I am posting the following question from [Math.Stackexchange](https://math.stackexchange.com/questions/172528/limit-connected-with-a-periodic-function):



> 
> 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.



Limit connected with a periodic function
No, convergence does not hold for all irrational $x$. I posted a full answer to the question at [math.stackexchange](https://math.stackexchange.com/a/173029/1321) 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](http://en.wikipedia.org/wiki/Meagre_set). 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 $x$.


