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A trigonometric approach to infinite series (involving Zeta and Dirichlet Beta functions)

DreamWeaver

Well-known member
Sep 16, 2013
337
In certain forms - including the logarithmic - a number of the trigonometric and hyperbolic functions can be used to sum series having Riemann Zeta and Dirichlet Beta functions (in the general series term). In this tutorial, we explore some of these connections, and present a variety of Zeta and Beta series.


---------------
Preliminaries:
---------------



The following series expansions for the trigonometric functions will be used throughout, where\(\displaystyle B_k\) and \(\displaystyle E_k\) are the Bernoulli respectively Euler numbers.


\(\displaystyle (1.0)\quad \log(\sin x) = \log x+ \sum_{k=1}^{\infty} (-1)^k \frac{2^{2k-1}B_{2k}}{k(2k)!}x^{2k}\)

Condition: \(\displaystyle 0 < x < \pi\)



\(\displaystyle (1.1)\quad \log(\cos x) = \sum_{k=1}^{\infty} (-1)^k \frac{ 2^{2k-1}(2^{2k}-1) B_{2k}}{k(2k)!}x^{2k}\)

Condition: \(\displaystyle -\pi/2 < x < \pi/2\)



\(\displaystyle (1.2)\quad \log(\tan x) = \log x+ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{2^{2k} (2^{2k}-1) B_{2k} }{k(2k)!}x^{2k}\)

Condition: \(\displaystyle 0 < x < \pi/2\)



\(\displaystyle (1.3)\quad \tan x = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{ 2^{2k}(2^{2k}-1) B_{2k} }{(2k)!} x^{2k-1}\)

Condition: \(\displaystyle -\pi/2 < x < \pi/2\)



\(\displaystyle (1.4)\quad \cot x = \frac{1}{x} + \sum_{k=1}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(2k)!} x^{2k-1}\)

Condition: \(\displaystyle -\pi < x < \pi\)



\(\displaystyle (1.5)\quad \csc x = \frac{1}{x} + 2\, \sum_{k=1}^{\infty}(-1)^{k+1}\frac{(2^{2k-1}-1) B_{2k}}{(2k)!} x^{2k-1}\)

Condition: \(\displaystyle -\pi < x < \pi\)



\(\displaystyle (1.6)\quad \sec x = 1+ \sum_{k=1}^{\infty} \frac{E_{2k}}{(2k)!} x^{2k}\)

Condition: \(\displaystyle -\pi/2 < x < \pi/2\)



The Riemann Zeta function \(\displaystyle \zeta(x)\) and Dirichlet Beta function \(\displaystyle \beta(x)\) are defined in the usual way:


\(\displaystyle (1.7)\quad \zeta(x) = 1+\frac{1}{2^x} +\frac{1}{3^x} +\frac{1}{4^x} + \cdots = \sum_{k=1}^{\infty}\frac{1}{k^x}\)


\(\displaystyle (1.8)\quad \beta(x) = 1-\frac{1}{3^x} +\frac{1}{5^x} -\frac{1}{7^x} + \cdots = \sum_{k=10}^{\infty}\frac{(-1)^k}{(2k+1)^x}\)


For Zeta functions of even index, \(\displaystyle \zeta(2k)\), and Beta functions of odd index, \(\displaystyle \beta(2k+1)\), have the following closed forms:


\(\displaystyle (1.9)\quad \zeta(2n) = (-1)^{n+1}\frac{(2\pi)^{2n} B_{2n}}{2(2n)!}\)


\(\displaystyle (1.10)\quad \beta(2n+1) = (-1)^n\frac{\pi^{2n+1}E_{2n}}{2^{2n+2}(2n)!}\)




Back in a bit... (Heidy)(Heidy)(Heidy)


http://mathhelpboards.com/commentar...tric-approach-infinite-series-quot-10871.html
 
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DreamWeaver

Well-known member
Sep 16, 2013
337
Rearranging (1.9) to express the Bernoulli number \(\displaystyle B_{2k}\) in terms of the Zeta function \(\displaystyle \zeta(2k)\) gives:


\(\displaystyle (2.0)\quad B_{2k}= (-1)^{k+1}\frac{2(2k)!\, \zeta(2k) }{(2\pi)^{2k}}\)


Substituting this into the series definition for \(\displaystyle \log(\sin x)\), and then replacing x with \(\displaystyle x \to \pi z\) yields:


