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I am trying to prove that the radius of convergence of a power series does not change after differentiating term by term. Let $\sum a_nx^n$ be a power series with radius of convergence $R$. Let $R_2$ be the radius of convergence of $\sum na_nx^{n-1}$. Then $$ \frac{1}{R_2} = \limsup\limits_{n\to\infty}\sqrt[n]{|na_n|} $$ Let $n_k$ be a strictly increasing sequence of natural numbers such that $\sqrt[n_k]{|n_ka_{n_k}|}$ converges to the limit superior. Then we have that $$ \frac{1}{R_2} = \lim\limits_{k\to\infty}\sqrt[n_k]{|n_ka_{n_k}|} = \lim\limits_{k\to\infty}\sqrt[n_k]{n_k}\lim\limits_{k\to\infty}\sqrt[n_k]{|a_{n_k}|} = \lim\limits_{k\to\infty}\sqrt[n_k]{|a_{n_k}|} $$ Since the limit superior of the sequence $\sqrt[n]{|na_n|}$ is a limit of a subsequence of $\sqrt[n]{|a_n|}$, we conclude that $$ \limsup\limits_{n\to\infty}\sqrt[n]{|na_n|} \leq \limsup\limits_{n\to\infty}\sqrt[n]{|a_n|} $$

Is the reverse inequality achieved in the same way?

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  • $\begingroup$ If the oringinal series is inside the radius of convergence then $\lim\limits_{n\to\infty}\sqrt[n]{|a_n|}<1$, Now show that if that is true then $\lim\limits_{n\to\infty}\sqrt[n]{n|a_n|} = \lim\limits_{n\to\infty}\sqrt[n]{n}\sqrt[n]{|a_n|}<1$ $\endgroup$
    – Doug M
    Jun 8, 2016 at 19:03
  • $\begingroup$ why $1/R_2=lim sup |na_n|^{1/n}$? instead of $1/R_2=lim sup |(n+1)a_{n+1}|^{1/n}$ ? since $\sum na_nx^{n-1}=\sum (n+1)a_{n+1}x^{n}$ $\endgroup$
    – xyz
    Jul 7, 2020 at 2:05

1 Answer 1

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The reverse inequality is easier: Since for all $n>1$ $$ \sqrt[n]{|na_n|} = \sqrt[n]{n} \sqrt[n]{|a_n|} > \sqrt[n]{|a_n|} $$ that is, since the sequence $\sqrt[n]{|na_n|}$ is term-by-term greater than $\sqrt[n]{|a_n|}$, we immediately know that $$ \limsup_{n\to\infty}\sqrt[n]{|na_n|} \geq \limsup_{n\to\infty}\sqrt[n]{|a_n|} $$

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