3
$\begingroup$

Problem:

I need to show that the power series $\sum_{n=1}^{\infty}z^{n!}$ diverges for infinitely many $z$ with $|z|=1$. I tried to prove it by contradiction by assuming that diverges for finitely many $z$'s, but I wasn't successful. Anyone can prove it?

$\endgroup$
  • 4
    $\begingroup$ It diverges for all $z$ with $\lvert z\rvert = 1$. The terms of the sum don't converge to $0$. $\endgroup$ – Daniel Fischer Feb 12 '14 at 22:24
  • 1
    $\begingroup$ You must mean diverges to $\infty$, since it does not converge anywhere on the unit circle. $\endgroup$ – André Nicolas Feb 12 '14 at 22:32
  • $\begingroup$ @AndréNicolas "You must mean diverges to $\infty$". Or: Does not converge to $\infty$. $\endgroup$ – Andrés E. Caicedo Feb 12 '14 at 23:00
2
$\begingroup$

I'd like to expand on @DanielFischer's comment because I think the following fact is often overlooked by students:

Fact. Let $\{\mathbf{x}_k\}$ be a sequence in $\mathbb R^n$. Then $\displaystyle\lim_{k\rightarrow\infty}\mathbf{x}_k=\mathbf0$ if and only if $\displaystyle\lim_{k\rightarrow\infty}\left|\mathbf{x}_k\right|=0$.

We can apply this fact to the question by using the so called $n$th term test for divergence. That is, let $z\in\mathbb C$ with $|z|=1$. Then $$ \lim_{n\rightarrow\infty}\left|z^{n!}\right|=\lim_{n\rightarrow\infty}|z|^{n!}=\lim_{n\rightarrow\infty}1^{n!}=\lim_{n\rightarrow\infty}1=1\neq0 $$ Hence $\displaystyle\sum_{n=1}^\infty z^{n!}$ diverges for all $z\in\mathbb C$ with $|z|=1$.

$\endgroup$
  • 2
    $\begingroup$ Note that this argument works for any exponent, $z^{f(n)}$ instead of $z^{n!}$. $\endgroup$ – Jesse Madnick Feb 12 '14 at 23:09
3
$\begingroup$

Take $z=e^{2 \pi i q}$ for $q$ rational.

$\endgroup$

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.