Suppose that $\sum_{k = 0}^\infty a_kx^k$ has radius of convergence of $R \in (0,\infty)$.

a) Find the radius of convergence of $\sum_{k = 0}^\infty a_kx^{2k}$

b) Find the radius of convergence of $\sum_{k = 0}^\infty a_k^2x^k$

Attempt: Given $$\left(\lim_{k \rightarrow \infty} \frac{\left|a_k\right|^\frac{1}{k}}{\left|a_{k+1}\right|^\frac{1}{k}}\right) = R$$

Then $$\left(\lim_{k \rightarrow \infty} \frac{\left|a_k\right|^\frac{1}{2k}}{\left|a_{k+1}\right|^\frac{1}{2k}}\right) = \left(\lim_{k \rightarrow \infty} \frac{\left|a_k\right|^\frac{1}{k}}{\left|a_{k+1}\right|^\frac{1}{k}}\right)^\frac{1}{2} = R^\frac{1}{2}$$


$$\left(\lim_{k \rightarrow \infty} \frac{\left|a_k^2\right|^\frac{1}{k}}{\left|a_{k+1}^2\right|^\frac{1}{k}}\right)= \left(\lim_{k \rightarrow \infty} \frac{\left|a_k\right|^\frac{1}{k}}{\left|a_{k+1}\right|^\frac{1}{k}}\right)^\frac{1}{2} =\left(\lim_{k \rightarrow \infty} \frac{\left|a_k\right|^\frac{1}{2k}}{\left|a_{k+1}\right|^\frac{1}{2k}}\right) = \left(\lim_{k \rightarrow \infty} \frac{\left|a_k \right|^\frac{1}{k}}{\left|a_{k+1}\right|^\frac{1}{k}}\right)^2 = R^2$$

Is this correct? Can anyone please help me? Any suggestion feedback would be really appreciate it. Thank you.


Let $a$ be the function in a), note that $a(x) = f(x^2)$. Hence the series is absolutely convergent iff $|x^2| < R$ iff $|x| < \sqrt{R}$. Hence the radius of convergence is $\sqrt{R}$.

Note that you can't assume that the ratio test applies. For example, if every third $a_n$ is zero then the limit is not defined.

For b), we have ${1 \over R} = \limsup_n \sqrt[n]{|a_n|}$. Then $\limsup_n \sqrt[n]{|a_n|^2} = \limsup_n (\sqrt[n]{|a_n|})^2 = (\limsup_n \sqrt[n]{|a_n|})^2 ={1 \over R^2}$. Hence the radius of convergence is $R^2$.


To justify the exchange of $\limsup$ and squaring, suppose $A \subset \mathbb{R}$ and $\phi: \overline{A} \to \mathbb{R}$ is non increasing and continuous. Then $\phi(\sup A) = \sup \phi(A)$.

Suppose $a \in A$, then $a \le \sup A$ and so $\phi(a) \le \phi(\sup A)$, which gives $\sup \phi(A) \le \phi(\sup A)$.

Now suppose $a_n \uparrow \sup A$, with $a_n \in A$. Then $\phi(a_n) \le \phi(\sup A)$. Continuity gives $\phi(\sup A) \le \sup \phi(A)$.

In the above, $A = \{ \sqrt[n]{|a_n|} \}_n$ and $\phi(x) = x^2$.

  • $\begingroup$ So the limit exists because the lim sup always exists so we can find the radius of convergence using the lim sup? $\endgroup$ – Mahidevran Mar 9 '15 at 23:31
  • $\begingroup$ Correct, the limit may be $0$ in which case we take $R= \infty$ as such. Unfortunately b) cannot be addressed in the same way as a), so using the $\limsup$ makes it manageable. $\endgroup$ – copper.hat Mar 10 '15 at 0:13
  • $\begingroup$ @copper.hat: Why does the equality $\limsup_n (\sqrt[n]{|a_n|})^2 = (\limsup_n \sqrt[n]{|a_n|})^2 $ hold? Can you share a reference or maybe elaborate in your post? $\endgroup$ – el_tenedor May 4 '15 at 12:53
  • $\begingroup$ @copper.hat: Ok, so you use continuity of $x \mapsto x^2 $ on a subsequence of $ \sqrt[n]{| a_n |} $ converging to the $ \limsup $? $\endgroup$ – el_tenedor May 4 '15 at 14:16
  • $\begingroup$ @el_tenedor: I added a more complete explanation to the answer. $\endgroup$ – copper.hat May 4 '15 at 15:17

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