2026-04-13 12:04:53.1776081893
I am trying to understand a proof regarding the derivative of a power function. Let ${ \sum_{k=0}^{ \infty }{a_k(x-x_0)^k} }$ be a real power series with a positive radius of convergence ${ R \le \infty }$. Then the radius of convergence of ${ \sum_{k=1}^{ \infty }{ka_k(x-x_0)^{k-1}} }$ is also R. The proof goes as follows:
WLOG Let ${ x_0 =0 }$ and ${ r \lt R }$. Then let
$${sup ( \left| a_k x^k\right| \vert x \in [-r,r])= \left| a_k\right|r^k }$$ If ${ r \gt R }$, then ${\left| a_k r^k\right|}$ isn't a zero sequence and this implicates that ${\left| ka_k r^{k-1}\right|}$ also isn't a zero sequence. Therefore the radius of convergence of ${\sum_{k=1}^{\infty} ka_k x^{k-1}}$ must be less than or equal to R. Now let ${ r \lt R }$.
Let ${ r_2 \gt 0}$ so that ${r \lt r_2 \lt R}$. It holds that ${ k \left| a_k \right| r^{k-1} \le \left| a_k \right| r_2^k}$ for almost all k. As a result ${\sum_{k=1}^{\infty} \left| a_k \right| r_2^k}$ is a convergent majorant for ${\sum_{k=1}^{\infty} ka_k r^{k-1}}$, which means that ${\sum_{k=1}^{\infty} ka_k r^{k-1}}$ converges absolutely.
Why is the implication enough to conclude that the radius of convergence must be less than or equal to R?Also I don't get this inequation ${ k \left| a_k \right| r^{k-1} \le \left| a_k \right| r_2^k}$. I simplified it to ${ \frac{k}{r_2} r^{k-1} \le r_2^{k-1}}$ but I don't see the how the right value must be greater.
The derivative of a power series has the same radius of convergence
3.1k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtI am trying to understand a proof regarding the derivative of a power function. Let ${ \sum_{k=0}^{ \infty }{a_k(x-x_0)^k} }$ be a real power series with a positive radius of convergence ${ R \le \infty }$. Then the radius of convergence of ${ \sum_{k=1}^{ \infty }{ka_k(x-x_0)^{k-1}} }$ is also R. The proof goes as follows:
WLOG Let ${ x_0 =0 }$ and ${ r \lt R }$. Then let
$${sup ( \left| a_k x^k\right| \vert x \in [-r,r])= \left| a_k\right|r^k }$$ If ${ r \gt R }$, then ${\left| a_k r^k\right|}$ isn't a zero sequence and this implicates that ${\left| ka_k r^{k-1}\right|}$ also isn't a zero sequence. Therefore the radius of convergence of ${\sum_{k=1}^{\infty} ka_k x^{k-1}}$ must be less than or equal to R. Now let ${ r \lt R }$.
Let ${ r_2 \gt 0}$ so that ${r \lt r_2 \lt R}$. It holds that ${ k \left| a_k \right| r^{k-1} \le \left| a_k \right| r_2^k}$ for almost all k. As a result ${\sum_{k=1}^{\infty} \left| a_k \right| r_2^k}$ is a convergent majorant for ${\sum_{k=1}^{\infty} ka_k r^{k-1}}$, which means that ${\sum_{k=1}^{\infty} ka_k r^{k-1}}$ converges absolutely.
Why is the implication enough to conclude that the radius of convergence must be less than or equal to R?Also I don't get this inequation ${ k \left| a_k \right| r^{k-1} \le \left| a_k \right| r_2^k}$. I simplified it to ${ \frac{k}{r_2} r^{k-1} \le r_2^{k-1}}$ but I don't see the how the right value must be greater.
1
Have you encountered the Cauchy-Hadamard theorem regarding the radius of a power series? Assuming a power series of the for $\sum\limits_{n=0}^{\infty}a_n(x-x_0)^n$ the radius of convergence is given by the formula: $$ \frac{1}{R}=\limsup\sqrt[n]{|a_n|} $$ Using this formula and the fact that the derivative series is $\sum\limits_{n=1}^{\infty}na_n(x-x_0)^{n-1}$, the new radius must be: $$ \frac{1}{\tilde{R}}=\limsup\sqrt[n-1]{|na_n|}=\limsup\sqrt[n-1]{n}\sqrt[n-1]{|a_n|} $$ And since $\sqrt[n-1]{n}$ converges to $1$, it is elementary to show that: $$ =\lim\limits_{n\to\infty}\sqrt[n-1]{n}\limsup\sqrt[n-1]{|a_n|}=\limsup\sqrt[n-1]{|a_n|}=\frac{1}{R} $$ And so we get the same radius.