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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.

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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
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
up vote 8 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 $1$.

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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.

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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

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 $1$.

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