The series $\sum 3^n\sin\big( \frac 1{4^n} x\big)$ converges absolutely and uniformly on $(a,\infty)$ where $a\gt 0$.
The utmost I can think of is to prove the uniform convergence on bounded intervals $(a,b]$ by using $|\sin x|\le |x|$ and the $M_n$ test for series.
I don't know how to pass on to unbounded interval. Also it's easy to see the (limit)sum function $s(x)$ satisfy $|s(x)|\le 3x$ though I couldn't get $s(x)$ explicitly.
Please help me with this problem. Thanks for your suggestion and time.
Absolute convergence isn't a problem to prove. Uniform convergence does not hold for an unbounded interval. If it would, there would be a $K\ge 1$ such that
$$\lvert s(x)-\sum_{n=1}^k 3^n\sin\left(\frac{x}{4^n}\right)\rvert < 1, \forall x, \forall k \ge K.$$
Select any $m\ge K$. Because the above inequality holds for both $k=m$ and $k=m+1$, we have
$$\lvert s(x)-\sum_{n=1}^m 3^n\sin\left(\frac{x}{4^n}\right)\rvert < 1, \forall x$$
and
$$\lvert s(x)-\sum_{n=1}^{m+1} 3^n\sin\left(\frac{x}{4^n}\right)\rvert = \lvert s(x)-\sum_{n=1}^{m} 3^n\sin\left(\frac{x}{4^n}\right) - 3^{m+1}\sin\left(\frac{x}{4^{m+1}}\right)\rvert < 1, \forall x.$$
If we abbreviate $s(x)-\sum_{n=1}^m 3^n\sin\left(\frac{x}{4^n}\right)$ as $d_m(x)$, this means
$$|d_m(x)| < 1\text{ and } |d_m(x) - 3^{m+1}\sin\left(\frac{x}{4^{m+1}}\right)| < 1,$$
so by the triangle inequality we get $$ \lvert 3^{m+1}\sin\left(\frac{x}{4^{m+1}}\right)\rvert < 2, \forall x, \forall m \ge K.$$
That's not true for $m=K, x=4^{m+1}\frac{\pi}2$, for example.
What makes me a bit suspicious is that there may be some mistake from transcribing the orignal problem to here is the part of $a$ in the problem statement. A statement like "uniformly convergent on $(a,\infty)$ for all $a>0$" usually signifies that $x=0$ is a problem case for the summation, usually a singularity. That's not the case here, the summation is totally well behaving around $x=0$.
Was the original problem talking about
$$\sum_{n=1}^k 3^n\sin\left(\frac{1}{4^nx}\right)$$
or something similar?