2
$\begingroup$

I'm revising for an exam at the moment and I'm struggling with part of a question.

I'm asked to find the radius of convergence of the series $\sum_{n=0}^{\infty }z^{n!}$ and then find where it converges on the boundary.

By the ratio test I have R=1. Clearly for z=1 and z=-1 the series diverges but for the complex values I can't determine the behaviour at all, could somebody give me a hint? I've tried writing z in polar form but I can't seem to get anywhere with that either.

$\endgroup$
  • 5
    $\begingroup$ If the summand doesn't go to zero, can the sum converge? $\endgroup$ – Cameron Williams May 24 '15 at 17:14
2
$\begingroup$

As you say, by the ratio test the radius of convergence as a power series about $0$ is indeed $R=1$, so the series must converge in $D=\{z \in \mathbb{C} : |z|<1 \}$. Lets see what happens if $z \in \partial D=\{ z: |z|=1\}$. Consider the set $$ E=\{ e^{2\pi i \cdot r} : r \in \mathbb{Q} \} \subset \partial D $$ For every $z \in E$, we have $$ \sum_{n=0}^{\infty} z^{n!} = \sum_{n=0}^{\infty} e^{2\pi i \cdot r n!} . $$ But clearly for each $r \in \mathbb{Q}$ there exist $N=N(r)\in \mathbb{N}$ such that $r n! \in \mathbb{N}$ for all $n\geq N$, and thus $e^{2\pi i \cdot r n!}=e^0=1$ for all $n\geq N$. Then if $z \in E$, $$ \sum_{n=0}^{\infty} z^{n!} = \sum_{n=0}^N z^{n!} + \sum_{n=N+1}^\infty 1 $$ thus the series diverges for all $z \in E$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, it follows that $E= \{ e^{2\pi i \cdot r} : r \in \mathbb{Q} \}$ is dense in $\partial D= \{ e^{2\pi i \cdot t} : t \in \mathbb{R} \}$, which gives that the series diverges for all $z \in \overline{E}=\partial D$. Then the series only converges in $D$ since the points on $\partial D$ are all singularities.

$\endgroup$
  • $\begingroup$ The last step uses uniform convergence, right? $\endgroup$ – abnry May 25 '15 at 1:10
  • $\begingroup$ @nayrb I do not see the use of uniform convergence on the proof. Precisely which step are you thinking it must be used? How and why ? $\endgroup$ – Alonso Delfín May 25 '15 at 1:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.