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  "body": "\n## Proof\n\nWe bound the distance from each partial sum to the sum by a tail of $\\sum M_n$, which does not depend on $x$.\n\n1. Let $S$, $f_n$ and $M_n$ be as in the statement. Let $s_N = \\sum_{n=0}^{N} f_n$ be the partial sums of the [[series of functions|0MRE6XRE5V]].\n2. Fix $x \\in S$. Since $0 \\le |f_n(x)| \\le M_n$ and $\\sum M_n$ converges, the [[comparison test|04Z5G0THN6]] shows that $\\sum |f_n(x)|$ converges.\n3. So $\\sum f_n(x)$ [[converges absolutely|0W3K3QPEWT]]. By [[absolutely convergent series converge|0Y123X48GA]], it converges. Let $s(x) = \\sum_{n=0}^{\\infty} f_n(x)$.\n4. Let $M = \\sum_{n=0}^{\\infty} M_n$ and $T_m = \\sum_{n=0}^{m} M_n$. Since $M_n \\ge 0$, the sequence $(T_m)$ is increasing, and $T_m \\to M$ by the definition of the [[sum of a series|0K0H5E3Z5N]].\n5. By the [[monotone convergence theorem|0FM6N4R0VB]], $M$ is the supremum of the $T_m$. So $T_N \\le M$ for all $N$.\n6. Let $\\tau_m = M - T_m$. By step 4, $\\tau_m \\to 0$. By step 5, $\\sum_{k=m+1}^{N} M_k = T_N - T_m \\le \\tau_m$ for all $N \u003e m$.\n7. Let $x \\in S$ and $N \u003e m$. By the [[triangle inequality|03HF5PMY5T]] and step 6,\n$$|s_N(x) - s_m(x)| = \\Big| \\sum_{k=m+1}^{N} f_k(x) \\Big| \\le \\sum_{k=m+1}^{N} |f_k(x)| \\le \\sum_{k=m+1}^{N} M_k \\le \\tau_m.$$\n8. So $-\\tau_m \\le s_N(x) - s_m(x) \\le \\tau_m$ for all $N \u003e m$. Letting $N \\to \\infty$ and using step 3, we get $|s(x) - s_m(x)| \\le \\tau_m$, since [[limits preserve non-strict inequalities|0N32V0E3XJ]].\n9. Let $\\varepsilon \u003e 0$. By step 6, there is $m_0$ with $\\tau_m \u003c \\varepsilon$ for all $m \\ge m_0$. By step 8, $|s_m(x) - s(x)| \u003c \\varepsilon$ for all $m \\ge m_0$ and all $x \\in S$.\n10. So $s_m \\to s$ [[uniformly|08FPABDQ4C]] on $S$, and the series converges uniformly on $S$.\n11. The functions $|f_n|$ satisfy the same hypotheses, with the same $M_n$. So steps 1 to 10 also show that $\\sum |f_n|$ converges uniformly on $S$.\n12. By steps 3, 10 and 11, the series $\\sum_{n=0}^{\\infty} f_n$ converges absolutely and uniformly on $S$.\n\n## Idea\n\nAt every point, the tail of the series is at most the tail $\\sum_{k \u003e m} M_k$ of a convergent series of numbers. That tail tends to $0$ and does not involve $x$. So one index $m_0$ works for all points at once, which is uniform convergence.\n",
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