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  "body": "\n## Proof\n\nWe apply the [[ratio formula for the radius of convergence|0JBMYAA9QH]].\n\n1. Let $\\alpha$ be a real number that is not a natural number. For each natural number $k$, let $c_k = \\binom{\\alpha}{k} = \\alpha (\\alpha - 1) \\cdots (\\alpha - k + 1) / k!$, the coefficients of the [[binomial series|0HB0XKDXHH]].\n2. For each natural number $j$, $\\alpha - j \\ne 0$, since $\\alpha$ is not a natural number. So $c_k$ is a product of nonzero numbers divided by $k!$, and $c_k \\ne 0$ for all $k$.\n3. From the formula in step 1, $c_{k+1} = c_k \\cdot (\\alpha - k)/(k + 1)$. So $\\abs{c_{k+1}} / \\abs{c_k} = \\abs{\\alpha - k} / (k + 1)$.\n4. For $k \u003e \\abs{\\alpha}$ we have $\\alpha - k \u003c 0$. So $\\abs{\\alpha - k}/(k + 1) = (k - \\alpha)/(k + 1) = 1 - (1 + \\alpha)/(k + 1)$.\n5. We show $1/(k + 1) \\to 0$. Let $\\varepsilon \u003e 0$. There is a natural number $n \\ge 1$ with [[$1/n \u003c \\varepsilon$|0WH916TJNY]].\n6. For $k \\ge n$, $0 \u003c 1/(k + 1) \u003c 1/n \u003c \\varepsilon$. By the [[epsilon-N form|09G6KHS6GY]], $1/(k + 1) \\to 0$.\n7. By the [[product rule|0734RSVF2V]] and the [[sum rule|0RAM4E07NR]] for limits, $1 - (1 + \\alpha)/(k + 1) \\to 1 - 0 = 1$.\n8. By steps 3 and 4, $\\abs{c_{k+1}} / \\abs{c_k} \\to 1$ as $k \\to \\infty$.\n9. By step 2 and step 8, the ratio formula applies with $L = 1$. So the radius of convergence is $R = 1/1 = 1$.\n\n## Idea\n\nThe ratio of consecutive coefficients is $(\\alpha - k)/(k + 1)$, which tends to $-1$. So its absolute value tends to $1$. The hypothesis on $\\alpha$ makes every coefficient nonzero, so the ratio formula applies. If $\\alpha$ were a natural number, the series would stop and be a polynomial.\n",
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