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While doing research on the Clausen function, I came across this nice identity, $$8\operatorname{Cl}_2\left(\frac{\pi}2\right)+3\operatorname{Cl}_2\left(\frac{\pi}3\right)=12\operatorname{Cl}_2\left(\frac{\pi}6\right)$$

The two addends on the LHS are Catalan's constant and Gieseking's constants. It made me wonder if there were similar relations. Define,

$$\begin{aligned} \text{Cl}_m\left(\frac{\pi}2\right) &= \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}2\right)}{k^m}=\sum_{n=0}^\infty\left(\frac1{(4n+1)^m}-\frac1{(4n+3)^m}\right)\\ \text{Cl}_m\left(\frac{\pi}3\right) &= \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}3\right)}{k^m}=\frac{2^{m-1}+1}{2^m}\sum_{n=0}^\infty\left(\frac{\sqrt3}{(3n+1)^m}-\frac{\sqrt3}{(3n+2)^m}\right)\\ \text{Cl}_m\left(\frac{\pi}6\right) &= \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}6\right)}{k^m} \end{aligned}$$

and, $$\begin{aligned} a &= 2^{m-1}(3^{m-1}+1)\\ b &= 3^{m-1}\\ c &= 2^m\,3^{m-1} \end{aligned}$$

Q: How do we prove, with $a,b,c$ as defined above, that, $$a\operatorname{Cl}_m\left(\frac{\pi}2\right)+b\operatorname{Cl}_m\left(\frac{\pi}3\right)=c\operatorname{Cl}_m\left(\frac{\pi}6\right)$$ or equivalently, $$a\sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}2\right)}{k^m}+b\sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}3\right)}{k^m} = c\sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}6\right)}{k^m}$$ for all integer $m>1$?

Tito Piezas III
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2 Answers2

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In general, assuming convergence, we have $$ \sum_{n=1}^\infty \sin\left(\frac{\pi}{6}\,n\right)\, a_{n} = \sum_{n=1}^\infty \sin\left(\frac{\pi}{3}\,n\right)\, a_{2n} + \sum_{n=1}^\infty \sin\left(\frac{\pi}{2}\,n\right)\, \left(\frac12 a_{n}+\frac32 a_{3n} \right)$$

To see this, we partition the naturals into evens ($2n$) and odds ($6n+1,6n+3,6n+5$).

The evens give $$\sum_{n=1}^\infty \sin\left(\frac{\pi}{3}\,n\right)\, a_{2n} $$ while the odds give $$\begin{align} &\sum_{n=0}^\infty \sin\left(\frac{\pi}{6}\,(6n+1)\right)\, a_{6n+1} +\sin\left(\frac{\pi}{6}\,(6n+3)\right)\, a_{6n+3} +\sin\left(\frac{\pi}{6}\,(6n+5)\right)\, a_{6n+5} \\&=\sum_{n=0}^\infty \frac12 (-1)^n \, a_{6n+1} +(-1)^n \, a_{6n+3} +\frac12 (-1)^n \, a_{6n+5} \\&=\sum_{n=1}^\infty \sin\left(\frac{\pi}{2}\,n\right)\left(\frac12 a_{3n-2}+a_{3n}+\frac12 a_{3n+2}\right) \end{align}$$ But, at the same time $$\sum_{n=1}^\infty \sin\left(\frac{\pi}{2}\,n\right)\left(a_{3n-2}-a_{3n}+ a_{3n+2}\right) =\sum_{n=1} ^\infty \sin\left(\frac{\pi}{2}\,n\right)\, a_{n},$$

from which the claim follows.

Yours is the case $a_n = n^{-m}$.

