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  "body": "\n## Proof\n\nWe compare consecutive terms $\\abs{a_n} r^n$ directly, as in the ratio test.\n\n1. Let $\\sum a_n (x - c)^n$ be a real [[power series|0VP56B4EMX]]. Let $n_0$ be such that $a_n \\ne 0$ for all $n \\ge n_0$.\n2. For $n \\ge n_0$ let $\\rho_n = \\abs{a_{n+1}} / \\abs{a_n}$. By hypothesis $\\rho_n \\to L \\in [0, \\infty]$. When $L = \\infty$, this means that for each $K$ there is $N$ with $\\rho_n \u003e K$ for all $n \\ge N$.\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 $L \u003c \\infty$, $r \u003e 0$ and $r L \u003c 1$, then $r \\in T$. To see this, pick $q$ with $r L \u003c q \u003c 1$. Then $L \u003c q / r$.\n6. By the [[epsilon-N form|09G6KHS6GY]] with $\\varepsilon = q/r - L$, there is $N \\ge n_0$ with $\\rho_n \u003c q / r$ for all $n \\ge N$.\n7. For $n \\ge N$, step 6 gives $\\abs{a_{n+1}} r^{n+1} = \\rho_n r \\abs{a_n} r^n \u003c q \\abs{a_n} r^n \\le \\abs{a_n} r^n$.\n8. By induction on $n$, step 7 gives $\\abs{a_n} r^n \\le \\abs{a_N} r^N$ for all $n \\ge N$. The finitely many terms with $n \u003c N$ are bounded by their maximum. So $r \\in T$.\n9. Claim B: if $L \u003e 0$, $r \u003e 0$ and $r L \u003e 1$, then $r \\notin T$. Here $r L \u003e 1$ is read as true for all $r \u003e 0$ when $L = \\infty$.\n10. Pick $q$ with $1 \u003c q \u003c r L$. When $L = \\infty$, take $q = 2$. Then $q / r \u003c L$.\n11. If $L \u003c \\infty$, the epsilon-N form with $\\varepsilon = L - q/r$ gives $N \\ge n_0$ with $\\rho_n \u003e q/r$ for $n \\ge N$. If $L = \\infty$, step 2 with $K = q/r$ gives such an $N$.\n12. For $n \\ge N$, as in step 7, $\\abs{a_{n+1}} r^{n+1} \u003e q \\abs{a_n} r^n$. By induction, $\\abs{a_n} r^n \\ge D q^{n - N}$ for all $n \\ge N$, where $D = \\abs{a_N} r^N \u003e 0$.\n13. By the [[binomial theorem|0EHWPWGHB2]], $q^j = (1 + (q - 1))^j \\ge 1 + j (q - 1)$ for all natural numbers $j$, since every term of the expansion is nonnegative.\n14. Let $C$ be a real number. The reals are [[Archimedean|0A3GWZ86AY]], so there is $j$ with $D j (q - 1) \u003e C$. Then $n = N + j$ gives $\\abs{a_n} r^n \\ge D (1 + j(q-1)) \u003e C$.\n15. So $(\\abs{a_n} r^n)$ is not bounded, and $r \\notin T$.\n16. Case $L = 0$. Claim A gives every $r \u003e 0$ in $T$. So $T$ is not bounded above, and $R = \\infty = 1/L$.\n17. Case $L = \\infty$. Claim B and step 4 give $T = \\set{0}$. So $R = 0 = 1/L$.\n18. Case $0 \u003c L \u003c \\infty$. By Claim B, $1/L$ is an upper bound of $T$. By Claim A, every $r$ with $0 \u003c r \u003c 1/L$ lies in $T$.\n19. So no number below $1/L$ is an upper bound of $T$. By the definition of the [[supremum|0BTMQ04556]], $R = 1/L$.\n20. In all cases $R = 1/L$, with the conventions $1/0 = \\infty$ and $1/\\infty = 0$.\n\n## Idea\n\nIf $\\abs{a_{n+1}} / \\abs{a_n}$ is close to $L$, then $\\abs{a_n} r^n$ is eventually multiplied by about $L r$ at each step. When $L r \u003c 1$ the terms shrink, and when $L r \u003e 1$ they grow like a geometric sequence with ratio above $1$. So the cutoff is $r = 1/L$.\n",
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