$\lim\limits_{s\to0^+}\sum_n\frac{\cos\left(\pi\frac{n}{m}\right)}{n^s}$ & $\lim\limits_{s\to0^+}\sum_n\frac{\sin\left(\pi\frac{n}{m}\right)}{n^s}$

Question

$(1).$ Show that: $$ \lim_{s\to0^+}\,\left[\sum_{n=1}^{\infty}\cos\left(\pi\frac{n}{m}\right)\frac{1}{n^s}\right]=\color{red}{-\frac{1}{2}} \quad\colon\space\forall\,m\in\mathbb{N}^{+}\tag{1} $$ $(2).$ Find a closed-form for: $$ \lim_{s\to0^+}\,\left[\sum_{n=1}^{\infty}\sin\left(\pi\frac{n}{m}\right)\frac{1}{n^s}\right]=\color{red}{\,\,\,\,?\,\,\,\,} \quad\colon\space\,\,\,\,m\in\mathbb{N}^{+}\tag{2} $$

Both series converge by Dirichlet's test for $\mathrm{Re}(s)>0$.
I could not find a good reason way the first series shall converge to the same constant!!
Thanks for you help.

Answer 1

Denote $\frac\pi m=x$, rewrite $$\lim_{s\to0^+}\left[\sum_{n=1}^{\infty}\cos\left(\pi\frac{n}{m}\right)\frac{1}{n^s}\right]=\lim_{s\to0^+}\sum_{n=1}^\infty\frac{e^{ixn}+e^{-ixn}}{2n^s}\\ =\lim_{s\to0^+}\sum_{n=1}^\infty\frac{e^{ixn}+e^{-ixn}}{2n^s}\\ =\lim_{s\to0^+}\frac12\left(\operatorname{Li}_s(e^{ix})+\operatorname{Li}_s(e^{-ix})\right)$$where $\operatorname{Li}_s(z)$ is the analytic continuation of $\sum_{n=1}^\infty\frac{z^n}{n^s}$ and is named polylogarithm function.
Notice that $\operatorname{Li}_s(z)$ is holomorphic in $s$ in a neighborhood of the origin whatever $z$ is, hence continuous in $s\in(0,\epsilon)$. We have $$\lim_{s\to0^+}\frac12\left(\operatorname{Li}_s(e^{ix})+\operatorname{Li}_s(e^{-ix})\right)=\frac12\left(\operatorname{Li}_0(e^{ix})+\operatorname{Li}_0(e^{-ix})\right)=\frac{1}{2} \left(\frac{e^{i x}}{1-e^{i x}}+\frac{e^{-i x}}{1-e^{-i x}}\right)=\frac12$$ Similarly, $$\lim_{s\to0^+}\,\left[\sum_{n=1}^{\infty}\sin\left(\pi\frac{n}{m}\right)\frac{1}{n^s}\right]=\frac1{2i}\left(\operatorname{Li}_0(e^{ix})-\operatorname{Li}_0(e^{-ix})\right)=\frac{1}{2} i \left(\frac{e^{-i x}}{1-e^{-i x}}-\frac{e^{i x}}{1-e^{i x}}\right)=\frac12\cot\frac x2$$ Evaluation of $\operatorname{Li}_0(z)$
$$\sum_{n=1}^\infty\frac{z^n}{n^0}=z\sum_{n=0}^\infty z^n=\frac{z}{1-z}(|z|<1)$$ Since $\operatorname{Li}_0(z)$ is the analytic continuation of it, $\operatorname{Li}_0(z)=\frac{z}{1-z}$.

Answer 2

$$f(z,s)=\sum_{n=1}^{\infty}\frac{z^n}{n^s}$$ can be analytically continue to all complex $z$ except $z\in \mathbf R \bigwedge z\ge1$ and for complex $s$ in the disk $|s|<1$. So we need only to evaluate $f(z,s=0)$ for $|z|<1$. That is $$f(z,s=0)=\sum_{n=1}^{\infty}z^n=\frac z{1-z},\, \forall |z|<1.$$ Analytically continue this fraction to $|z|=1\bigwedge z\ne 1$, we have $\forall\theta\in\mathbf R\bigwedge\theta\ne0$ $$\lim_{s\rightarrow 0}f(e^{i\theta},s)= \lim_{z\rightarrow e^{i\theta}\atop s\rightarrow 0} f(z,s)=\lim_{z\rightarrow e^{i\theta}} f(z,s=0)=\frac {e^{i\theta}}{1- e^{i\theta}}=-\frac12+\frac i2\cot\frac \theta 2.$$ Separating out the real and imaginary parts of the above equation, we obtain the desired results.