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$(1).$ Show that:
$$ lim_{sto0^+},left[sum_{n=1}^{infty}cosleft(pifrac{n}{m}right)frac{1}{n^s}right]=color{red}{-frac{1}{2}} quadcolonspaceforall,minmathbb{N}^{+}tag{1} $$
$(2).$ Find a closed-form for:
$$ lim_{sto0^+},left[sum_{n=1}^{infty}sinleft(pifrac{n}{m}right)frac{1}{n^s}right]=color{red}{,,,,?,,,,} quadcolonspace,,,,minmathbb{N}^{+}tag{2} $$

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.

$$f(z,s)=sum_{n=1}^{infty}frac{z^n}{n^s}$$ can be analytically continue to all complex $z$ except $zin mathbf R bigwedge zge1$ 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 $|z|=1bigwedge zne 1$, we have $forallthetainmathbf Rbigwedgethetane0$
$$lim_{srightarrow 0}f(e^{itheta},s)= lim_{zrightarrow e^{itheta}atop srightarrow 0} f(z,s)=lim_{zrightarrow e^{itheta}} f(z,s=0)=frac {e^{itheta}}{1- e^{itheta}}=-frac12+frac i2cotfrac theta 2.$$
Separating out the real and imaginary parts of the above equation, we obtain the desired results.

Denote $fracpi m=x$, rewrite $$lim_{sto0^+}left[sum_{n=1}^{infty}cosleft(pifrac{n}{m}right)frac{1}{n^s}right]=lim_{sto0^+}sum_{n=1}^inftyfrac{e^{ixn}+e^{-ixn}}{2n^s}\
=lim_{sto0^+}sum_{n=1}^inftyfrac{e^{ixn}+e^{-ixn}}{2n^s}\
=lim_{sto0^+}frac12left(operatorname{Li}_s(e^{ix})+operatorname{Li}_s(e^{-ix})right)$$where $operatorname{Li}_s(z)$ is the analytic continuation of $sum_{n=1}^inftyfrac{z^n}{n^s}$ 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 $sin(0,epsilon)$. We have
$$lim_{sto0^+}frac12left(operatorname{Li}_s(e^{ix})+operatorname{Li}_s(e^{-ix})right)=frac12left(operatorname{Li}_0(e^{ix})+operatorname{Li}_0(e^{-ix})right)=frac{1}{2} left(frac{e^{i x}}{1-e^{i x}}+frac{e^{-i x}}{1-e^{-i x}}right)=frac12$$
Similarly, $$lim_{sto0^+},left[sum_{n=1}^{infty}sinleft(pifrac{n}{m}right)frac{1}{n^s}right]=frac1{2i}left(operatorname{Li}_0(e^{ix})-operatorname{Li}_0(e^{-ix})right)=frac{1}{2} i left(frac{e^{-i x}}{1-e^{-i x}}-frac{e^{i x}}{1-e^{i x}}right)=frac12cotfrac x2$$
Evaluation of $operatorname{Li}_0(z)$
$$sum_{n=1}^inftyfrac{z^n}{n^0}=zsum_{n=0}^infty z^n=frac{z}{1-z}(|z|<1)$$
Since $operatorname{Li}_0(z)$ is the analytic continuation of it, $operatorname{Li}_0(z)=frac{z}{1-z}$.