\(\displaystyle \log(\sin \pi z) = \log \pi z - \sum_{k=1}^{\infty}\frac{\zeta(2k)}{k}z^{2k}\)


Hence:


\(\displaystyle (2.1)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{k}z^{2k} = \log\left( \frac{\pi z}{\sin \pi z} \right)\)


Which is valid for \(\displaystyle 0 < z < 1\). Here are a few basic examples [find relevant Sine values at Wolfram Functions - Sine: Specific values (subsection 03/02) ]:


\(\displaystyle (2.2)\quad z=1/2 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{4^kk} = \log \pi\)



\(\displaystyle (2.3)\quad z=1/3 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{9^kk} =

\log \pi - \frac{3}{2}\log 3 + \log 2 \)



\(\displaystyle (2.4)\quad z=1/4 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{16^kk} = \log\pi - \frac{3}{2}\log 2 \)



\(\displaystyle (2.5)\quad z=1/5 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{25^kk} =

\log\pi - \log 5 + \frac{3}{2}\log 2 - \frac{1}{2}\log( 5-\sqrt{5} ) \)



\(\displaystyle (2.6)\quad z=1/6 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{36^kk} =

\log \pi - \log 3\)




More in a bit... (Heidy)(Heidy)(Heidy)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
As I've both defined and proved various properties for the Clausen functions in pretty much every tutorial I've posted on here, I'll forego the usual here. [If in doubt, the reader is, as I say, referred to pretty much any other tutorial I've posted here].

From proposition (2.1), which is valid for \(\displaystyle 0 < x < 1\), we have:

\(\displaystyle (2.1)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{k}x^{2k} = \log\left( \frac{\pi x}{\sin \pi x} \right)\)


Given the uniform convergence of the series on the L.H.S., there's nothing to stop us integrating the entire relation, term by term, over the range \(\displaystyle x=0\) to \(\displaystyle x=z\) (such that \(\displaystyle 0<z<1\)). Integration of the L.H.S. gives:


\(\displaystyle \int_0^z \Bigg\{ \sum_{k=1}^{\infty}\frac{\zeta(2k)}{k}x^{2k} \Bigg\}\, dx =

\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k(2k+1)}z^{2k+1} \)


Conversely, integration of the R.H.S. gives:


\(\displaystyle \int_0^z\log\left( \frac{\pi x}{\sin \pi x} \right)\, dx = \)


\(\displaystyle z\log \pi + \Bigg[x\log x-x\Bigg]_0^z - \int_0^z\log(\sin \pi x)\, dx = \)


\(\displaystyle z\log \pi z -z -\int_0^z\log(\sin \pi x)\, dx\)


Next, in a way that should be mind-numbingly familiar to folk who've read my other tutorials, we set \(\displaystyle \pi x = y/2\) in that last integral to obtain:


\(\displaystyle \int_0^z\log(\sin \pi x)\, dx = \)

\(\displaystyle \frac{1}{2\pi}\, \int_0^{2\pi z}\log\left( \sin \frac{y}{2} \right)\, dy = \frac{1}{2\pi}\, \int_0^{2\pi z}\log\left(2 \sin \frac{y}{2} \right)\, dy - \frac{z}{2\pi}\log 2 = \)


\(\displaystyle -\frac{\text{Cl}_2(2\pi z)}{2\pi} - \frac{z}{2\pi}\log 2\)


And so, after division of both sides by \(\displaystyle z\), we have


\(\displaystyle (3.0)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{k(2k+1)}z^{2k} = \frac{\text{Cl}_2(2\pi z)}{2\pi z} + \frac{1}{2\pi}\log 2 + \log \pi z - 1 \)





Finally, given that


\(\displaystyle \text{Cl}_2\left( \frac{\pi}{2} \right) = G \quad \quad \) [Catalan's constant]


\(\displaystyle m\in \mathbb{Z} \Rightarrow \text{Cl}_2(\pi m) = 0\)


and


\(\displaystyle \text{Cl}_2\left( \frac{2\pi}{3} \right) = \frac{2}{3}\, \text{Cl}_2\left( \frac{\pi}{3} \right)\)


We have the following special cases (to name but a few):


\(\displaystyle (3.1)\quad z=1/2 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{4^kk(2k+1)} = \log \pi + \left(\frac{1}{2\pi}-1\right) \log 2 -1\)



\(\displaystyle (3.2)\quad z=1/3 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{9^kk(2k+1)} = \frac{ \text{Cl}_2(\pi/3) }{\pi} +\frac{\log 2}{2\pi}+\log \pi -\log 3 -1\)



\(\displaystyle (3.3)\quad z=1/4 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{16^kk(2k+1)} = \frac{ 2G}{\pi} +\frac{\log 2}{2\pi}+\log \pi -2\log 2 -1\)



\(\displaystyle (3.4)\quad z=1/5 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{25^kk(2k+1)} = \frac{ 5\, \text{Cl}_2(2\pi/5) }{2\pi} +\frac{\log 2}{2\pi}+\log \pi -\log 5 -1\)



\(\displaystyle (3.5)\quad z=1/6 \Rightarrow\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{36^kk(2k+1)} = \frac{ 3\, \text{Cl}_2(\pi/3) }{\pi} +\frac{\log 2}{2\pi}+\log \pi -\log 6 -1\)
 
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DreamWeaver

Well-known member
Sep 16, 2013
337
The previous few examples should, hopefully, serve to illustrate the usefulness of basic trigonometric series. We could certainly apply a similar approach to the remaining trig series - some of which we will do later - but a little creative combination of these series can yield yet more interesting results. Here's a simple yet effective example.

Inspired by the basic trigonometric identity


\(\displaystyle \tan x = \cot x - 2\cot 2x\)


Let's ignore the L.H.S., and generalize the R.H.S. In other words, let's consider the cotangential difference:

\(\displaystyle (4.0)\quad \cot x - m\cot mx \quad\quad\quad\quad [\, m\in \mathbb{Z}^{+} \ge 2\, ] \)


Just to re-cap, we have:


\(\displaystyle (1.4)\quad \cot x = \frac{1}{x} + \sum_{k=1}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(2k)!} x^{2k-1}\)

Condition: \(\displaystyle -\pi < x < \pi\)


Applying this series to (4.0) we get:


\(\displaystyle \frac{1}{x} + \sum_{k=1}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(2k)!} x^{2k-1} - m\, \left[ \frac{1}{mx} + \sum_{k=1}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(2k)!} (mx)^{2k-1} \right]=\)


\(\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1}\frac{2^{2k}(m^{2k-1}-1) B_{2k}}{(2k)!} x^{2k-1}\)


Amended condition: \(\displaystyle -\pi < mx < \pi\)


By (2.0), we can then convert this into the Zeta sum:



\(\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1}\frac{2^{2k}(m^{2k-1}-1) }{(2k)!} \left[ (-1)^{k+1} \frac{2(2k)!\, \zeta(2k)}{2^{2k}\pi^{2k}} \right] x^{2k-1}=\)


\(\displaystyle 2\, \sum_{k=1}^{\infty} \frac{ (m^{2k-1}-1) \zeta(2k)}{\pi^{2k}} x^{2k-1}= \cot x - m\cot mx\)


Upon setting \(\displaystyle x=\pi z\), with \(\displaystyle -1 < z < 1\), we obtain the Zeta series:


\(\displaystyle (4.1)\quad \sum_{k=1}^{\infty} (m^{2k-1}-1) \zeta(2k) z^{2k}= \frac{\pi z}{2}\, \Bigg[ \cot \pi z - m\cot \pi mz \Bigg] \)


Condition: \(\displaystyle -1 < mz < 1\)




For example, let \(\displaystyle m=5\) and \(\displaystyle z=1/6\) to obtain


\(\displaystyle \sum_{k=1}^{\infty} \frac{ (5^{2k-1}-1) \zeta(2k)}{36^k} = \frac{\pi}{12}\, \left( \cot \frac{\pi}{6} - 5\cot \frac{5\pi}{6} \right)=\)


\(\displaystyle \frac{\pi}{12}\Bigg( \sqrt{3} - 5(-\sqrt{3})\Bigg) = \frac{\pi\sqrt{3}}{2}\)



\(\displaystyle \therefore \quad \sum_{k=1}^{\infty} \frac{ (5^{2k-1}-1) \zeta(2k)}{36^k} = \frac{\pi\sqrt{3}}{2}\)




Similarly, setting \(\displaystyle m=4\) and \(\displaystyle z=1/5\) we obtain


\(\displaystyle \sum_{k=1}^{\infty} \frac{ (4^{2k-1}-1) \zeta(2k)}{25^k} = \frac{\pi}{10}\, \left( \cot \frac{\pi}{5} - 4\cot \frac{4\pi}{5} \right)=\)


