Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am posting the following question from Math.Stackexchange:

Let $f$ be a $1$-periodic function, i.e., $f(x+1)=f(x)$, defined on the interval $(0, 1)$ by the formula $$ f(x)=2x-1. $$ For a real number $x$ consider the series $$ \sum_{n=1}^\infty\frac{f(nx)}{n}. $$ It is easily seen that the series converges if $x$ is rational and if $f$ is assumed to vanish at all integers.

My question is: does the convergence hold for irrational $x$?

I would also be grateful for references to any results about kindred objects.

I am also interested in a description of the set of all $x$ for which the series converges, and in properties of the function determined by the series.

share|improve this question

3 Answers 3

up vote 3 down vote accepted

No, convergence does not hold for all irrational $x$. I posted a full answer to the question at math.stackexchange and I'll summarize the result here.

There are uncountably many values of $x$ for which the partial sums $\sum_{n=1}^N\frac{f(nx)}n$ are unbounded. In fact, the set of values of $x$ for which the partial sums are bounded is meagre. Consequently, the set of $x$ for which the sum diverges is nonempty and, furthermore, is uncountable in the neighbourhood of any point. We can say rather more than this though. Choosing any function $\theta\colon\mathbb{Z}^+\to\mathbb{R}^+$ such that $\theta(n)=o(\log n)$ (i.e., $\theta(n)/\log n\to0$ as $n\to\infty$) then the set of $x$ such that the partial sums are bounded by a multiple of $\theta$ is meagre. This statement is, in a sense, optimal, because it can be shown that the partial sums do grow at a rate $o(\log N)$ for every $x$.

share|improve this answer

In the paper

J. Rosenblatt, Convergence of Series of Translations, Math. Ann. 230 (1977) 245-272

it is proved a general Theorem (Theorem 2.5) that in particular proves the convergence a. e. . Also the unconditional convergence in L^2[0,1].

The result in this particular case maybe was known before. Rosenblatt points to works of Hardy Littlewood and Khinchin.

share|improve this answer
    
Many thanks. I would like to see the proof, I plan to visit our library on Monday. –  kap44 Jul 19 '12 at 12:08

This type of question can be answered (typically) by applying (first) summation by parts, and then estimating sums of fractional parts. The sums of fractional parts can be handled by the Fourier series of a sawtooth function, and then the fact that irrationals have their multiples equidistributed mod 1 (in terms of Weyl's criterion). You don't say exactly why you are interested, but these standard techniques from analytic number theory may suffice.

share|improve this answer
    
As far as I understand, you say that for any $x$ the series converges. This expression appeared in my calculations, and I would like to see what is already known. –  kap44 Jul 19 '12 at 12:07
1  
I didn't express an opinion on the answer. –  Charles Matthews Jul 19 '12 at 14:12
    
Okay. I misunderstood you because you posted this as an answer. –  kap44 Jul 19 '12 at 14:34
1  
The problem seems a bit more subtle than I originally gave it credit for. You may need more than just equidistribution. –  Charles Matthews Jul 19 '12 at 15:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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