20
$\begingroup$

After numerical analysis it seems that

$$ \frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} $$

Could someone prove the validity of such identity?

$\endgroup$
7
  • 2
    $\begingroup$ There are lots and lots of identities of this type. Have you read answers to your previous question? This one can be solved in exactly the same manner. What have you tried anyway? $\endgroup$
    – Grigory M
    Dec 19, 2013 at 18:05
  • $\begingroup$ You must be right, there are lots of this kind of identities. And I was wondering if there is a more general formula. For exemple we have the following $$ \frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}} $$ and $$ \frac{\pi}{3}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{6^{2k}} $$ and in this line of thought how coud we express $\pi/5$ or $\pi/7$? $\endgroup$
    – Neves
    Dec 19, 2013 at 18:10
  • $\begingroup$ @Grigory, I can prove some of this formulas but not all of them so when I can't I ask here. $\endgroup$
    – Neves
    Dec 19, 2013 at 18:17
  • $\begingroup$ @Grigori, can you provide a formula for $\pi/7$ involving $\zeta$ or $\eta$ whatever you like? $\endgroup$
    – Neves
    Dec 19, 2013 at 18:21
  • $\begingroup$ I'm not sure I understand — certainly you won't be satisfied $\frac27(1+2\sum\eta(2k)2^{-2k})$ — but what's the question exactly then? Anyway, first question is, what is $\sum\zeta(2k)x^{2k}$ and $\sum\eta(2k)x^{2k}$ — do you happen to know answers? $\endgroup$
    – Grigory M
    Dec 19, 2013 at 18:33

4 Answers 4

29
$\begingroup$

Using the formula for a geometric series, $$ \begin{align} \sum_{k=1}^\infty\frac1{x^{2k}} &=\frac1{x^2-1}\\ &=\frac12\left(\frac1{x-1}-\frac1{x+1}\right)\tag{1} \end{align} $$ Differentiating $(1)$ twice, $$ \sum_{k=1}^\infty\frac{2k(2k+1)}{x^{2k+2}} =\frac1{(x-1)^3}-\frac1{(x+1)^3}\tag{2} $$ Changing the order of summation and applying $(2)$, $$ \begin{align} 1-\sum_{k=1}^\infty\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} &=1-\sum_{k=1}^\infty\sum_{j=1}^\infty\frac{2k(2k+1)}{(4j)^{2k+2}}\\ &=1-\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{2k(2k+1)}{(4j)^{2k+2}}\\ &=1-\sum_{j=1}^\infty\left(\frac1{(4j-1)^3}-\frac1{(4j+1)^3}\right)\\ &=1-\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)^3}\\ &=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\tag{3} \end{align} $$ The sum in $(3)$ can be generalized as $$ \beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\tag{4} $$ and is known as the Dirichlet beta function. As shown below, $\beta(3)=\dfrac{\pi^3}{32}$.


