Expansion and convergence of $\sum_{m=1}^{\infty}\frac{\sin(2\pi n x^{1/m})}{ m}$ Consider the series:
$$\sum_{m=1}^{\infty}\frac{\sin(2\pi n x^{1/m})}{ m}\;\;\;\;n\in\mathbb{N}$$
Other than formal manipulation of the Taylor series of the $\sin$ function, is there a way to expand it in terms of any functions besides the roots of $x$ ? A truncated version looks like this. It seems to me that a Fourier-Neumann expansion could be obtained, but i wasn't able to prove that. 
The series converges -at least heuristically- for all positive-real values of $x$. How can we prove that rigorously ? 
 A: Note that
$$
x^{1/m}=1+\frac{\ln x}{m}+\mathcal O\left(\frac{1}{m^2}\right).
$$
Thus by Mean Value Theorem, there is a $\xi$, such that
$$
\sin \left(2\pi nx^{1/m}\right)-\sin(2\pi n)=\cos (\xi)\left(\frac{\ln x}{m}+\mathcal O\left(\frac{1}{m^2}\right)\right),
$$
amd hence
$$
\left|\sin \left(2\pi nx^{1/m}\right)\right|\le \frac{c}{m}, 
$$
for some $c>0$. This makes the series absolutely convergent, using for example the comparison test.
A: We have $\sin(2\pi n x^{1/m}) = \sin(2 \pi n \{ x^{1/m} \})$, where $\{x\}$ takes the fractional part of $x$. Since $x^{1/m}$ approaches $1$ monotonically decreasing if $x>1$, then eventually $\{x^{1/m}\}=x^{1/m}-1$ and $\{x^{1/m}\} \to 0$ as $m \to \infty$. If $0<x<1$ then $\{x^{1/m}\} = x^{1/m}$.
In the first case we get $\sin(2\pi n x^{1/m}) = (2\pi n) \; O(x^{1/m}-1)$. But $x^{1/m}-1= \frac{1}{m} \ln(x) + \cdots = O(1/m)$. 
Thus, for $x>1$, because $\sin(2\pi n x^{1/m}) = O(1/m)$, the series does converge.
When $0<x<1$, we note the symmetry $\sin x = - \sin x$ and its periodicity. Hence, we can write $\sin(2 \pi n x^{1/m}) = \sin(2 \pi n \{1-x^{1/m}\})$ do the same thing. Again, for $0<x<1$, the series does converge.
(For $x=1$ or $x=0$ the convergence is obvious and $x<0$ will not work.)

EDIT:
We showed in the process that for appropriately small $x$ that $\sin(2\pi n x^{1/m}) = (2\pi n) (x^{1/m}-1) + \cdots = 2\pi n \frac{1}{m} \ln(x) +\cdots.$
So we get that the sum is
$$S(x) = \frac{n\pi^3}{3} \ln(x) +O(ln(x)^2)$$
