Prove that the radius of derived series $\sigma'$ is the same of $\sigma$.

Let be $\sigma$ the power series given by the equation $$ \sigma:=\sum_{k=0}^\infty a_k x^k $$ so that we call radius convergence of $\sigma$ the quantity $\rho$ given by the equation $$ \rho:=\sup\{x\in\Bbb R:\sigma(x)\text{ converges}\} $$ So let be now $\sigma'$ the series obtained by $\sigma$ deriving each term, that is $$ \sigma':=\sum_{k=1}^\infty ka_kx^{k-1} $$ so I'd like to understand better the following theorem proving that the radius of convergence of $\sigma$ is the same of $\sigma'$.

Theorem

The radius of convergence of $\sigma$ is the same of $\sigma'$

Proof. So let be $\rho$ and $\rho'$ the radius of convergence of $\sigma$ and $\sigma'$ respectively. So fist of all we let to prove that if $\sigma$ converges at $x_0$ then $\sigma'$ converges for any $x$ such that $|x|<|x_0|$ so that $\rho\le\rho'$. So if the series $\sigma$ converges at $x_0$ then there exist $L>0$ such that $$ |a_kx_0^k|\le L $$ for all $k\in\Bbb N$ so that by the comparision test and by the inequality $$ |k a_kx^{k-1}|=\frac{k|a_kx_0^k|}{|x_0|}\cdot\Biggl|\frac{x}{x_0}\Biggl|^{k-1}\le\frac{kL}{x_0}\cdot\Biggl|\frac{x}{x_0}\Biggl|^{k-1} $$ we conclude that the series $\sigma'$ converges.

Conversely we let to prove that if the series $\sigma'$ converges at $x_1$ then it converges at any $x$ such that $|x|<|x_1|$. So if the series $\sigma'$ converges at $x_1$ then then there exist $M>0$ such that $$ |ka_kx_1^{k-1}|\le M $$ for any $k\in\Bbb N$ so that by the comparison ratio and by the inequality $$ |a_k x^k|=\frac{1}k|ka_kx^{k-1}|\cdot|x_1|\cdot\Biggl|\frac x {x_1}\Biggl|^k\le\frac{M|x_1|}k\cdot\Biggl|\frac x{x_1}\Biggl|^k $$ we conclude that the series $\sigma$ converges.

So I do not understand why the implication $$ \sigma\text{ converges at }x_0\Rightarrow\sigma'\text{ converges at any }x\,\text{such that }|x|<|x_0| $$ implies that $\rho\le\rho'$ and anagously I do not understand also why the implication $$ \sigma'\text{ converges at }x_1\Rightarrow\sigma\text{ converges at any }x\,\text{such that }|x|<|x_1| $$ implies $\rho'\le\rho$.

So could someone explain this, please?

Solution 1:

Given the implication

$$\sigma\text{ converges at }x_0\Rightarrow\sigma'\text{ converges at any }x\text{ such that }|x|<|x_0|,$$

we see that $$(-|x_0|, |x_0|) \subseteq \{x \in \Bbb{R} : \sigma' \text{ converges at } x\}$$ whenever $\sigma$ converges at $x_0$. Taking the supremum of both sides, $$|x_0| \le \rho' \implies x_0 \le \rho'.$$ Note that $x_0$ is an arbitrary element of $\{x \in \Bbb{R} : \sigma \text{ converges at } x\}$, so we see that $\rho'$ is an upper bound of this set. The least upper bound (i.e. the supremum) of this set is $\rho$, so $\rho \le \rho'$.

The other implication follows analogously.