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$(1).$ 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.

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  • $\begingroup$ The limit in $(2)$ equals $\dfrac12 \cot\left(\dfrac \pi{2m}\right)$. $\endgroup$ – Tianlalu Oct 10 '18 at 0:36
  • $\begingroup$ @Tianlalu: Thanks, but how to show it (if it is the correct answer)? $\endgroup$ – Hazem Orabi Oct 10 '18 at 0:39
  • $\begingroup$ If we can prove $\lim_{s\to0^+}\operatorname{Li}_s(z)= \operatorname{Li}_0(z)$, then the question is trivial. $\endgroup$ – Kemono Chen Dec 14 '18 at 0:54
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$$f(z,s)=\sum_{n=1}^{\infty}\frac{z^n}{n^s}$$ 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.

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  • $\begingroup$ ... can be analytically continue to the whole complex plane except at $z\in \mathbf R \bigwedge z\ge1 \bigwedge s=1$ No, consider $s=2$, it is not even meromorphic in the complex plane. And the imaginary part is missing a factor $\frac12$. Just to be rigorous, the general idea of your answer is correct. $\endgroup$ – Kemono Chen Dec 14 '18 at 3:37
  • $\begingroup$ @KemonoChen: You are right. Thank you. I have made the correction. I will add even more details later. $\endgroup$ – Hans Dec 14 '18 at 9:20
  • $\begingroup$ I wish if I can split the bounty, many thanks. $\endgroup$ – Hazem Orabi Dec 18 '18 at 9:37
  • $\begingroup$ @HazemOrabi: Haha. No worries. $\endgroup$ – Hans Dec 18 '18 at 21:12
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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, $\operatorname{Li}_0(z)=\frac{z}{1-z}$.

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  • $\begingroup$ According to this development, $m$ need not be restricted to the natural numbers. $\endgroup$ – Mark Viola Dec 14 '18 at 2:08
  • $\begingroup$ It is slightly surreal to see a solution jumping to fine properties of polylogarithms to "help" an OP whose background is obviously lacking these. $\endgroup$ – Did Dec 15 '18 at 6:51

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