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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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The more interesting question is what might happen as you move out, within the unit disk, towards a point on the boundary, and evaluate the function on such points. – Lubin May 26 '12 at 0:25
Thanks Lubin, I'm currently thinking over this question, hope to make some progress soon. – Derrick May 26 '12 at 0:46
This is a lacunary function: – deoxygerbe May 26 '12 at 5:09

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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Actually convergence is impossible on the boundary, for the reason Gerry states (at least for this particular series - I understand nothing so strong can be said in the general case scenario). Perhaps there are points where the partial sums are bounded, though. (If there was a monotonic decreasing coefficient present like $n^{-1/2}$ or whatever, I would point out roots of unity as an infinite number of cases where the series diverges.) – anon May 26 '12 at 0:35
But didn't he just ask for the meaning of the question? AS I understood it, that as the most important part of his question – Egbert May 26 '12 at 0:42
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. – anon May 26 '12 at 0:49

As the comment pointed out this is a lacunary series,there is corresponding criterion for such seriesOstrowski-Hadamard gap theroem

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