We can develop a recurrence for $\beta(2k+1)$. First, the generating function for $\beta(2k+1)$ is $$ \begin{align} f(x) &= \sum_{k=0}^\infty \beta(2k+1) x^{2k+1}\\ &= \sum_{n=0}^\infty \sum_{k=0}^\infty (-1)^n\left(\frac{x}{2n+1}\right)^{2k+1}\\ &= \sum_{n=0}^\infty (-1)^n\frac{\frac{x}{2n+1}}{1-\left(\frac{x}{2n+1}\right)^2}\\ &= \frac{x}{2} \sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+x}+\frac{1}{2n+1-x}\right)\\ &= \frac{x}{2} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{2n+1+x}\\ &= \frac{x}{4} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{n+\tfrac{1+x}{2}}\\ &= \frac{x}{4} \pi \csc\left(\pi\tfrac{1+x}{2}\right)\\[3pt] &= \frac{\pi}{4} x \sec\left(\frac{\pi}{2}x\right)\tag{5} \end{align} $$ where we use $(7)$ from this answer to get $$ \begin{align} \sum_{n=-\infty}^{+\infty}\frac{(-1)^n}{n+z} &=\sum_{n=-\infty}^{+\infty}\frac2{2n+z}-\sum_{n=-\infty}^{+\infty}\frac1{n+z}\\[3pt] &=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[9pt] &=\pi\csc(\pi z)\tag{6} \end{align} $$ We can use equation $(5)$ to develop a recurrence relation: $$ \begin{align} \frac{\pi}{4} x &= \cos\left(\frac{\pi}{2} x\right) f(x)\\ &= \sum_{n=0}^\infty\sum_{k=0}^n (-1)^k \frac{(\frac{\pi}{2} x)^{2k}}{(2k)!}\;\beta(2n-2k+1)x^{2n-2k+1}\\ &= \sum_{n=0}^\infty\sum_{k=0}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)x^{2n+1}\tag{7} \end{align} $$ For $n=0$, we can use the arctan series to get $$ \beta(1) = \frac{\pi}{4}\tag{8} $$ and for $n\gt0$, $(7)$ gives $$ \beta(2n+1) = -\sum_{k=1}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)\tag{9} $$ Recursion $(9)$ yields $$ \begin{align} \beta(1)&=\frac{\pi}{4}\\ \beta(3)&=\frac{\pi^3}{32}\\ \beta(5)&=\frac{5\pi^5}{1536}\\ \beta(7)&=\frac{61\pi^7}{184320}\\ \beta(9)&=\frac{277\pi^9}{8257536}\\ \beta(11)&=\frac{50521\pi^{11}}{14863564800}\\ \beta(13)&=\frac{540553\pi^{13}}{1569592442880}\\ \beta(15)&=\frac{199360981\pi^{15}}{5713316492083200}\\ \beta(17)&=\frac{3878302429\pi^{17}}{1096956766479974400}\\ \beta(19)&=\frac{2404879675441\pi^{19}}{6713375410857443328000} \end{align} $$

$\endgroup$
5
  • $\begingroup$ Excellent job! (+1) $\endgroup$ Mar 19, 2014 at 10:19
  • 2
    $\begingroup$ would the downvoter care to comment? $\endgroup$
    – robjohn
    Apr 3, 2014 at 11:08
  • $\begingroup$ by (7), I get $$\frac{\pi}{4}x=\beta(1)x+\sum_{n=1}^{\infty}\left[x^{2n+1}\sum_{k=0}^{n}(-1)^k\frac{(\pi x/2)^{2k}}{(2k)!}\beta(2n-2k+1)\right]=\frac{\pi}{4}x+\sum_{n=1}^{\infty}\left[x^{2n+1}\left[\beta(2n+1)+\sum_{k=1}^{n}\frac{(-\frac{\pi^2}{4})^k}{(2k)!}\beta(2n-2k+1)\right]\right]$$ so $$\sum_{n=1}^{\infty}\left[x^{2n+1}\left[\beta(2n+1)+\sum_{k=1}^{n}\frac{(-\frac{\pi^2}{4})^k}{(2k)!}\beta(2n-2k+1)\right]\right]=0.$$ then how to get (9) please? maybe it is: if $\sum_{n=1}^{\infty}x^{2n+1}g(n)=0$, then we have $g(n)=0$. I can not prove it, could you explain it to me in detail please? $\endgroup$
    – xunitc
    Jun 28, 2017 at 0:10
  • $\begingroup$ @xunitc: The sum on the left is a polynomial which is equal to $0$. That means that each coefficient of $x^{2n+1}$ must be $0$. $(9)$ follows by using this fact. $\endgroup$
    – robjohn
    Jun 28, 2017 at 8:51
  • $\begingroup$ Thank you, I know it now, it converges to $0$, and can make differentiation of x, then the coefficient is $0$. $\endgroup$
    – xunitc
    Jun 28, 2017 at 9:35
17
$\begingroup$

Yes, we can prove it. We can change the order of summation in

$$\begin{align} \sum_{k=1}^\infty \frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} &= \sum_{k=1}^\infty \frac{2k(2k+1)}{4^{2k+2}}\sum_{n=1}^\infty \frac{1}{n^{2k+2}}\\ &= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{2k(2k+1)}{(4n)^{2k+2}}\\ &= \sum_{n=1}^\infty r''(4n), \end{align}$$

where, for $\lvert z\rvert > 1$, we define

$$r(z) = \sum_{k=1}^\infty \frac{1}{z^{2k}} = \frac{1}{z^2-1} = \frac12\left(\frac{1}{z-1} - \frac{1}{z+1}\right).$$

