# Proof of the unbounded-ness of ∑n≥11nsinxn\sum_{n\geq 1}\frac{1}{n}\sin\frac{x}{n}

For any $x\in\mathbb{R}$, the series

is trivially absolutely convergent. It defines a function $f(x)$ and I would like to show that $f(x)$ is unbounded over $\mathbb{R}$. Here there are my thoughts/attempts:

1. is a function with no secrets. It behaves like $\frac{\pi}{2s}$ in a right neighbourhood of the origin, like $\frac{\pi^2}{6s^2}$ in a left neighbourhood of $+\infty$. The origin is an essential singularity and there are simple poles at each $s$ of the form $\pm\frac{i}{m}$ with $m\in\mathbb{N}^+$. These facts do not seem to rule out the possibility that $f$ is bounded;
2. For any $N\in\mathbb{N}^+$ there clearly is some $x\ll e^N$ such that $\sin(x),\sin\left(\frac{x}{2}\right),\ldots,\sin\left(\frac{x}{N}\right)$ are all positive and large enough, making a partial sum of $\sum_{n\geq 1}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right)$ pretty close to $C\log N$. On the other hand I do not see an effective way for controlling $\sum_{n>N}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right)$ – maybe by summation by parts, by exploiting the bounded-ness of the sine integral function?
3. Some probabilistic argument might be effective. For any $n\geq 3$ we may define $E_n$ as the set of $x\in\mathbb{R}^+$ such that $\sin\left(\frac{x}{n}\right)\geq \frac{1}{\log n}$. The density of any $E_n$ in $\mathbb{R}^+$ is close to $\frac{1}{2}$, so by a Borel-Cantellish argument it looks reasonable that the set of points such that $|f(x)|\geq \frac{\log x}{100}$ is unbounded, but how to make it really rigorous?
4. To compute $\lim_{x\to x_0}f(x)$ through convolutions with approximate identities seems doable but not really appealing.

Thanks to MartinR and HJol for referencing a classical result of Hardy and Littlewood, which I am going to outline. The final result is

Proof:

Let $q$ be a number of the form $\prod p^\alpha$, where each prime $p$ is of the form $4n+1$. Since $\mathbb{Z}[i]$ is a UFD, the set of $q$s is the set of odd numbers which can be represented as a sum of two coprime squares. Let $K=\prod_{q\leq 4k+1}q$, $x=\frac{\pi}{2}K$ and $x_j=(4j+1)x$ for any $j\in[1,K]$. By introducing

it is enough to approximate $Q^*$ evaluated at each $x_j$. If $n\mid K$ then $\frac{K}{n}$ is a number of the form $4n+1$ and $\sin\left(\frac{x_j}{n}\right)=1$, hence

and

In the inner sum on the right, $\frac{2x}{n}$ differs from $\pi$ by at least $\frac{C}{n}$, such that $\left|\sum_{j=1}^{K}\sin\frac{(4j+1)x}{n}\right|\ll\frac{1}{\left|\sin\frac{2x}{n}\right|}=O(n)$ and the LHS of the above equation is $\geq \lambda(k)(1-\varepsilon)$. On the other hand it is trivial that

Let $N(n)$ be the number of $q$ which do not exceed $n$. By sieve methods it is well-known that $N(n)\sim\frac{Dn}{\sqrt{\log n}}$, hence the inequality

finishes the proof.

Similarly it can be proved that

Alternative approaches: to replace $\sum_{n=m}^{M}\frac{1}{n}\sin\frac{x}{n}$ with a similarly shaped integral and to carefully estimate the error terms. A classical tool for performing such manipulations is the Denjoy-Koksma inequality. It can be combined with Van Der Corput’s trick and known inequalities in order to produce interesting generalizations, as shown by Flett and Codecà.