What is the radius of convergence of:

$$ \sum_{n = 0}^{\infty}a_n^3z^n $$

I know that the formal calculation of the radius is by Cauchy-Hadamard:

$$ R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{a_n}} $$

So I don't understand why the answers show $2$ radiuses:

$$ R = \frac{1}{\limsup\sqrt[n]{|a_n|}} $$


$$ R' = \frac{1}{\limsup \sqrt[n]{|a_n|^3}} $$

Why are there $2$ radiuses? What is this $R$, it's not exactly as the formula is...?

  • $\begingroup$ Can you provide 'the answer' you have? $\endgroup$ May 24 '20 at 1:04
  • $\begingroup$ But first of all, $R'$ is the radius of convergent for the given series, according to the Cauchy-Hadamard theorem. $\endgroup$ May 24 '20 at 1:06
  • $\begingroup$ Its in hebrew and long and thats the start of the proof, is it be of any value to post a long answer in hebrew? And yes i thought that $R'$ is the radius, but where that $R$ came from? $\endgroup$
    – Alon
    May 24 '20 at 1:07
  • $\begingroup$ Ok so maybe its a mistake or used not aas the radius of the serie.. $\endgroup$
    – Alon
    May 24 '20 at 1:09
  • $\begingroup$ In both expressions of $ R$, replace $a_n$ by $u_n$. $R'$ is correct. $\endgroup$ May 24 '20 at 1:13

I don't know the literature, but I guess the first $R$ is from Cauchy-Hadamard theorem :

Consider $f(z)=\sum_{n=0}^\infty c_nz^n$. Then the radius of convergence for $f$ is $R=1/\limsup_{n\rightarrow\infty}|c_n|^{1/n}$.

Then the author of your 'the answer' did not want to use same notation for the radius of convergence of the given problem, so used $R'$ notation. Anyway, if you have $f(x)=\sum_{n=0}^\infty a_n^3z^n$, then it's radius of convergence is (note that $c_n=a_n^3$) : $$R'=\frac{1}{\limsup|a_n|^{3/n}}.$$


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