The function $$f(x)= \sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$$ is odd, uniformly convergent on all intervals $[\epsilon,\pi]$ for $0 < \epsilon < \pi$. Hence $f$ is continuous on $(0,\pi]$. Moreover $f(0)=0$.
Is there an asymptotic expansion for $f$ at the origin?
Your function $f$ may be rewritten as a Clausen function or the imaginary part of a polylogarithm : $$\tag{1}f(x)=\operatorname{S}_{1/2}(x)=\Im{\,\operatorname{Li}_{\,1/2}(e^{ix})}$$ (since the polygarithm is defined as $\;\displaystyle\operatorname{Li}_{s}(z)=\sum_{n=1}^\infty \frac {z^n}{n^s}\;$)
We may then use the DLMF expansion in $\,\log(z)=ix$ applied at $s=\frac 12$ : \begin{align} \operatorname{Li}_{s}(e^{ix})&=\Gamma(1-s)(-ix)^{s-1}+\sum_{n=0}^\infty\zeta(s-n)\frac{(ix)^n}{n!}\\ \tag{2}\operatorname{Li}_{1/2}(e^{ix})&=\sqrt{\pi}(-ix)^{-1/2}+\sum_{n=0}^\infty\zeta\left(\frac 12-n\right)\frac{(ix)^n}{n!}\\ \end{align} to get the equivalence (as obtained by Winther) by taking the imaginary part as $x\to 0$ : $$\tag{3}f(x)\sim\sqrt{\frac{\pi}{2\,x}},\quad x\to 0$$ and the expansion (keeping only the terms $n=2k+1$) : $$\tag{4}f(x)=\sqrt{\frac{\pi}{2\,x}}+\sum_{k=0}^\infty\zeta\left(-\frac 12-2k\right)\frac{(-1)^k\,x^{2k+1}}{(2k+1)!}$$
