A summable uniform majorant gives absolute and uniform convergence.Suggest a slogan
Let \(\Omega\) be a set, let \(f_n\colon\Omega\to\CC\) for \(n\in\NN\), and suppose there are real numbers \(M_n\) with \[ \sup_{x\in\Omega}\abs{f_n(x)}\le M_n\quad\text{for all }n,\qquad\sum_{n\in\NN}M_n<\infty. \] Then \(f(x)\coloneqq\sum_{n\in\NN}f_n(x)\) converges absolutely and uniformly on \(\Omega\). If moreover \(\Omega\) is a topological space and every \(f_n\) is continuous, then \(f\) is continuous, by the uniform limit theorem.
For each \(x\), \(\sum_n\abs{f_n(x)}\le\sum_nM_n<\infty\), so the series converges absolutely. For \(N\ge1\) and all \(x\in\Omega\), \[ \abs{f(x)-\sum_{n\le N}f_n(x)}\le\sum_{n>N}M_n, \] and the right side is independent of \(x\) and tends to \(0\) as \(N\to\infty\).