Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know how to find the radius of convergence of a power series $\sum\limits_{n=0}^{\infty} a_nz^n$, but how does this apply to the power series $\sum\limits_{n=0}^{\infty} z^{3^n}$? Would the coefficients $a_n=1$, so that one may apply D'Alembert's ratio test to determine the radius of convergence? I would appreciate any input that would be helpful.

share|improve this question
4  
Careful: you have $a_{3^n} = 1$ and $a_{k} = 0$ for $k \neq 3^n$, so the ratio test doesn't work. Try the root test. –  t.b. Oct 26 '11 at 20:41
    
@t.b.: OK, but will that make any difference? Still trying, maybe I will get it soon enough. –  Libertron Oct 26 '11 at 20:46
3  
Well, it makes a difference if you get the correct result the right way or the wrong way, but maybe that's a matter of taste :) –  t.b. Oct 26 '11 at 20:48
    
Almost a duplicate, see this. –  Did Oct 26 '11 at 20:57
add comment

3 Answers

up vote 7 down vote accepted

What do you do for the cosine and sine series? There, you cannot use the Ratio Test directly because every other coefficient is equal to $0$. Instead, we do the Ratio Test on the subsequence of even (resp. odd) terms. You can do the same here. We have $a_{3^n}=1$ for all $n$, and $a_j=0$ for $j$ not a power of $3$. Define $b_k = z^{3^k}$. Then we are trying to determine the convergence of the series $\sum b_k$, so using the Ratio Test we have: $$\lim_{n\to\infty}\frac{|b_{n+1}|}{|b_{n}|} = \lim_{n\to\infty}\frac{|z|^{3^{n+1}}}{|z|^{3^{n}}} = \lim_{n\to\infty}|z|^{3^{n+1}-3^n} = \lim_{n\to\infty}|z|^{2\times3^{n}} = \left\{\begin{array}{cc} 0 & \text{if }|z|\lt 1\\ 1 & \text{if }|z|=1\\ \infty &\text{if }|z|\gt 1 \end{array}\right.$$ So the radius of convergence is $1$.

share|improve this answer
add comment

In a nutshell, my advice would be to forget that this is a power series and that you learned something called the ratio test, and to remember that a series $\sum\limits_nx_n$ cannot converge unless $x_n\to0$. In your case, $x_n$ is a power of $z$ hence $(x_n)$ does not converge to zero if $|z|\geqslant1$. In the other direction, a simple comparison such as $|x_n|\leqslant|z|^n$ for every $|z|\leqslant1$ yields the result.

And you might wish to read this.

share|improve this answer
6  
And of course, of course... to avoid accepting answers to your posts less than 20 minutes after you asked them. –  Did Oct 26 '11 at 21:04
add comment

The root test related formula for the radius of convergence $R$ is $\frac{1}{R}=\limsup\limits_{n\to\infty}\sqrt[n]{|a_n|}$. Here $a_n$ is either $0$ or $1$ for each $n$. Since it is $1$ infinitely many times, we see that the $\limsup$ is $1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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