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.