Evaluate $\lim_{x\to0+}\sum_{n=0}^{\infty}2^{-n}\frac{\sin (2^nx)}{x}$. 
Suppose $\displaystyle f(x)=\sum_{n=0}^{\infty}2^{-n}\sin(2^nx)$, evaluate
  $$
\lim_{x\to0+}\frac{f(x)-f(0)}{x}.
$$

$f(x)$ converges uniformly on $\mathbb R$. But I don't know how to evaluate
$$
\lim_{x\to0+}\sum_{n=0}^{\infty}2^{-n}\frac{\sin (2^nx)}{x}.
$$
 A: After defining
$$g_n(x) = \frac{\sin\left(2^nx\right)}{2^nx},$$
and
$$g(x) = \sum_{n=0}^{+\infty}g_n(x),$$
observe that in the interval
$$0<x<\frac{\pi}{2^N},$$


*

*For $0\leq j\leq m\leq N-1$, $g_j(x)\geq g_m(x)$.

*$g_n(x)$ is a monotonically decreasing function.


Consider a sequence $\{a_k\}$, with
$$\{a_k\}\to 0^+.$$
For large enough $k$,
$$a_k< \frac{\pi}{2^N}$$
so that
\begin{eqnarray}
g\left(a_k\right) &=& \sum_{n=0}^{N-1}g_n\left(a_k\right) + \sum_{N}^{+\infty} g_n\left(a_k\right)\geq\\
&\geq& N\cdot g_{N-1}\left(a_k\right) - \sum_{n=N}^{+\infty}\frac1{2^n\pi}>\\
&>& N\cdot g_{N-1}\left(\frac{\pi}{2^N}\right) -\frac1{2^{N-1}\pi}=\\
&=&\frac{2N}{\pi} -\frac1{2^{N-1}\pi},
\end{eqnarray}
where we took advantage of fact 1 in the first inequality, and of fact 2. in the second inequality. From the above chain we conclude that $g\left(a_k\right)$ can be made arbitrarily large by a suitable choice of $k$, thus
$$\left\{g\left(a_k\right)\right\} \to +\infty,$$
and therefore
$$\lim_{x\to 0^+} \sum_{n=0}^{+\infty} \frac{\sin\left(2^nx\right)}{2^nx} = +\infty.$$

Edit
Prompted by Jake's comment I want to make facts 1. and 2. above more explicit.


*

*This fact is just a consequence of $$\left|\frac{\sin\left(2^m\alpha\right)}{2^m\alpha}\right| = \left|\frac{\sin\left(2^{m-1}\alpha\right)\cos\left(2^{m-1}\alpha\right)}{2^{m-1}\alpha}\right| \leq \left|\frac{\sin\left(2^{m-1}\alpha\right)}{2^{m-1}\alpha}\right|$$

*This fact can be derived by differentiation, since for $0 <\alpha<\frac{\pi}2$, $$\alpha < \tan \alpha$$ and therefore, in such interval we get
$$\frac{d\left(\frac{\sin\alpha}{\alpha}\right)}{d\alpha}=\frac{\alpha\cos\alpha-\sin\alpha}{\alpha^2}<0$$
I do hope someone can give a simpler proof of the limit in OP's question, since I could not find one, so far.