\(\displaystyle \frac{\pi}{10}\, \left[ \sqrt{1+ \frac{2}{\sqrt{5}}} - 4 \, \left( - \sqrt{1+ \frac{2}{\sqrt{5}}} \right) \right] = \frac{\pi}{2}\, \sqrt{1+ \frac{2}{\sqrt{5}}} \)



\(\displaystyle \therefore \quad \sum_{k=1}^{\infty} \frac{ (4^{2k-1}-1) \zeta(2k)}{25^k} = \frac{\pi}{2}\, \sqrt{1+ \frac{2}{\sqrt{5}}} \)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Returning briefly to the Zeta series (3.1) through (3.5), note that the general series can be split as follows:


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{m^kk(2k+1)} = \sum_{k=1}^{\infty}\frac{[(2k+1)-2k]\zeta(2k)}{m^kk(2k+1)} =\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{\zeta(2k)}{m^kk} - 2\, \sum_{k=1}^{\infty}\frac{\zeta(2k)}{m^k(2k+1)}\)


Rearranging these series we get


\(\displaystyle (5.0)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{m^k(2k+1)} = \)


\(\displaystyle \frac{1}{2}\, \sum_{k=1}^{\infty}\frac{\zeta(2k)}{m^kk}
-\frac{1}{2}\, \sum_{k=1}^{\infty}\frac{\zeta(2k)}{m^kk(2k+1)}\)


On the second line of (5.0), the series on the left is of the form (2.2) through (2.6), while the series on the right matches (3.1) through (3.5). The following series then result:



\(\displaystyle (5.1)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{4^k(2k+1)} = \frac{(2\pi-1)}{4\pi}\log 2+\frac{1}{2}\)



\(\displaystyle (5.2)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{9^k(2k+1)} = -\frac{ \text{Cl}_2(\pi/3) }{2\pi} -\frac{1}{4}\log 3 + \frac{(2\pi-1)}{4\pi}\log 2+\frac{1}{2}\)



\(\displaystyle (5.3)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{16^k(2k+1)} = -\frac{ G }{\pi} + \frac{(\pi-1)}{4\pi}\log 2+\frac{1}{2}\)



\(\displaystyle (5.4)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{25^k(2k+1)} = -\frac{ 5\, \text{Cl}_2(2\pi/5) }{4\pi} + \frac{(3\pi-1)}{4\pi}\log 2- \frac{1}{4}\log( 5-\sqrt{5} ) +\frac{1}{2}\)



\(\displaystyle (5.5)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{36^k(2k+1)} = -\frac{ 3\, \text{Cl}_2(\pi/3) }{2\pi} + \frac{(2\pi-1)}{4\pi}\log 2 +\frac{1}{2}\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Next, let's use (1.4) in integrated form to derive a new Zeta series. Just to re-cap, we have:


\(\displaystyle (1.4)\quad \cot x = \frac{1}{x} + \sum_{k=1}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(2k)!} x^{2k-1}\)

Condition: \(\displaystyle -\pi < x < \pi\)


The integral we are - or at least I am (lol) - interested in is:


\(\displaystyle \int_0^zx^2\cot x\, dx = z^2\log(\sin z) - 2\, \int_0^z x\log(\sin x)\, dx = \)


\(\displaystyle z^2\log(\sin z) - \frac{1}{2}\, \int_0^{2z} y \log \left( \sin \frac{y}{2} \right)\, dy =\)


\(\displaystyle z^2\log(\sin z) - \frac{1}{2}\, \left[ \int_0^{2z} x \log \left( 2\sin \frac{x}{2} \right)\, dx - \log 2\, \int_0^{2z}x\, dx \right] =\)


\(\displaystyle z^2\log(2\sin z) - \frac{1}{2}\, \int_0^{2z} x \log \left( 2\sin \frac{x}{2} \right)\, dx =\)


\(\displaystyle z^2\log(2\sin z) + \frac{1}{2}\, \left[ 2z\text{Cl}_2(2z) - \int_0^{2z} \text{Cl}_2(x)\, dx \right] =\)


\(\displaystyle z^2\log(2\sin z) + z\, \text{Cl}_2(2z) - \frac{1}{2}\, \sum_{k=1}^{\infty}\frac{1}{k^2}\, \int_0^{2z} \sin kx\, dx =\)


