# Radius of convergence for $\sum\limits_{n=0}^\infty2^{-n}z^{n^2}$ [duplicate]

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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## marked as duplicate by Did, t.b., Jennifer Dylan, sdcvvc, Michael Greinecker♦Aug 29 '12 at 16:08

See this. –  Did May 24 '12 at 13:23
Ok, thanks for your help Didier! –  Derrick May 24 '12 at 13:33

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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Thanks for your help, understood it now! –  Derrick May 24 '12 at 13:34
One quick question, from $\lim_{n\to\infty} \frac{z^{2n+1}}{2}$, how did you proceed from there? I can't see what you did after that. –  Derrick May 24 '12 at 13:41
I replaced $z$ by $r=|z|$ (and took the complex absolute value). –  bgins May 24 '12 at 13:43
Ahh okay, thanks! –  Derrick May 24 '12 at 13:45

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! –  Derrick May 24 '12 at 13:35

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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Thanks Julian, didnt realise I could use the ratio test to the series and 'forget' it was a power series! –  Derrick May 24 '12 at 13:35
Working out the limit, what would I do after I reach $\lim_{n\to\infty} \frac{z^{2n+1}}{2}$? –  Derrick May 24 '12 at 13:42
First of all: the ratio test is used in series of positive terms, so you must use $|z|$ and not $z$. You find the value of the limit as a function of $|z|$. In this particular case there are three possibilities deppending on the value of $|z|$: $|z|<1$, $|z|=1$ and $|z|>1$. Th series will converge when the limit is $<1$. –  Julián Aguirre May 24 '12 at 13:49
Okay, thanks Julian, most appreciated! –  Derrick May 24 '12 at 16:22