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

Question

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:

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

Answer 1

[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.