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Suppose the radius of convergence of $\sum_n a_n x^n$ is $r$ ($r$ is a positive number). Prove that the radius of convergence of $\sum_n a_n^2 x^n$ is $r^2.$

I've tried to use Cauchy–Hadamard theorem, but I think it's illegal to use arithmetic of limits when we talk about limsup.

Thank you.

mfl
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Shirly Geffen
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1 Answers1

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You can use the ratio test. The ratio test says that if the limit as n goes to infinity of the absolute value of the ratio of successive terms in a series is less than 1, then the series converges. If greater than 1, then it does not.

The radius of convergence of the first series is r, so this means that the first series converges for all $x$ with $|x|<r$. In terms of the ratio test this means that for any $x$ with $|x|<r$ and no x with $|x|>r$ we have: $$ \lim_{n \to \infty} \left| \frac{a_{n+1}x^{n+1}}{a_n x^n} \right| < 1\\ \lim_{n\to \infty} \left| \frac{a_{n+1}x}{a_n} \right| <1\\ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| |x| <1\\ |x| \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| <1 $$

Since this is true for all $x$ with $|x|<r$ and no $x$ with $|x|>r$, it must be that $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{r}$.

Now consider the ratio test for the second sequence: $$ \lim_{n \to \infty} \left| \frac{(a_{n+1})^2x^{n+1}}{(a_n)^2 x^n} \right| \\ = \lim_{n \to \infty} \left| \frac{(a_{n+1})^2x}{(a_n)^2 } \right|\\ = \lim_{n \to \infty} \left| \frac{(a_{n+1})^2}{(a_n)^2} \right||x|\\ = |x| \lim_{n \to \infty} \left| \frac{(a_{n+1})^2}{(a_n)^2} \right|\\ = |x| \left( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\right)^2\\ = |x| \left(\frac{1}{r}\right)^2\\ = \frac{|x|}{r^2} $$ The series will converge if this is more than 1, and diverge if it's less than 1. It will be less than 1 when $|x| < r^2$ and more than 1 when $|x| > r^2$. So the radius of convergence of the second series must be $r^2$.

DavidButlerUofA
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    I think we need the Root Test since we can have convergence with the limit of the ratio not existing. – André Nicolas Jul 01 '14 at 04:18
  • @DavidButlerUofA What allows you to go from $= |x| \lim_{n \to \infty} \left| \frac{(a_{n+1})^2}{(a_n)^2} \right|$ to $|x| \left( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\right)^2$ ? – makansij May 27 '19 at 23:35
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    @nundo Squaring can be brought out of absolute value, and also squaring is a continuous function so limits can be brought inside squaring. – DavidButlerUofA May 27 '19 at 23:46
  • Thank you, now I understand that step. But, from a larger perspective, the ratio test is only a one-way IF, right? I thought that the ratio test basically says if $\lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} < 1 $ then the series converges. But you seem to be using the IF in the other direction: if the series converges then $\lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} < 1 $? is that right? – makansij May 28 '19 at 03:18
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    The ratio test also says that if the limit is more than 1 then it definitely does NOT converge. So if it converges then that limit will not be more than 1. The radius of convergence is the furthest distance from the centre you can go before there are values of x that produce a sequence that doesn’t converge. So you don’t really need to worry about the very edge of the interval if you are only concerned with the radius. – DavidButlerUofA May 28 '19 at 03:24
  • I see so you just applied the contrapositive! Very clever thank you. – makansij May 28 '19 at 05:47
  • Okay, one more thing @DavidButlerUofA , and then I'll let you go: I understand that we used the ratio test in order to prove the claim $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{r}$. But wouldn't that have been a lot easier using Cauchy Hadamard, since in this case? – makansij May 30 '19 at 20:33
  • I have never heard of Cauchy Hadamard. – DavidButlerUofA May 30 '19 at 20:59
  • Looking back at the original question, the OP did mention Cauchy Hadamard, but I didn’t know what that was, so I answered using techniques I knew. – DavidButlerUofA May 30 '19 at 21:05