Is $f(x)=\sum_{k\in\mathbb N}\frac1k\sin\frac x{2^k}$ bounded? $$f(x)=\sum_{k\in\mathbb N}\frac1k\sin\frac x{2^k}$$Is this function bounded?


So obviously this converges because $|\frac1k\sin\frac x{2^k}|&lt;|\frac x{2^k}|$ and $\sum\frac x{2^k}$ converges by the integral test.

Now I need to show that there exists a $N$ for all $y\in\text{range}\,f$ such that $|y|&lt;N$.

So I think:

Let $a_k=\frac1k$ and $b_k=\sin\frac x{2^k}$. From the Schwarz inequality we get $|\sum a_kb_k|\leq\sqrt{\sum|a_k|^2\sum|b_k|^2}$. Since $a_k&gt;0$ for all $k$ we have $a_k=|a_k|$ and thus $|a_k|^2=\frac1{k^2}$. $\sum\frac1{k^2}=\frac{\pi^2}6$ from the Reimann-Zeta. Now all we have left to prove is that $\sum|b_k|^2$ is bounded as well.
The function is unbounded. In fact, take $x_m = \frac{2^{3m}}7 2\pi$; then the sequence $f(x_m)$ tends to infinity. To see this, note that
$$
\sin\Big( \frac n72\pi \Big) \approx \begin{cases}
0.781831, &amp;\text{if }n\equiv1\pmod 7, \\
0.974928, &amp;\text{if }n\equiv2\pmod 7, \\
-0.433884, &amp;\text{if }n\equiv4\pmod 7.
\end{cases}
$$
Therefore
\begin{align*}
f(x_m) &amp;= \sum_{j=1}^m \bigg( \frac1{3j-2}\sin\Big(\frac{2^{3m-(3j-2)}}7 2\pi\Big) + \frac1{3j-1}\sin\Big(\frac{2^{3m-(3j-1)}}7 2\pi\Big) + \frac1{3j}\sin\Big(\frac{2^{3m-3j}}7 2\pi\Big) \bigg) \\
&amp;\qquad{}+ \sum_{k=3m+1}^\infty \frac1k\sin\Big( \frac\pi{7\cdot 2^{k-3m-1}} \Big) \\
&amp;\approx \sum_{j=1}^m \bigg( \frac{0.974928}{3j-2} - \frac{0.433884}{3j-1} + \frac{0.781831}{3j} \bigg) + \sum_{k=3m+1}^\infty \frac1k\sin\Big( \frac\pi{7\cdot 2^{k-3m-1}} \Big) \\
&amp;&gt; \sum_{j=1}^m \frac{0.44}j + \sum_{k=3m+1}^\infty 0 &gt; 0.44 \ln m.
\end{align*}
Note that for $x\in[0,2\pi]$
$$
\left\lvert\,\sin\frac x{2^k}\right\rvert\le \frac{\lvert x\rvert}{2^k}\le \frac{2\pi}{2^k},
$$
and hence
$$\lvert\, f(x)\rvert\le \sum_{k\in\mathbb N}\left|\frac1k\sin\frac x{2^k}\right|\le
\sum_{k=1}^\infty \frac{2\pi}{k2^k}&lt;2\pi.
$$