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Consider a power series $$ \sum_{n=0}^{\infty}a_nz^n $$ where $a_n$ and $z$ are complex numbers. There is radius $R$ of convergence. Let us assume that is a positive real number. It is well known that for $|z|<R$ the series converges absolutely; for $|z|>R$ it does not converge.

On the other hand, when $|z|=R$, the series can have very different behaviors. This has been discussed in many posts, e.g. Behaviour of power series on their circle of convergence , Seeking a Geometric Proof of a Generalized Alternating Series' Convergence .

I am looking for some relatively easy explicit examples of power series with funny behavior at the boundary. The only I know is $$ \sum_n\frac{1}{n}z^n $$ or small variations, e.g. replacing $z$ with $z^k$.

thanks

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    $\begingroup$ This is hard to answer without knowing what you mean by "funny". One remark: given $f(z) = \sum_n z^n/n$, which has its only singularity on the unit circle at $z=1$, you can form linear combinations $\sum_k c_k f(e^{i\theta_k}z)$ to get singuarities at any finite number of boundary points, or even an infinite number if $\sum |c_k|$ converges. You could also consider $\sum_m z^{2^m}$, which has a singularity at every $2^j$th root of unity. $\endgroup$
    – Greg Martin
    Oct 3, 2014 at 7:43
  • $\begingroup$ Funny means just that the set $S$ of singular points is not the all circle and maybe it is dense or I do not know. I would like just explicit examples. How do you show that $\sum_m z^{2^m}$ converges outside the $2^j$ roots of unity??? This would be the sort of example I am looking for. $\endgroup$
    – Giulio
    Oct 3, 2014 at 7:50
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    $\begingroup$ It looks like $\sum z^{2^k}$ does not converge at any root of unity. At a $2^j$th root it diverges to infinity. At other roots the partial sums are periodic (try $e^{2\pi i/3}.$) I think that would converge every place else. Suppose $\sum c_k$ diverges with the $c_i$ positive and decreasing to $0.$ Then $\sum z^{2^k}c_k$ would diverge to infinity at a $2^j$th root of unity but should converge everywhere else. $\endgroup$
    – Aaron Meyerowitz
    Oct 3, 2014 at 8:48
  • $\begingroup$ what I am thrilled by is how to prove convergence everywhere else. Say at a point whose phase is irrational. $\endgroup$
    – Giulio
    Oct 3, 2014 at 9:00
  • $\begingroup$ I wonder if a method as in the post "seeking..." I linked in the question could work. $\endgroup$
    – Giulio
    Oct 3, 2014 at 9:04

2 Answers 2

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The series $$f(z) = \sum_{n=1}^\infty \dfrac{z^{2^n}}{n}$$ converges almost everywhere on the unit circle by Carleson's theorem (it is the Fourier series of an $L^2$ function). However, it diverges on a dense set, including all the $2^k$'th roots of unity: in fact at each of those points the real parts of the partial sums $S_k = \sum_{n=1}^k z^{2^n}/n$ are unbounded above.

Now for each positive integer $N$, the set $U_N$ of points $z$ such that $\text{Re}(S_k) > N$ for some $k$ is open and dense in the unit circle. The intersection $G$ of the $U_N$ is a dense $G_\delta$ by the Baire category theorem, and the series diverges at every point of $G$.

Putting these facts together, we find that the set of points of the unit circle where the series diverges is negligible in the sense of Lebesgue measure but the set where it converges is negligible in the sense of Baire category. I'd call that funny...

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  • $\begingroup$ What are you talking about? It doesn't diverge at every non-root of $2^j$, and neither does $\sum_n \frac{z^n}{n}$ (which is also an $L^2$ function). $\endgroup$
    – Robert Israel
    Oct 5, 2014 at 6:28
  • $\begingroup$ Sorry, it is taking me a while to understand. Let me try to rephrase your argument to see if I get it. $\endgroup$
    – Giulio
    Oct 5, 2014 at 15:07
  • $\begingroup$ Convergence. The sequence of S_k is a Fourier series and it is convergent sequence for the L2 norm on the circle. Because of Carleson's theorem it converges point-wise almost everywhere for the Lebesgue measure. $\endgroup$
    – Giulio
    Oct 5, 2014 at 17:14
  • $\begingroup$ Divergence. The series diverges at every $2^j$ root of unity. Let $U_N$ be the subset of the circle where $|S_k|>N$ for some k (or also $Re(S_k)$). Since $S_k$ are continuos functions the $U_N$ are open. Each $U_N$ contains the $2^j$ roots of unity, so they are dense. Call G the intersection of all $U_N$. The series can not converge on G because it is not Cauchy. By Baire's theorem G is dense in the circle; moreover it is a $G_{\delta}$, so, for instance, it is not countable. $\endgroup$
    – Giulio
    Oct 5, 2014 at 18:17
  • $\begingroup$ Does this make sense?? $\endgroup$
    – Giulio
    Oct 5, 2014 at 18:17
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Here is a funny result. $\newcommand{\ve}{\varepsilon}$ Consider the random series

$$ \sum_{n\geq 1}\frac{\ve_n}{n} z^n, $$

where the signs $\ve_n=\pm 1$ are chosen randomly and independently with ${\rm Prob}\;(\ve_n=\pm 1)=\frac{1}{2}$, $\forall n$.

Then, almost surely, the above series cannot be continued across the circle of convergence $|z|=1$; see Chapter 4 of Kahane's book on random series of functions.

On the other hand, Kolmogorov's three-series theorem implies that for any fixed $z$ with $|z|=1$ the above series converges for almost any choice of random signs $\ve_n$.

Put these two facts together to deduce that, with probability $1$, the above series cannot be extended across $|z|=1$, yet it converges for any $z$ of the form $z=\exp( 2\pi \sqrt{-1} r)$, $r\in \mathbb{Q}$.

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