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What would be the radius of convergence of $\sum\limits_{n=0}^{\infty} z^{3^n}$?

How would I find the radius of convergence for $\displaystyle\sum_{n=0}^\infty2^{-n}z^{n^2}$? Im not sure how to deal with the $z^{n^2}$ term.

I know the ratio test = $\displaystyle\limsup_{n\rightarrow \infty}|c_n|^\frac{1}{n}$ for $\displaystyle\sum_{n=0}^\infty c_n(z-z_0)^n$, but since I have the $z^{n^2}$ term, how would I deal with it? Is the ratio test I should use now be $\displaystyle\limsup_{n\rightarrow \infty}|c_{n^2}|^\frac{1}{n^2}$? If so, what exactly would $c_{n^2}$ be?

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    Ok, thanks for your help Didier!2012-05-24

3 Answers 3

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As you say, you cannot find the radius of convergence using the ratio test, since the quotient $c_{n+1}/c_n$ is not defined whenever $c_n=0$. However you can apply the ratio test to the series, forgetting it is a power series, and obtain a condition on $z$ that guarantees convergence. To do this, you must find $ \lim_{n\to\infty}\frac{2^{-(n+1)}|z|^{(n+1)^2}}{2^{-n}|z|^{n^2}} $ and impose the condition that it be less than $1$.

You can use the root test in the same fashion, or as you suggest. In that case $c_{n^2}$ is, by definition, the coefficient of $z^{n^2}$.

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    Okay, thanks Julian, most appreciated!2012-05-24
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You are right. First, $c_{n^2}$ would be $2^{-n}$. Hence $\lvert c_{n^2}\rvert^{1/n^2}$ is $2^{-1/n}$.

Since this goes to $1$ the convergence radius is $1$ because $\limsup\lvert c_{n^2}\rvert^{1/n^2}=1$.

By the way, for the sequence $\lvert c_{n}\rvert^{1/n}$ the limit does not exist.

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    Okay thanks, yes the main thing I was not sure was whether $c_{n^2} = 2^{-n}$ or not!2012-05-24
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You have the root test in your OP. The $\limsup$ in the definition allows you to ignore the "missing terms" (because they are lesser) in $ c_n=\left\{\matrix{ 2^{-\sqrt{n}}&&\sqrt{n}\in\mathbb{N}\\ 0&&\text{otherwise}}\right. $ and reparametrize $n$ at will, i.e. in this case, to avoid vanishing terms; hence we take the $n^2$th rather than the $n$th root: $ C \quad=\quad \limsup_{n\to\infty}\,\left(c_n\cdot|z^n|\right)^{1/n} \quad\underset{n\to n^2}=\quad |z| \cdot \lim_{n\to\infty}\left(\frac1{2^n}\right)^{1/n^2} \quad=\quad |z| $ which converges for $|z|\le r=1$ since $ \lim_{n\to\infty} 2^{-n/n^2}=1 \,. $ If you use the ratio test, you proceed with the inequality $ 1 > \lim_{n\to\infty} \frac{2^{-(n+1)}z^{(n+1)^2}}{2^{-n}z^{n^2}} = \lim_{n\to\infty} \frac{z^{2n+1}}{2} $ to still get $ r = \lim_{n\to\infty} 2^{-1/(2n+1)}=1 \,. $

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    Ahh okay, thanks!2012-05-24