Setting: Let $\sum a_n z^n$ have radius of convergence $R$. We have that

$$ R = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} $$

via Hadamard's formula for the radius of convergence.

Question: What are the radii of convergence for (i) $\sum a_n z^{2n}$ and (ii) $\sum a_n^2 z^n$?

Attempt for $\sum a_n^2 z^n$:

  1. For ease of notation, let $R_1$ and $R_2$ denote the radii of convergence for power series $\sum a_n z^{2n}$ and $\sum a_n^2 z^n$ respectively.

  2. From Hadamard's formula, we then have

    $$ R_2 = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n^2 \right|^{1/n}} = \left( \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} \right)^2 = R^2 $$

But what about $R_1$?

  • $\begingroup$ Consider using \limsup to get $\limsup$ instead of $limsup$. $\endgroup$ – Fly by Night Mar 4 '14 at 18:15
  • $\begingroup$ TooOldForMath -- I just corrected Hadamard's formula. $\endgroup$ – user125104 Mar 4 '14 at 18:22
  • $\begingroup$ Use the D'Alembert property again! $\endgroup$ – mookid Mar 4 '14 at 18:22
  • 1
    $\begingroup$ Quote from a comment on your previous quite receznt question: "there are much simpler and shorter approaches than Hadamard's formula (a vastly overrated tool, based on the questions on this site)." $\endgroup$ – Did Mar 4 '14 at 18:49

Your calculation for the second is correct. For the first one note that $$\sum_{n=0}^\infty a_n z^{2n} = \sum_{n=0}^\infty b_n z^n$$

with $b_n=a_{k}$ if $n=2k$ for $k\in\mathbb{N}$ and $b_n=0$ otherwise.


$$R_1=\frac{1}{\limsup_{n\rightarrow\infty}|b_n|^{1/n}}=\frac{1}{\limsup_{n\rightarrow\infty} |a_n|^{1/(2n)}}=\left(\frac{1}{\limsup_{n\rightarrow\infty} |a_n|^{1/n}}\right)^{1/2}=R^{1/2}$$


Let $w=z^2$. Then the first series converges for $|w|<R$, thus...


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