One of the past comp question

Suppose $\sum a_n z^n$ has a radius of convergence $R_1$ with $0< R_1 < \infty$, and $\sum b_n z^n$ has a radius of convergence $R_2$ with $0< R_2 < \infty$. Prove that $\sum \frac {a_n}{b_n} z^n$ has a radius of convergence $R_3$ satisfying $ R_3<=\frac {R_1} {R_2}$

I think the idea is to prove the series $\sum \frac {a_n}{b_n} z^n$ diverges when $|z|> \frac {R_1} {R_2} $

For that we use rational density theorem and manipulate the terms to get the desired result. I don't think this method is standard way of doing it. I was wondering if someone like to give me another mind blowing approach. Thanks in advance.

  • $\begingroup$ Do you mean $\sum a_n z^n$? $\endgroup$
    – Ron Gordon
    Dec 26 '12 at 21:22
  • $\begingroup$ @rlgordonma Thanks for pointing out. $\endgroup$
    – Deepak
    Dec 26 '12 at 21:26
  • $\begingroup$ I just edited.. $\endgroup$
    – Deepak
    Dec 26 '12 at 21:27

This is straightforward using the formula for the radius of convergence of power series, which is a standard result in analysis:

$$R^{-1} = \limsup_n \sqrt[n]{|a_n|}$$

Applying the formula in our case, we get:

$$R_3^{-1} = \limsup_n \sqrt[n]{\left|\frac{a_n}{b_n}\right|} \geq \frac{\limsup_n \sqrt[n]{|a_n|}}{\limsup_n \sqrt[n]{|b_n|}} = \frac{R_1^{-1}}{R_2^{-1}}$$

So $R_3 \leq \frac{R_1}{R_2}$.

For the inequality above we use $\limsup x_ny_n \leq (\limsup x_n)(\limsup y_n)$, which follows, for example, from sub-additivity ($\limsup (x_n + y_n) \leq \limsup x_n + \limsup y_n$) applied to the logarithms of the sequences.

  • $\begingroup$ But you want $\limsup x_n/y_n$ and wouldn't this be $ \leq (\limsup x_n)/(\liminf y_n)$? $\endgroup$ Dec 26 '12 at 23:55
  • 2
    $\begingroup$ @martycohen: take $x_n = \sqrt[n]{\frac{|a_n|}{|b_n|}}$ and $y_n = \sqrt[n]{|b_n|}$, then apply the inequality I said and divide both sides by $\limsup y_n$. $\endgroup$ Dec 27 '12 at 8:08

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