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I have calculated using the root test that the radius of convergence of $\displaystyle\sum_{n=0}^\infty z^{n!}$ is $1$.

But how would I show that there is an infinite number of $z \in \mathbb{C}$ with $|z|=1$ for which the series diverge? I don't really understand what is meant in the above question, could someone explain to me what the question means?

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    This is a lacunary function: http://en.wikipedia.org/wiki/Lacunary_function2012-05-26

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Terms not going to zero implies series not converging.

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You have already found out that the radius of convergence is $1$. This means that for $z$ with $|z|<1$, the series converges and for $|z|>1$, the series diverges. On the radius of convergence itself, i.e. for $|z|=1$, pretty much anything can happen.

Note that the set $\{z\in \mathbb{C}:|z|=1\}$ is a circle around $(0,0)$ with radius one. You have to prove that the series diverges for infinitely many points on the circle. One example of such a point is $z=1$, but apparently there are infinitely many more. One possible way (but not necessarily the correct way) is to show that there are only finitely, or countably, many points on the circle for which the series does converge.

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    It wasn't specified in the first few sentences that the general case was being discussed, rather than this particular case, so I wanted to clear up ambiguity just in case it were to be an issue.2012-05-26
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As the comment pointed out this is a lacunary series,there is corresponding criterion for such seriesOstrowski-Hadamard gap theroem