Define the Polylogarithm functions $$\newcommand{\Li}{\operatorname{Li}} \Li_s(x)=\sum_{k=1}^\infty\frac{x^k}{k^s}\tag1 $$ For $s\gt0$, Summation by Parts and the Mean Value Theorem say $$ \begin{align} (1-x)\Li_s(x) &=\sum_{k=1}^\infty\frac{x^k-x^{k+1}}{k^s}\tag{2a}\\ &=x+\sum_{k=2}^\infty x^k\!\left(\frac1{k^s}-\frac1{(k-1)^s}\right)\tag{2b}\\ &=x-s\sum_{k=2}^\infty\frac{x^k}{(k-\theta_{k,s})^{s+1}}\tag{2c} \end{align} $$ Explanation:
$\text{(2a)}$: multiply $(1)$ by $1-x$
$\text{(2b)}$: Summation by Parts
$\text{(2c)}$: Mean Value Theorem where $0\lt\theta_{k,s}\lt1$
Thus, $\frac1{(k-\theta_{k,s})^{s+1}}$ is decreasing from something less than $1$ to $0$, and, as usual, we have the bound $\left|\,\sum\limits_{k=1}^ne^{ikx}\,\right|\le\frac1{|\sin(x/2)\,|}$. Dirichlet's Test then says that $$ \left(1-e^{ix}\right)\Li_s\left(e^{ix}\right)=e^{ix}+O\!\left(\frac s{|\sin(x/2)\,|}\right)\tag3 $$ This means that for $e^{ix}\ne1$, $$ \begin{align} \lim_{s\to0^+}\Li_s\left(e^{ix}\right) &=\frac{e^{ix}}{1-e^{ix}}\tag{4a}\\ &=-\frac{e^{ix/2}}{e^{ix/2}-e^{-ix/2}}\tag{4b}\\ &=-\frac{\cos(x/2)+i\sin(x/2)}{2i\sin(x/2)}\tag{4c}\\ &=-\frac12+\frac i2\cot(x/2)\tag{4d} \end{align} $$ Explanation:
$\text{(4a)}$: take the limit of $(3)$
$\text{(4b)}$: multiply numerator and denominator by $-e^{-ix/2}$
$\text{(4c)}$: write things in terms of $\sin$ and $\cos$
$\text{(4d)}$: simplify
Taking the imaginary parts of $\text{(4d)}$, we get $$ \bbox[5px,border:2px solid #C0A000]{\lim_{s\to0^+}\sum_{k=1}^\infty\frac{\sin(kx)}{k^s}=\frac12\cot(x/2)}\tag5 $$ The original sum, for $s=0$, still does not converge, but taking the limit of this regularization gives a value.
