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How could you find out the radius of convergence of $\displaystyle\sum z^{n!}$? I'm used to applying the ratio test to power series of the form $\displaystyle\sum a_{n}z^{n}$, but for a different power of $z$ I am a bit stumped? What about $\displaystyle\sum z^{2n+a}$ for another example? Where $a\in \mathbb{R}$.

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Well, $\sum z^{2n+a} = z^a \sum(z^2)^n$ so that's not too difficult. – Chris Taylor Oct 16 '11 at 14:39
up vote 10 down vote accepted


If $|z| = 1$, we know its behavior. If $|z| > 1$, then it explodes. And if $|z| < 1$, then $z^{n!} < z^n$, where I assumed $z$ was positive.

For the other one, note that you can factor out the $z^a$ to the outside of the sum, and are left just with $(z^2)^n$ on the inside.

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Recall that the radius of convergence $R$ of the series $\sum\limits_na_nz^n$ is such that for every positive real number $r>R$, the real valued sequence $(x^{(r)}_n)$ defined by $x^{(r)}_n=|a_n|r^n$ is unbounded and for every positive real number $r<R$, this same sequence $(x^{(r)}_n)$ is bounded.

This dichotomy determines $R$ uniquely but of course much more is true since, for every $r<R$, $x^{(r)}_n\to0$ exponentially fast. On the other hand, the possible behaviours of $(x^{(R)}_n)$ (that is, at the critical value $r=R$) are more diverse since one can observe anything between (non exponential) convergence to zero and (non exponential) unboundedness.

Application: Consider any complex valued sequence $(a_n)$ such that $|a_n|\in\{0,1\}$ for every $n$ and introduce the set of indices $N=\{n\mid|a_n|=1\}$. Then the radius of convergence of the series $\sum\limits_na_nz^n$ is $R=+\infty$ if $N$ is finite, and $R=1$ if $N$ is infinite.

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