I have to make a study on this series $\sum_{i=o}^\infty z^{i!}$. Whatever special I could find about it I should write it down. So I started with convergence. Almost every gap series I met, I can relate it to some function such as $sin , cos ,exp, log,$ etc and find where it converges. But on this one I have an obstacle to do that. The obstacle is $i!$ cause it is not ascending exponentialy with a steady step and i can't find a series-function to relate it to.What I understand so far is that something happens when $|z|=1$ so i tries relating the series to $f(x)=e^z$. Any help, hint or a more general analysis?

  • $\begingroup$ Not sure if you can say anything general, but considering $z$ to be a complex number, e.g., $z=|z|e^{jx}$ then for $|z|=1$ the same can be written as $\sum_{i=0}^\infty\cos(i!x)+j\sum_{i=0}^\infty\sin(i!x)$. But that doesn't help much anyway. $\endgroup$ – pisoir Sep 20 '17 at 21:01
  • $\begingroup$ This function is well-known to have a natural boundary on the circle $|z|=1$. I'm not sure what else you can say about it. A "gap" (lacunary) series one can say rather more about is $\sum z^{n^2}$. $\endgroup$ – Lord Shark the Unknown Sep 20 '17 at 21:14
  • $\begingroup$ yeah but the natural boundary can be caused either for the inability of the limit to exist or for the limit to be infinite in certain points of a disc. And also from which points the inability of the series to expand to this disc comes from $\endgroup$ – Pookaros Sep 20 '17 at 21:28
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    $\begingroup$ Suggestion: Don't use $i$ for an index in the context of complex analysis. $\endgroup$ – zhw. Sep 20 '17 at 22:42

Ok I found out what is going on. $f(z)=\sum_{i=0}^\infty z^{i!}$ $f(z)$ holomorphic when $|z|<1$ since $R=1/\lim\sup|z|^{\frac{n!}{n!}}=1$. Then use $\sum_{i=0}^\infty r^{n!}e^{i\frac{2πp}{q}n!}$, we have to observe that when $n\geq q$ (cause $n!/q \in Z$) then $e^{i\frac{2πp}{q}n!}=1$ so the series will become $\sum_{i=0}^\infty r^{n!}e^{i\frac{2πp}{q}n!}=\sum_{i=0}^{q-1} r^{n!}e^{i\frac{2πp}{q}n!} + \sum_{i=q}^{\infty} r^{n!}e^{i\frac{2πp}{q}n!}=\sum_{i=0}^{q-1} r^{n!}e^{i\frac{2πp}{q}n!} + \sum_{i=q}^{\infty} r^{n!} $ and $\sum_{i=q}^{\infty} r^{n!} \rightarrow \infty$ when $r\rightarrow 1$. I may be wrong somewhere till here but the idea is on circle $|z| =1$ the series will have anomalies in every multiple of $2πi$


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