Consider the function $$ f(x):=\sum_{n=1}^\infty \frac {\sin (x/2^n)}{2^n},\quad x\in \mathbb R. $$ Is it true that $\lim_{x\to \infty}f(x)=0$?
The series converges uniformly and absolutely on $\mathbb R$. We know that for any partial sum $S_N$ of the series defining $f$, $S_N$ is periodic, so it cannot have a limit at infinity. But the function $f$ is not periodic. Any suggestions?
On
In fact, $f(x)$ does not have a limit as $x \to \infty$: consider $x=(8k\pm 1)\pi$. Then the first two terms are $$ \frac{1}{2}\sin{(4k \pm 1/2)\pi}+\frac{1}{4}\sin{(2k \pm 1/4)\pi} = \pm \left( \frac{1}{2} + \frac{1}{4\sqrt{2}} \right), $$ while the other terms are bounded by $$ \sum_{n=3}^{\infty} 2^{-n} = \frac{1}{4}. $$ So we find $$ f((8k+1)\pi) > \frac{1}{4}(1+2^{-1/2}), \\ f((8k-1)\pi) < -\frac{1}{4}(1+2^{-1/2}), $$ so $f$ can't have a limit at $\infty$ since it oscillates over a finite interval.
On
More simply, $$ \lim_{x\to +\infty} f(x) = 0 $$ cannot be true because that would imply $$ \lim_{x\to +\infty} \left[2 f(2x)-f(x)\right] = 0 $$ while $2 f(2x)-f(x) = \sin x$.
It might be interesting to prove that the Laplace transform of $f(x)$ behaves like $-K\log(s)$ in a right neighbourhood of the origin. For sure, the origin is a point of accumulation of simple poles for $(\mathcal{L} f)(s)$.
On the one hand the first term $\frac12 \sin \frac x2$ takes values $1\over2$ for every $x=\pi+4k\pi$ where $k\in \Bbb Z$.
On the other hand, the rest of the series, $\sum_{n\geq 2} \frac{1}{2^n} \sin \frac {x}{2^n}$, lies in the interval $[-\frac12, \frac12]$ for all $x$. Therefore, the terms $n\geq2$ need a lot of coordination to compensate the term $n=1$, and this feels unlikely.
More precisely, have a look at the next term (for $x=\pi+4k\pi$): $$\frac14 \sin (\frac{\pi}{4}+k\pi)$$
Whenever $k$ is even, this term is equal to $\sqrt2\over 8$, where you were hoping for something close to $-\frac14$ as $x\to \infty$. Therefore: