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  "body": "\n## Proof\n\nWe work from the definition of the radius of convergence and compare $\\abs{a_n}$ with powers $t^n$.\n\n1. Let $\\sum a_n (x - c)^n$ be a real [[power series|0VP56B4EMX]]. For $n \\ge 1$ let $x_n = \\abs{a_n}^{1/n} \\ge 0$.\n2. Let $\\ell = \\limsup_{n \\to \\infty} x_n$, the [[limit superior|0P8BA7T7S5]]. If $(x_n)$ is not bounded above, then $\\ell = \\infty$ by convention. Otherwise $(x_n)$ is [[bounded|05FQ3VGAR9]], since $x_n \\ge 0$, and $0 \\le \\ell \u003c \\infty$.\n3. Let $T$ be the set of $r \\ge 0$ for which $(\\abs{a_n} r^n)$ is bounded. By the [[definition|0660QX1VPR]], $R = \\sup T$, with $R = \\infty$ when $T$ is not bounded above.\n4. The number $0$ lies in $T$, since $\\abs{a_n} 0^n = 0$ for $n \\ge 1$.\n5. Claim A: if $\\ell \u003c \\infty$, $r \u003e 0$ and $r \\ell \u003c 1$, then $r \\in T$. To see this, pick $t$ with $\\ell \u003c t \u003c 1/r$.\n6. When $\\ell \u003c \\infty$, let $s_n = \\sup \\set{x_k : k \\ge n}$. The sequence $(s_n)$ decreases and [[converges to its infimum|02V2GYAC8P]], which is $\\ell$.\n7. Since $t \u003e \\ell$, the [[epsilon-N form|09G6KHS6GY]] gives $N \\ge 1$ with $s_N \u003c t$. So $x_k \u003c t$ for all $k \\ge N$.\n8. For $k \\ge N$, step 7 gives $\\abs{a_k} \u003c t^k$. So $\\abs{a_k} r^k \u003c (t r)^k \\le 1$, since $0 \u003c t r \u003c 1$.\n9. The finitely many terms with $k \u003c N$ are bounded by their maximum. So $(\\abs{a_k} r^k)$ is bounded, and $r \\in T$. This proves Claim A.\n10. Claim B: if $r \u003e 0$ and $r \\ell \u003e 1$, then $r \\notin T$. Here $r \\ell \u003e 1$ is read as true for all $r \u003e 0$ when $\\ell = \\infty$.\n11. Pick $t$ with $1/r \u003c t \u003c \\ell$. When $\\ell = \\infty$, take $t = 2/r$. We show that for each $n \\ge 1$ there is $k \\ge n$ with $x_k \u003e t$.\n12. If $\\ell \u003c \\infty$, then $s_n \\ge \\ell \u003e t$, because $\\ell$ is the infimum of the $s_n$. The [[approximation property of the supremum|0M8TGGP9JP]] gives $k \\ge n$ with $x_k \u003e t$.\n13. If $\\ell = \\infty$, then $(x_k)$ is not bounded above. So some $x_k$ exceeds $\\max(t, x_1, \\dots, x_n)$. Such a $k$ is not in $\\set{1, \\dots, n}$, so $k \u003e n$ and $x_k \u003e t$.\n14. Let $q = t r \u003e 1$. If $x_k \u003e t$, then $\\abs{a_k} \u003e t^k$ and $\\abs{a_k} r^k \u003e q^k$.\n15. By the [[binomial theorem|0EHWPWGHB2]], $q^k = (1 + (q - 1))^k \\ge 1 + k (q - 1)$, since all the terms of the expansion are nonnegative.\n16. Let $C$ be a real number. The reals are [[Archimedean|0A3GWZ86AY]], so there is $n \\ge 1$ with $n (q - 1) \u003e C$.\n17. Steps 11 to 13 give $k \\ge n$ with $x_k \u003e t$. Steps 14 and 15 give $\\abs{a_k} r^k \u003e 1 + k (q - 1) \u003e C$.\n18. So $(\\abs{a_k} r^k)$ is not bounded, and $r \\notin T$. This proves Claim B.\n19. Case $\\ell = 0$. Claim A gives every $r \u003e 0$ in $T$. So $T$ is not bounded above, and $R = \\infty = 1/\\ell$.\n20. Case $\\ell = \\infty$. Claim B gives $T = \\set{0}$, with step 4. So $R = 0 = 1/\\ell$.\n21. Case $0 \u003c \\ell \u003c \\infty$. By Claim B, every element of $T$ is at most $1/\\ell$, so $1/\\ell$ is an upper bound of $T$.\n22. In this case, let $b \u003c 1/\\ell$. The number $r = (\\max(b, 0) + 1/\\ell)/2$ satisfies $b \u003c r \u003c 1/\\ell$, and $r \\in T$ by Claim A. So no number below $1/\\ell$ is an upper bound.\n23. By steps 21 and 22 and the definition of the [[supremum|0BTMQ04556]], $R = 1/\\ell$.\n24. In all cases $R = 1/\\ell$. With the conventions $1/0 = \\infty$ and $1/\\infty = 0$, this is $1/R = \\limsup_{n \\to \\infty} \\abs{a_n}^{1/n}$.\n\n## Idea\n\nThe number $\\ell$ measures the exponential growth rate of $\\abs{a_n}$: eventually $\\abs{a_n} \u003c t^n$ for each $t \u003e \\ell$, and $\\abs{a_n} \u003e t^n$ infinitely often for each $t \u003c \\ell$. So $\\abs{a_n} r^n$ stays bounded when $r \\ell \u003c 1$ and grows without bound when $r \\ell \u003e 1$. The cutoff is $r = 1/\\ell$.\n",
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