Is this sine series bounded? Let $$f(x)=\sum_{k=1}^{\infty}\sin\left(\frac{x}{2^k}\right)$$ 
Is $f(x)$ bounded? 
EDIT:
I'm asking if there is $M \in R$ (M constant), s.t. $\forall x $ $|f(x)|<M$
EDIT2:
I deleted parts that caused confusion.
 A: No, it is unbounded.
Let $f_N(x) = \sum_{k=1}^N \sin(x/2^k)$.  Then $|f(x) - f_N(x)| \le \sum_{k=N+1}^\infty |x|/2^k = |x|/2^N$.  Now on $[0, 2^N \pi]$, $\sin(x/2^k)$ for $k=1,\ldots,N$ are orthogonal, so $\int_0^{2^N \pi} f_N(x)^2 = \sum_{k=1}^N \int_0^{2^N \pi} \sin^2(x/2^k)\ dx = N 2^{N-1} \pi$.  Thus there must be some $x_N \in [0, 2^N \pi]$ with $|f_N(x_N)| > \sqrt{N/2}$, and 
$|f(x_N)| > \sqrt{N/2} - \pi$.    
A: There is no bound, because
$$ f(2^{3n}\cdot\frac{2\pi}{7}) = f(\frac{2\pi}{7}) + n\biggl(\sin(\tfrac17 2\pi) + \sin(\tfrac27 2\pi) + \sin(\tfrac47 2\pi)\biggr)$$
and the factor of $n$ on the right-hand side is nonzero (it is about 1.32).
A: Later edit:
I see you've edited your question so that the part of my answer that's in quotation marks is no longer a verbatim quote from the question, and maybe the whole intent of the question has changed.
end of later edit
The statement that $\lim\limits_{x\to0}\sin x = x$ doesn't make sense if taken literally with the usual definitions.  In the expression $\lim\limits_{x\to0}\sin x$, the variable $x$ is a bound variable, so no "$x$" should appear when the expression is evaluated.
"We can treat the series as $\sum\limits_k (x/2^k)$" can make sense if you're talking about whether it converges, but not if you're talking about what the sum is.
Convergence or divergence is not affected by changing finitely many terms.  Those early terms where $\sin(x/2^k)$ is not close to $x/2^k$ therefore don't affect convergence or divergence.  But they do affect the value of the sum.
