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  "body": "\n## Definition\n\nLet $S$ be a set, and let $f_n \\colon S \\to \\C$ be functions for $n \\ge 0$.\nLet $s_N = \\sum_{n=0}^{N} f_n$ be the partial sums of the [[series of functions|0MRE6XRE5V]] $\\sum_{n=0}^{\\infty} f_n$, and let $f \\colon S \\to \\C$ be a function.\nThe series *converges uniformly* to $f$ on $S$ if the sequence $(s_N)$ converges [[uniformly|08FPABDQ4C]] to $f$ on $S$.\nThat is, for every $\\varepsilon \u003e 0$ there is an integer $N_0$ such that $\\abs{s_N(x) - f(x)} \u003c \\varepsilon$ for all $N \\ge N_0$ and all $x \\in S$.\n\n## Remarks\n\nEquivalently, the series converges at each point of $S$ and $\\sup_{x \\in S} \\abs{\\sum_{n \u003e N} f_n(x)} \\to 0$ as $N \\to \\infty$.\nA Cauchy form also holds: the series converges uniformly on $S$ if and only if for every $\\varepsilon \u003e 0$ there is $N_0$ with $\\abs{\\sum_{n=M+1}^{N} f_n(x)} \u003c \\varepsilon$ for all $N \u003e M \\ge N_0$ and all $x \\in S$.\nFor a subset of a larger domain, this is [[uniform convergence on a set|08HJRTG5N7]] of the partial sums.\n\nExample: let $0 \u003c r \u003c 1$ and $f_n(x) = x^n$ on $S = [0, r]$, with $f(x) = 1/(1 - x)$.\nThe tail is $\\sum_{n \u003e N} x^n = x^{N+1}/(1 - x) \\le r^{N+1}/(1 - r)$ for all $x \\in S$, and this bound tends to $0$.\nSo the series converges uniformly on $[0, r]$.\n\nNon-example: the same series on $S = [0, 1)$ converges at each point to $1/(1 - x)$, but not uniformly.\nFor each $N$, the tail $x^{N+1}/(1 - x)$ tends to $\\infty$ as $x \\to 1^-$, so its supremum over $S$ is infinite.\n\nThe [[Weierstrass M-test|0NX5NWAFC9]] is the usual way to prove uniform convergence.\n\n## Explanations\n\n### L1 human\n\nAt each point $x$, the series is an ordinary series of numbers, and its partial sums approach $f(x)$.\nThe series converges uniformly when one number of terms is enough to make the error small at every point at once.\nFor $\\sum x^n$, this works on $[0, r]$ with $r \u003c 1$, but not on $[0, 1)$, where points near $1$ need more and more terms.\n",
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