Proof of the unbounded-ness of $\sum_{n\geq 1}\frac{1}{n}\sin\frac{x}{n}$

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

$$ \sum_{n\geq 1}\frac{1}{n}\,\sin\frac{x}{n}=\Omega\left(\sqrt{\log\log x}\right)\text{ as }x\to +\infty.\tag{FR}$$

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 $$ Q^*(x) = \sum_{n=1}^{K}\frac{1}{n}\,\sin\frac{x}{n}$$ 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 $$ Q^*(x_j) = \sum_{n\mid K}\frac{1}{n}+\sum_{\substack{1\leq n \leq K\\ n\nmid K}}\frac{1}{n}\,\sin\left(\frac{x_j}{n}\right)=\lambda(k)+R(x_j)$$ and $$ \frac{1}{K}\sum_{j=1}^{K}Q^*(x_j) = \lambda(k)+\frac{1}{K}\sum_{\substack{1\leq n \leq K\\ n\nmid K}}\frac{1}{n}\sum_{j=1}^{K}\sin\frac{(4j+1)x}{n}.$$ 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 $$ \lambda(k)\geq \sum_{q\leq 4k+1}\frac{1}{q}.$$ 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

$$ \sum_{q\leq 4k+1}\frac{1}{q}=\sum_{n\leq 4k+1}\frac{N(n)-N(n-1)}{n}\geq \sum_{n\leq 4k+1}\frac{N(n)}{n(n+1)} $$ finishes the proof.

Similarly it can be proved that

$$ \sum_{n\geq 1}\frac{1}{n}\,\cos\frac{x}{n}=\Omega\left(\log\log x\right)\text{ as }x\to +\infty.\tag{CR}$$

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.

[Edit: according to https://math.stackexchange.com/q/182491 Hardy and Littlewood originally showed that this function was unbounded]

A probabilistic argument on the head $\sum_{n\leq N}\tfrac1n\sin(x/n)$ shows that it must take a $O(1)$ value in any interval of length $\Omega(N).$ The tail $\sum_{n>N}\tfrac1n\sin(x/n)$ is $O(1/N)$-Lipschitz so varies by $O(1)$ over this interval. So if the head is unbounded above but $f$ is bounded above, then $f$ is unbounded below.

This doesn't rule out the possibility that $f$ is bounded above or below as $x\to+\infty,$ though it must be unbounded above and below on $\mathbb R$ by oddness.

In more detail: define

$$f_{\leq N}(x)=\sum_{n\leq N}\frac1n\sin\left(\frac xn\right),$$ $$f_{> N}(x)=\sum_{n> N}\frac1n\sin\left(\frac xn\right).$$

For any $x,$ $$\frac1{2N}\int_{x-N}^{x+N}f_{\leq N}(t)dt=\frac1{2N}\sum_{n\leq N}\left[-\cos\left(\frac tn\right)\right]_{x-N}^{x+N}\leq 1.$$ So there is some $t\in[x-N,x+N]$ with $\sum_{n\leq N}\frac1n\sin\left(\frac tn\right)\leq 1.$ To bound the Lipschitz constant of the tail we compute $$\left|\frac{d}{dx}f_{>N}(x)\right|=\left|\sum_{n>N}\frac1{n^2}\cos\left(\frac xn\right)\right|\leq 1/N.$$

Assume $x$ satisfies $f_{\geq N}(x)\geq C\log N$ and $f(x)\leq\tfrac 12C\log N.$ Then $f_{>N}(x)\leq -\tfrac 12C\log N.$ We showed there is a $t\in[x-N,x+N]$ with $f_{\leq N}(t)\leq 1,$ and the Lipschitz bound gives $f_{>N}(t)\leq 1-\tfrac 12C\log N.$ So $f(t)\leq 2-\tfrac 12C\log N$ which is unbounded below.