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  "body": "\n## Proof\n\nFirst we show that the derived series has the same radius of convergence. Then we apply term-by-term differentiation on a closed interval around each point.\n\n*Part 1: the derived series has radius $R$.*\n\n1. Let $f(x) = \\sum_{n=0}^{\\infty} a_n (x - c)^n$ have [[radius of convergence|0660QX1VPR]] $R \u003e 0$. For $m \\ge 0$, let $b_m = (m + 1) a_{m+1}$. The derived series is $\\sum_{n=1}^{\\infty} n a_n (x - c)^{n-1} = \\sum_{m=0}^{\\infty} b_m (x - c)^m$.\n2. Let $A = \\set{ t \\ge 0 : (|a_n| t^n) \\text{ is bounded} }$ and $B = \\set{ t \\ge 0 : (|b_m| t^m) \\text{ is bounded} }$. Then $R = \\sup A$, and the derived series has radius $R' = \\sup B$.\n3. We show $B \\subseteq A$. The constant sequence $(|a_n| 0^n)$ is bounded, so $0 \\in A$. Let $t \\in B$ with $t \u003e 0$, and let $D$ bound $|b_m| t^m$. Then $|a_{m+1}| t^{m+1} = \\frac{t \\, |b_m| t^m}{m + 1} \\le t D$ for all $m \\ge 0$. So $(|a_n| t^n)$ is bounded, and $t \\in A$.\n4. By step 3, $R' = \\sup B \\le \\sup A = R$.\n5. Now let $0 \u003c r \u003c R$. As $r \u003c R = \\sup A$, $r$ is not an upper bound of $A$, by the definition of the [[supremum|0BTMQ04556]]. So there are $s \\in A$ with $s \u003e r$, and $C \\ge 0$ with $|a_n| s^n \\le C$ for all $n$.\n6. Let $q = \\frac{r}{s} \\in (0, 1)$. For $m \\ge 0$, step 5 gives\n$$|b_m| r^m = (m + 1) |a_{m+1}| s^{m+1} \\frac{r^m}{s^{m+1}} \\le \\frac{C}{s} (m + 1) q^m.$$\n7. Let $\\rho = \\frac{1}{q} \u003e 1$. Then $(m + 1) q^m = \\frac{m}{\\rho^m} + \\frac{1}{\\rho^m}$. Both terms tend to $0$, since [[exponentials dominate powers|0MVTCFW42K]]. By the [[sum rule|0RAM4E07NR]], $(m + 1) q^m \\to 0$.\n8. A [[convergent sequence is bounded|0S4JF619V7]]. So by steps 6 and 7, $(|b_m| r^m)$ is bounded, and $r \\in B$.\n9. Suppose that $R' \u003c R$. Choose $r$ with $\\max(R', 0) \u003c r \u003c R$. By step 8, $r \\in B$, so $r \\le \\sup B = R'$. This is a contradiction. So $R' \\ge R$, and with step 4, $R' = R$.\n\n*Part 2: differentiability.*\n\n10. Fix $x \\in (c - R, c + R)$. Choose $r$ with $|x - c| \u003c r \u003c R$, and let $J = [c - r, c + r]$.\n11. For $n \\ge 0$ and real $t$, let $f_n(t) = a_n (t - c)^n$. Apply the [[derivative of a polynomial|0MH1WK89PP]] to $u \\mapsto a_n u^n$, and the [[chain rule|0RXX2PT1MQ]] with $u = t - c$. So $f_n$ is differentiable on $\\R$ with $f_n'(t) = n a_n (t - c)^{n-1}$ for $n \\ge 1$, and $f_0' = 0$.\n12. In particular, at the endpoints of $J$ the [[one-sided derivatives|0HXEX0DPRA]] of $f_n$ exist and equal $f_n'$. So each $f_n$ is differentiable on $J$.\n13. At $t = c$, every term $f_n(c)$ with $n \\ge 1$ is $0$. So $\\sum_n f_n(c)$ converges, with sum $a_0$.\n14. By step 11, $\\sum_n f_n'$ is the derived series. By Part 1 it has radius $R \u003e 0$. Since $0 \u003c r \u003c R$, it converges uniformly on $J$, because [[power series converge uniformly on smaller intervals|0BWXV7GADW]].\n15. By steps 12 to 14, [[term-by-term differentiation|0TJ8QBJX5K]] applies on $J$. So $\\sum_n f_n$ converges uniformly on $J$, its sum is differentiable on $J$, and the derivative is $\\sum_n f_n'$.\n16. The sum of $\\sum_n f_n$ on $J$ is $f$ restricted to $J$. Since $x$ lies in the interior of $J$, the derivative of $f$ on $J$ at $x$ is the ordinary two-sided derivative of $f$ at $x$.\n17. So $f$ is differentiable at $x$, and $f'(x) = \\sum_{n=1}^{\\infty} n a_n (x - c)^{n-1}$.\n18. Since $x \\in (c - R, c + R)$ was arbitrary, $f$ is differentiable on $(c - R, c + R)$ with this derivative. By Part 1, the derived series has radius of convergence $R$.\n\n## Idea\n\nMultiplying the coefficients by $n$ does not change the radius of convergence, because $n$ grows more slowly than any geometric factor $\\rho^n$ with $\\rho \u003e 1$. So the derived series converges uniformly on each $[c - r, c + r]$ with $r \u003c R$. The theorem on term-by-term differentiation then applies around each point.\n",
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