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Thread: Region of convergence of an infinite series

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    MHF Contributor arbolis's Avatar
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    Region of convergence of an infinite series

    Find the region of convergence of the power series $\displaystyle \sum _{n=0}^{+\infty} a^{n^2} z^n$ where $\displaystyle a \in \mathbb{C}$.
    According to my class notes, if I have a power series of the form $\displaystyle \sum _{n=0}^{+\infty} a_n z^n$, the radius of convergence is equal to $\displaystyle R=\lim _{n \to \infty} \frac{|a_n|}{|a_{n+1}|}$. So I applied the formula and I got that $\displaystyle R= \lim _{n \to \infty} \frac{1}{|a^{2n+1}|}$.
    I wrote $\displaystyle a=re^{i \theta} \Rightarrow a^{2n+1}=r^{(2n+1)} e^{i(2n+1)\theta}$. So I think that $\displaystyle \lim _{n \to \infty} |a^{2n+1}|$ depends on $\displaystyle |a|=r$. If $\displaystyle |a|<1$ the series converges while if $\displaystyle |a|>1$ the series diverges, $\displaystyle \forall z \in \mathbb{C}$. So the radius of convergence would be 1, although it seems totally wrong to me since I think I should have had a condition on z, not only on a. For instance if $\displaystyle a=1$ I fall in the case when I have $\displaystyle \sum _{n=0}^{+\infty} z^n$ and the radius of convergence is indeed 1 but because of a condition on z.
    So I'd like to know what I'm doing wrong.
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    Quote Originally Posted by arbolis View Post
    Find the region of convergence of the power series $\displaystyle \sum _{n=0}^{+\infty} a^{n^2} z^n$ where $\displaystyle a \in \mathbb{C}$.
    According to my class notes, if I have a power series of the form $\displaystyle \sum _{n=0}^{+\infty} a_n z^n$, the radius of convergence is equal to $\displaystyle R=\lim _{n \to \infty} \frac{|a_n|}{|a_{n+1}|}$. So I applied the formula and I got that $\displaystyle R= \lim _{n \to \infty} \frac{1}{|a^{2n+1}|}$.
    I wrote $\displaystyle a=re^{i \theta} \Rightarrow a^{2n+1}=r^{(2n+1)} e^{i(2n+1)\theta}$. So I think that $\displaystyle \lim _{n \to \infty} |a^{2n+1}|$ depends on $\displaystyle |a|=r$. If $\displaystyle |a|<1$ the series converges while if $\displaystyle |a|>1$ the series diverges, $\displaystyle \forall z \in \mathbb{C}$. So the radius of convergence would be 1, although it seems totally wrong to me since I think I should have had a condition on z, not only on a. For instance if $\displaystyle a=1$ I fall in the case when I have $\displaystyle \sum _{n=0}^{+\infty} z^n$ and the radius of convergence is indeed 1 but because of a condition on z.
    So I'd like to know what I'm doing wrong.
    The formula $\displaystyle R= \lim _{n \to \infty} \frac{1}{|a^{2n+1}|}$ is correct. If |a| > 1 then this limit is 0. In other words, the radius of convergence is then 0 (which means that the series does not converge for any value of z except 0). Similarly, if |a| < 1 then the limit is infinite. In other words, the radius of convergence is then infinity (which means that the series converges for all z). Finally, if |a| = 1 then you are correct to say that the radius of convergence is 1.
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