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

$$(1).$$ 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.

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

$$f(z,s)=\sum_{n=1}^{\infty}\frac{z^n}{n^s}$$ 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.

• ... 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. – Kemono Chen Dec 14 '18 at 3:37
• @KemonoChen: You are right. Thank you. I have made the correction. I will add even more details later. – Hans Dec 14 '18 at 9:20
• I wish if I can split the bounty, many thanks. – Hazem Orabi Dec 18 '18 at 9:37
• @HazemOrabi: Haha. No worries. – Hans Dec 18 '18 at 21:12

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, $$\operatorname{Li}_0(z)=\frac{z}{1-z}$$.

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