\(\displaystyle z^2\log(2\sin z) + z\, \text{Cl}_2(2z) - \frac{1}{2}\, \sum_{k=1}^{\infty}\frac{1}{k^2}\, \left(\frac{1}{k} - \frac{\cos 2kz}{k} \right) = \)


\(\displaystyle z^2\log(2\sin z) + z\, \text{Cl}_2(2z) + \frac{1}{2}\text{Cl}_3(2z) - \frac{\zeta(3)}{2}\)


Conversely, by the series expansion (1.4) for the cotangent,


\(\displaystyle \int_0^zx^2\cot x\, dx = \int_0^z x\, dx + \sum_{k=1}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(2k)!} \, \int_0^z x^{2k+1}\, dx = \)


\(\displaystyle \frac{z^2}{2} + \sum_{k=1}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(2k+2)(2k)!} z^{2k+2}=\)


\(\displaystyle \frac{z^2}{2} + \frac{z^2}{2}\, \sum_{k=1}^{\infty} (-1)^k\frac{2^{2k}B_{2k}}{(k+1)(2k)!} z^{2k}=\)


\(\displaystyle \frac{z^2}{2} + \frac{z^2}{2}\, \sum_{k=1}^{\infty} (-1)^k\frac{2^{2k}}{(k+1)(2k)!} z^{2k} \left[ (-1)^{k+1} \frac{2(2k)! \zeta(2k)}{(2\pi)^{2k}} \right] = \)


\(\displaystyle \frac{z^2}{2} - z^2\, \sum_{k=1}^{\infty} \frac{\zeta(2k)}{\pi^{2k} (k+1)} z^{2k} = \)


\(\displaystyle z^2\log(2\sin z) + z\, \text{Cl}_2(2z) + \frac{1}{2}\text{Cl}_3(2z) - \frac{\zeta(3)}{2}\)



A modest rearrangement of terms gives the following Zeta series:


\(\displaystyle (6.0)\quad \sum_{k=1}^{\infty} \frac{\zeta(2k)}{(k+1)} z^{2k} = \)


\(\displaystyle \frac{1}{2}- \log(2\sin \pi z) - \frac{\text{Cl}_2(2\pi z) }{\pi z} - \frac{\text{Cl}_3(2\pi z) }{2\pi^2 z^2} + \frac{\zeta(3)}{2\pi^2z^2}\)


To evaluate this Zeta series at, say, \(\displaystyle z=1/2,\, 1/3,\, 1/4,\, 1/5\), and \(\displaystyle 1/6\), we will need a few special values for the third order Clausen function \(\displaystyle \text{Cl}_3(2\pi z)\). These are calculated below, where the Eta function is also used:


\(\displaystyle (6.1)\quad \eta(x) = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^x} = \left( 1- \frac{2}{2^x} \right) \zeta(x)\)



============================
============================


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Case A:
----------


\(\displaystyle \text{Cl}_3(\pi) = \sum_{k=1}^{\infty}\frac{\cos \pi k}{k^3} = \sum_{k=1}^{\infty} \frac{(-1)^k}{k^3} = -\eta(3) = -\left(1- \frac{2}{2^3}\right) \zeta(3) = - \frac{3\, \zeta(3)}{4}\)


\(\displaystyle \therefore \quad \text{Cl}_3(\pi) = - \frac{3\, \zeta(3)}{4}\)



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Case B:
----------


\(\displaystyle \text{Cl}_3\left( \frac{2\pi}{3} \right) = \sum_{k=1}^{\infty} \frac{\cos (2\pi k/3)}{k^3} =\)


\(\displaystyle \frac{\cos (2\pi /3)}{1^3} +
\frac{\cos (4\pi /3)}{2^3} +
\frac{\cos (2\pi)}{3^3} +
\frac{\cos (2\pi /3)}{4^3} +
\frac{\cos (4\pi /3)}{5^3} +
\frac{\cos (2\pi)}{6^3} +
\cdots = \)


\(\displaystyle -\frac{1}{2\cdot 1^3}
-\frac{1}{2\cdot 2^3}
+\frac{1}{3^3}
-\frac{1}{2\cdot 4^3}
-\frac{1}{2\cdot 5^3}
+\frac{1}{6^3} - \cdots = \)


\(\displaystyle -\frac{1}{2}\, \sum_{k=1}^{\infty}\frac{1}{k^3} + \frac{3}{2}\, \sum_{k=1}^{\infty}\frac{1}{(3k)^3} = \)