Differentiating yields $r''(z) = \frac{1}{(z-1)^3} - \frac{1}{(z+1)^3}$, so

$$1 - \sum_{k=1}^\infty \frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} = \sum_{\nu = 0}^\infty \frac{(-1)^\nu}{(2\nu+1)^3},$$

and the latter sum is by an earlier answer using the partial fraction decomposition of $\dfrac{1}{\cos z}$:

$$\sum_{\nu=0}^\infty \frac{(-1)^\nu}{(2\nu+1)^3} = - \frac{\pi^3}{32} E_2 = \frac{\pi^3}{32}.$$

$\endgroup$
2
  • 2
    $\begingroup$ I took too long to move my recursion for $\beta(2k+1)$ from my pdf, and I didn't take time to realize that the first part of my answer is almost the same as your answer. I hope you don't mind if I leave my answer; I would like to have the recursion for $\beta(2k+1)$ on site. $\endgroup$
    – robjohn
    Dec 19, 2013 at 21:01
  • $\begingroup$ Not at all, au contraire. Thanks for the link, that'll come handy to look the values and recursions up. $\endgroup$ Dec 19, 2013 at 21:11
8
$\begingroup$

Starting from the Laurent series of the cotangent function: $$\pi z\cot \left( \pi z \right) =1-2\,\sum _{k=0}^{\infty }\zeta \left( 2\,k+2 \right) {z}^{2k+2} \tag{1}$$ apply the differential operator: $$\hat{D}=z^2\dfrac{d^2}{dz^2}-2z\dfrac{d}{dz}+2 \tag{2}$$ to get: $${z}^{3}{\pi }^{3}\cot \left( \pi z \right) \left( 1+ \cot \left( \pi z \right) ^{2} \right) =1-\sum _{k=0}^{\infty }2k \left( 2k+1 \right) \,\zeta \left( 2\,k+2 \right) {z}^{2k+2}\tag{3}$$ which, by the ratio test, has a radius of convergence of $|z|<1$. Then from: $$z=\dfrac{1}{4}, \quad \cot\left(\dfrac{\pi}{4}\right)=1 \tag{4}$$ we have: $$\dfrac{{\pi }^{3}}{32}=1-\sum _{k=0}^{\infty }{\frac {2k \left( 2\,k+1 \right) \zeta \left( 2\,k+2 \right) }{{4}^{2k+2}}}\tag{5}$$

$\endgroup$
1
$\begingroup$

$$ \begin{align} S\, &=\,\sum_{k=1}^{\infty}\frac{2k\,(2k+1)\,\zeta(2k+2)}{4^{2k+2}} \,=\,\sum_{k=1}^{\infty}\frac{(2k+1)!\,\zeta(2k+2)}{(2k-1)!\,4^{2k+2}} \\[4mm] &=\,\sum_{k=1}^{\infty}\frac{(1/4)^{2k+2}}{(2k-1)!}\,\int_{0}^{\infty}\frac{x^{2k+1}}{e^x-1}\,dx \,=\,4^{-3}\int_{0}^{\infty}\frac{x^2}{e^x-1}\,\left(\sum_{k=1}^{\infty}\frac{(x/4)^{2k-1}}{(2k-1)!}\right)\,dx \\[4mm] &=\,4^{-3}\int_{0}^{\infty}\frac{x^2}{e^x-1}\,\sinh\left(\frac{x}{4}\right)\,dx \,=\,\frac{4^{-3}}{2}\int_{0}^{\infty}\frac{x^2}{1-e^{-x}}\,\left({\Large e}^{-\frac34x}\,-\,{\Large e}^{-\frac54x}\right)\,dx \\[4mm] &=\,4^{-3}\left[\zeta\left(3,\frac34\right)\,-\,\zeta\left(3,\frac54\right)\right] \,=\,1-4^{-3}\left[\zeta\left(3,\frac14\right)\,-\,\zeta\left(3,\frac34\right)\right] \\[4mm] &=\,1-\beta(3) \,=\,{\color\red{1-\frac{\pi^3}{32}}} \\ \end{align} $$


$\,\zeta(s,q) \,\,$ : Hurwitz Zeta Function
$\,\beta(s)\quad\,$ : Dirichlet Beta Function

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .