0
$\begingroup$

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

and

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

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

$\endgroup$
1
$\begingroup$

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 $R^2$.

Addendum:

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 $\phi(x) = x^2$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.