# Absolute convergence when all the rotated series converge

The question here might be standard in some textbook. Let $a_n, n\ge1$ be a series of real numbers. It is evident that

• if $\displaystyle \sum_{n\ge 1} |a_n|<+\infty$, then $\displaystyle \sum_{n\ge1} e^{2n\pi i t}a_n$ converges for all $0\le t< 1$.

What about the converse implication? That is,

• Assume $\displaystyle \sum_{n\ge1} e^{2n\pi i t}a_n$ converges for all $0\le t< 1$. Does this imply $\displaystyle \sum_{n\ge 1} |a_n|<+\infty$?
• Very interesting question. The answer to the first question is a simple "yes": an absolutely convergent series is always convergent. The second one is more difficult. If the exponential factor were not present, the answer would be "no": the alternating harmonic series converges, but the harmonic series does not. Again go very interesting! This will probably keep me up tonight thinking about it--thanks a lot ;)
– MPW
Apr 15, 2014 at 1:03
• It might be helpful to notice that this is a power series evaluated at a point on the unit circle. The terms are $a_nz^n$ with $z=e^{2\pi i t}$...
– MPW
Apr 15, 2014 at 1:05
• Relevant question Apr 15, 2014 at 1:07
• I expect the answer is no. A counterexample might be found by taking $a_n = \frac1n e^{i\theta(n)}$, where $\theta(n)$ are randomly chosen between $0$ and $2\pi$; alternatively, where $\theta(n)=n^2$. Apr 15, 2014 at 1:32

Here is my thought according to MPW's viewpoint as a power/Fourier series.

(a). $\displaystyle \sum_{n\ge1} e^{2n\pi i t}a_n$ converges for all $0\le t< 1$.

Condition (a) implies that we have a well defined function, say $f(t)$.

(b). $\displaystyle \sum_{n\ge 1} |a_n|<+\infty$.

Condition (b) implies that the series in (a) converges to $f$ absolutely and uniformly.

According to Wikipedia, there exists some function, whose "Fourier series converges pointwise, but not uniformly; see Antoni Zygmund, Trigonometric Series, vol. 1, Chapter 8, Theorem 1.13, p. 300." This should be a theoretically counterexample.

For each $n\ge 1$, and for each $2^{n-1}\le k< 2^n$, let $\displaystyle a_k=\frac{(-1)^n}{n\cdot 2^n}$. Consider the series $\displaystyle \sum_{k\ge 1}a_k z^k$.
• This series converges (uniformly) on the closed disk $|z|\le 1$.
• $\displaystyle\sum_{k\ge 1}|a_k|=\sum_{n\ge 1}\sum_{2^{n-1}\le k< 2^n}\frac{1}{n\cdot 2^n}=\sum_{n\ge 1}\frac{1}{2n}$ diverges.