Noam Shalev
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  • Thanks. However, if the main post is the case $a_n = n^{-m}$, how come there is no relation for the case $a_n = (2n+1)^{-m}$ hence $$p\sum_{k=1}^\infty\frac{\sin\left(k,\frac{\pi}2\right)}{(2k+1)^m}+q\sum_{k=1}^\infty\frac{\sin\left(k,\frac{\pi}3\right)}{(2k+1)^m} \overset{\color{red}?}= r\sum_{k=1}^\infty\frac{\sin\left(k,\frac{\pi}6\right)}{(2k+1)^m}$$ for rationals $p,q,r$? – Tito Piezas III Jul 27 '19 at 02:55
  • @TitoPiezasIII Because if $a_n=(2n+1)^{-m}$, we can't factor $a_n$ out of $\frac12 a_n + \frac32 a_{3n}$ as easily as we did with $a_n=n^{-m}$. The relation is still true though. – Noam Shalev Jul 27 '19 at 10:31
  • Nice formula. (+1) – user90369 Oct 01 '19 at 15:21
  • @NoamShalev-nospoon This was my last question for 2019 before I became active again in 2022. Don't remember why I didn't click on the button. – Tito Piezas III Dec 08 '23 at 16:45
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Let $~~m,n\in\mathbb{N}~~$ and $~~0<x<1~$ .

$\displaystyle \text{E}_m(x):=\text{Li}_m(e^{i2\pi x})~ , ~~ \text{Cl}_m(2\pi x)=\Im \text{E}_m(x)~ , ~~ \overline{\text{E}_m(x)}~$ is conjugated complex to $~\text{E}_m(x)$

We have:

$$ \text{E}_m(1-x) = \overline{\text{E}_m(x)}~~ , ~~~~ \sum\limits_{k=0}^{n-1}\text{E}_m \left(\frac{x+k}{n}\right) = n^{1-m}~\text{E}_m(x)$$

It follows

$\displaystyle (n,x):=\left(2,\frac{1}{6}\right) : \hspace{0.5cm} \text{E}_m \left(\frac{1}{12}\right) + \text{E}_m \left(\frac{7}{12}\right) = 2^{1-m}~\text{E}_m \left(\frac{1}{6}\right) \hspace{2.8cm}(\text{I})$

$\displaystyle (n,x):=\left(3,\frac{1}{4}\right) : \hspace{0.5cm} \text{E}_m \left(\frac{1}{12}\right) + \text{E}_m \left(\frac{5}{12} \right)+ \text{E}_m \left(\frac{3}{4}\right) = 3^{1-m}~\text{E}_m \left(\frac{1}{4}\right)$

$\hspace{4cm}$ and conjugated we get

$\displaystyle \hspace{4.2cm} \overline{\text{E}_m \left(\frac{1}{12}\right)} + \text{E}_m \left(\frac{7}{12}\right) + \text{E}_m \left(\frac{1}{4}\right) = 3^{1-m}~\overline{\text{E}_m \left(\frac{1}{4}\right)} \hspace{0.6cm}(\text{II})$

such that the difference $~\text{(I)}-\text{(II)}~$ of the equations yields the following relationship:

$$\text{E}_m \left(\frac{1}{12}\right) - \overline{\text{E}_m \left(\frac{1}{12}\right)} - \text{E}_m \left(\frac{1}{4}\right) = 2^{1-m}~\text{E}_m \left(\frac{1}{6}\right) - 3^{1-m}~\overline{\text{E}_m \left(\frac{1}{4}\right)}$$

We take the imaginary part of this equation, add $\displaystyle ~\Im\text{E}_m \left(\frac{1}{4}\right)~$, multiply with $~6^{m-1}~$ and receive the desired result:

$\displaystyle 2\Im\text{E}_m \left(\frac{1}{12}\right) - \Im\text{E}_m \left(\frac{1}{4}\right) = 2^{1-m} \Im\text{E}_m \left(\frac{1}{6}\right) + 3^{1-m} \Im\text{E}_m \left(\frac{1}{4}\right)$

$\displaystyle 2^{m-1} (3^{m-1}+1) \Im\text{E}_m \left(\frac{1}{4}\right) + 3^{m-1} \Im\text{E}_m \left(\frac{1}{6}\right) = 2^{m}3^{m-1} \Im\text{E}_m \left(\frac{1}{12}\right)$

$\displaystyle 2^{m-1} (3^{m-1}+1)\text{Cl}_m \left(\frac{\pi}{2}\right) + 3^{m-1}\text{Cl}_m \left(\frac{\pi}{3}\right) = 2^{m}3^{m-1}\text{Cl}_m \left(\frac{\pi}{6}\right)$

user90369
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