The series $\sum_{n=1}^\infty nc_n (x-a)^{n-1}$ converges uniformly to $f'$ on the interval $[a-r, a+r]$

Question

(Tao Vol.2, P.77) Let $\sum_{n=0}^\infty c_n(x-a)^n$ be a power series, and let $R$ be the radius of convergence. Suppose that $R>0$ (the series converges at least one other point than $x= a$). Let $f : (a- R, a+ R) \to \mathbb{R}$ be the function $f(x) : = \sum_{n=0}^\infty c_n (x-a)^n$.

I need to prove the following statement: (Differentiation of power series) The function $f$ is differentiable on $(a- R, a+R)$, and for any $0<r <R$, the series $\sum_{n=1}^\infty nc_n (x-a)^{n-1}$ converges uniformly to $f'$ on the interval $[a-r, a+r]$.

The author suggests to use the following theorem: Theorem 3.7.1. Let $[a,b]$ be an interval, and for every integer $n \ge 1$, let $f_n : [a,b] \to \mathbb{R}$ be a differentiable function whose derivative $f'_n:[a,b] \to \mathbb{R}$ is continuous. Suppose that the derivatives $f'_n$ converge uniformly to a function $g: [a,b] \to \mathbb{R}$. Suppose also that there exists a point $x_0$ such that the limit $\lim_{n \to \infty} f_n(x_0)$ exits. Then the function $f_n$ converge uniformly to a differentiable function $f$, and the derivative of $f$ equals $g$.

Let $F_N(x) = \sum_{n=0}^N c_n(x-a)^n$ for $x\in[a-r, a+r]$. $F'_N(x) = \sum_{n=1}^{N} nc_n(x-a)^{n-1}$ is continuous. I need to show that $(F'_N (x))_N$ converges uniformly to $g$. Then, by Theorem 3.7.1, $(F_N(x))_N$ converges uniformly to $f$, and $f$ is differentiable and $\sum_{n=1}^{\infty} nc_n(x-a)^{n-1} =f'(x) = g(x)$.

To show uniform convergence, it suffices to show that $\sum_{n=1}^\infty ||nc_n(x-a)^{n-1}||_{\infty}$ converges. $||nc_n(x-a)^{n-1}|| _\infty = \sup\{|nc_n||x-a|^{n-1} :x \in [a-r, a+r]\} = |nc_n||r|^{n-1}.$ I am trying to prove the convergence of $\sum_{n=1}^\infty|nc_n||r|^{n-1}$ using the root test. Thus, to finish the proof, I have to show that $$\lim \sup_{n\to \infty}|n|^{1/n}|c_n|^{1/n} |r|^{(n-1)/n} <1.$$

I don't know how to prove this inequality. I know that $|n|^{1/n} \to 1$ and $\lim \sup |c_n|^{1/n} = 1/R$.

Am I using the right approach? How can I finish the proof?

I would appreciate if you give some help.

Answer 1

Let $r<r'<R$. Then $\lim \sup |c_n|^{n} =\frac 1 R<\frac 1 {r'}$ so there exist $n_0$ such that $|c_n|^{n} <\frac 1 {r'}$ for all $n \geq n_0$. Hence $\lim \sup n^{1/n}|c_n|^{1/n}r^{1-\frac 1 n} =\lim \sup |c_n|^{1/n}r^{1-\frac 1 n} \leq \frac 1 {r'} r <1$.