Proof of the unbounded-ness of $\sum_{n\geq 1}\frac{1}{n}\sin\frac{x}{n}$ For any $x\in\mathbb{R}$, the series
$$ \sum_{n\geq 1}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) $$
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:


*

*$$(\mathcal{L} f)(s) = \sum_{n\geq 1}\frac{1}{1+n^2 s^2} = \frac{-s+\pi\coth\frac{\pi}{s}}{2s}=\sum_{m\geq 1}\frac{(-1)^{m+1}\,\zeta(2m)}{s^{2m}}$$
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;

*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?

*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?

*To compute $\lim_{x\to x_0}f(x)$ through convolutions with approximate identities seems doable but not really appealing.

 A: 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.
A: [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.
