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  "body": "\n## Proof\n\nWe dominate the terms by a convergent geometric series and apply the Weierstrass M-test.\n\n1. Let $\\sum a_n (x - c)^n$ be a [[power series|0VP56B4EMX]] with [[radius of convergence|0660QX1VPR]] $R \u003e 0$. Let $A = \\set{ t \\ge 0 : (|a_n| t^n) \\text{ is bounded} }$, so that $R = \\sup A$.\n2. Let $0 \u003c r \u003c R$. Choose a real number $s$ with $r \u003c s \u003c R$. For example, take $s = \\frac{r + R}{2}$ if $R \u003c \\infty$, and $s = r + 1$ if $R = \\infty$.\n3. Since $s \u003c R = \\sup A$, the number $s$ is not an upper bound of $A$, by the definition of the [[supremum|0BTMQ04556]]. So there is $t \\in A$ with $t \u003e s$.\n4. Since $t \\in A$, there is $C \\ge 0$ with $|a_n| t^n \\le C$ for all $n$. Since $0 \u003c s \u003c t$, also $|a_n| s^n \\le |a_n| t^n \\le C$ for all $n$.\n5. Let $q = \\frac{r}{s}$. Then $0 \u003c q \u003c 1$.\n6. Let $x \\in [c - r, c + r]$. Then $|x - c| \\le r$, so by step 4,\n$$|a_n (x - c)^n| \\le |a_n| r^n = |a_n| s^n q^n \\le C q^n.$$\n7. Let $M_n = C q^n$. Since $|q| \u003c 1$, the [[geometric series|01X5CVV1Q5]] $\\sum_{n=0}^{\\infty} M_n$ converges.\n8. By steps 6 and 7, the [[Weierstrass M-test|0NX5NWAFC9]] applies to the functions $x \\mapsto a_n (x - c)^n$ on $[c - r, c + r]$.\n9. So the power series converges absolutely and uniformly on $[c - r, c + r]$.\n\n## Idea\n\nSlightly beyond $r$, at some $s \u003c R$, the terms $|a_n| s^n$ stay bounded. On $[c - r, c + r]$, each term is then at most a constant times $(r/s)^n$. These bounds form a convergent geometric series that does not depend on $x$, so the M-test gives uniform convergence.\n",
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    "name": "Proof that power series converge uniformly on smaller intervals inside the radius of convergence",
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        "locator": "Section 2.2, Power series",
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  "name": "Proof that power series converge uniformly on smaller intervals inside the radius of convergence",
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