\(\displaystyle -\frac{\zeta(3)}{2} + \frac{\zeta(3)}{18} = - \frac{4\, \zeta(3)}{9} \)


\(\displaystyle \therefore \quad \text{Cl}_3\left( \frac{2\pi}{3} \right) = - \frac{4\, \zeta(3)}{9} \)



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Case C:
----------


\(\displaystyle \text{Cl}_3\left( \frac{\pi}{2} \right) = \sum_{k=1}^{\infty} \frac{\cos(\pi k/2)}{k^3} = \)


\(\displaystyle -\frac{1}{2^3} + \frac{1}{4^3} -\frac{1}{6^3} + \frac{1}{8^3} - \cdots = -\frac{1}{2^3}\, \left( 1 - \frac{1}{2^3} + \frac{1}{3^3} - \cdots \, \right) = \)


\(\displaystyle -\frac{\eta(3)}{8} = -\frac{1}{8}\cdot \frac{3\, \zeta(3)}{4} = - \frac{3\, \zeta(3) }{32} \)


\(\displaystyle \therefore\quad \text{Cl}_3\left( \frac{\pi}{2} \right) = - \frac{3\, \zeta(3) }{32}\)



----------
Case D:
----------


\(\displaystyle \text{Cl}_3\left( \frac{\pi}{3} \right) = \)


\(\displaystyle \frac{\cos (\pi /3)}{1^3} +
\frac{\cos (2\pi /3)}{2^3} +
\frac{\cos (\pi)}{3^3} +
\frac{\cos (4\pi /3)}{4^3} +
\frac{\cos (5\pi /3)}{5^3} +
\frac{\cos (2\pi)}{6^3} +
\cdots = \)


\(\displaystyle \frac{1}{2\cdot 1^3} - \frac{1}{2\cdot 2^3} - \frac{1}{3^3} -
\frac{1}{2\cdot 4^3} + \frac{1}{2\cdot 5^3} + \frac{1}{6^3} + \cdots = \)


\(\displaystyle \frac{1}{2}\, \left( 1-\frac{1}{2^3} + \frac{1}{3^3} - \frac{1}{4^3} + \frac{1}{5^3} - \frac{1}{6^3} + \cdots \, \right) \, -\frac{3}{2}\, \left( \frac{1}{3^3} - \frac{1}{6^3} + \frac{1}{9^3} - \cdots \, \right) = \)


\(\displaystyle \frac{\eta(3)}{2} - \frac{3}{2} \cdot \frac{\eta(3)}{3^3} = \frac{4\, \zeta(3)}{9} \)


\(\displaystyle \therefore\quad \text{Cl}_3\left( \frac{\pi}{3} \right) = \frac{4\, \zeta(3)}{9} \)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Using (6.0) above, and the explicit evaluations of \(\displaystyle \text{Cl}_3(x)\) given, we have the following Zeta sums:



\(\displaystyle (7.0)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{4^k(k+1)} = \frac{1}{2} - \log 2 +\frac{7\, \zeta(3)}{2\pi^2}\)



\(\displaystyle (7.1)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{9^k(k+1)} = \frac{1}{2} - \frac{1}{2}\log 3 - \frac{2\, \text{Cl}_2(\pi/3)}{\pi} +\frac{13\, \zeta(3)}{2\pi^2}\)



\(\displaystyle (7.2)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{16^k(k+1)} = \frac{1}{2} - \frac{1}{2}\log 2 - \frac{4\, G}{\pi} +\frac{35\, \zeta(3)}{4\pi^2}\)



\(\displaystyle (7.3)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{25^k(k+1)} = \)


\(\displaystyle \frac{1}{2} + \frac{1}{2}\log 2 - \frac{1}{2}\, \log(5-\sqrt{5}) - \frac{5\, \text{Cl}_2(2\pi/5)} {\pi} - \frac{25\, \text{Cl}_3(2\pi/5)}{2\pi^2} +\frac{25\, \zeta(3)}{2\pi^2}\)



\(\displaystyle (7.4)\quad \sum_{k=1}^{\infty}\frac{\zeta(2k)}{36^k(k+1)} = \frac{1}{2} - \frac{ 6\, \text{Cl}_2(\pi/3) }{\pi} + \frac{ 80\, \zeta(3) }{9 \pi^2}\)
 
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