Rudin: Prove that if $f(x)=\sum_{n=0}^{\infty}c_nx^n$ converges for $|x|<R$, then $f'(x)=\sum_{n=1}^{\infty}nc_nx^{n-1}$ in $(-R, R)$.

Question

The Problem: If $f(x)=\sum_{n=0}^{\infty}c_nx^n$ converges for $|x|<R$, then $f'(x)=\sum_{n=1}^{\infty}nc_nx^{n-1}$ in $(-R, R)$.

The problem is a theorem from Walter Rudin's PMA:

Here is part of Rudin's proof:

My Question: I don't quite follow the texts highlighted in yellow. The Root Test-which I suppose is what he is applying here-says that:

Thus, (3) converges implies only that $\underset{n\to\infty}{\lim}\sup\sqrt[n]{\vert c_nx^n\vert}\leq1$ by (b) of Theorem 3.33, from which we at most can deduce that $\underset{n\to\infty}{\lim}\sup\sqrt[n]{|n c_n x^{n-1}|}\leq 1$ (because we have $\underset{n\to\infty}{\lim}\sup\sqrt[n]{n\vert c_n\vert}=\underset{n\to\infty}{\lim}\sup\sqrt[n]{\vert c_n\vert}$, thus it seems $\underset{n\to\infty}{\lim}\sup\sqrt[n]{\vert c_nx^n\vert}=\underset{n\to\infty}{\lim}\sup\sqrt[n]{|n c_n x^{n-1}|}$). But what happens if $\underset{n\to\infty}{\lim}\sup\sqrt[n]{|n c_n x^{n-1}|}=1$? The same conundrum exists when (3) diverges.

Any help would be greatly appreciated.

Answer 1

You are looking at the root test in the wrong way. Instead of Theorem 3.3, you should take a look at Theorem 3.39 in Rudin, which tells you that the radius of convergence is determined by the limsup of the absolute value of the coefficients. Therefore, the identity $$ \limsup \sqrt[n]{n|c_n|}=\limsup \sqrt[n]{|c_n|} $$ tells you that the (3) converges if and only if (5) converges. Note that $R$ in the theorem you quote could be any number not bigger than the radius of convergence (the $R$ in Theorem 3.39).

Answer 2

Observe the radius of convergence is defined as $R = \dfrac{1}{\limsup (|c_n|^{1/n})}$ so if you take $x \in (-R,R)$, you will have $\left|x \times {\limsup (|c_n|^{1/n})}\right| < 1$, now ...