to \(\sum_{n\geq 0} f_n\).
Equivalently, the conclusion holds whenever there are real numbers \(M_n\) with \(\abs{f_n(x)} \leq M_n\) for all \(x\in A\) and \(n\geq 0\) and \(\sum_{n\geq 0} M_n < \infty\).
For each \(x\), \(\sum_n\abs{f_n(x)}\leq\sum_n M_n<\infty\), so \(S(x)\coloneqq\sum_{n\geq 0}f_n(x)\) exists. For \(x\in A\) and \(N\geq 0\), \(\abs{S(x) - \sum_{n=0}^{N}f_n(x)}\leq\sum_{n>N}M_n\), which tends to \(0\) as \(N\to\infty\) independently of \(x\). The first form is the case \(M_n = \norm{f_n}_{\infty,A}\), and conversely \(\norm{f_n}_{\infty,A}\leq M_n\).
Conversely, if \(\sum_n f_n\) converges uniformly on \(A\), then \(\sup_{x\in A}\abs{f_n(x)}\to0\); this necessary condition does not imply \(\sum_n\sup_{x\in A}\abs{f_n(x)}<\infty\), as the constant functions \(f_n\coloneqq(-1)^n/n\) show.