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?

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    $\begingroup$ 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 '13 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 '13 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 '13 at 18:17
  • $\begingroup$ @Grigori, can you provide a formula for $\pi/7$ involving $\zeta$ or $\eta$ whatever you like? $\endgroup$ – Neves Dec 19 '13 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 '13 at 18:33

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 $\dfrac{1}{\cos z}$:

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

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    $\begingroup$ 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 '13 at 21:01
  • $\begingroup$ Not at all, au contraire. Thanks for the link, that'll come handy to look the values and recursions up. $\endgroup$ – Daniel Fischer Dec 19 '13 at 21:11

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 $$ \begin{align} \beta(1)&=\frac{\pi}{4}\\ \beta(3)&=\frac{\pi^3}{32}\\ \beta(5)&=\frac{5\pi^5}{1536}\\ \beta(7)&=\frac{61\pi^7}{184320}\\ \beta(9)&=\frac{277\pi^9}{8257536}\\ \beta(11)&=\frac{50521\pi^{11}}{14863564800}\\ \beta(13)&=\frac{540553\pi^{13}}{1569592442880}\\ \beta(15)&=\frac{199360981\pi^{15}}{5713316492083200}\\ \beta(17)&=\frac{3878302429\pi^{17}}{1096956766479974400}\\ \beta(19)&=\frac{2404879675441\pi^{19}}{6713375410857443328000} \end{align} $$

  • $\begingroup$ Excellent job! (+1) $\endgroup$ – user 1591719 Mar 19 '14 at 10:19
  • 1
    $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn Apr 3 '14 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 '17 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 '17 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 '17 at 9:35

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: $$\dfrac{{\pi }^{3}}{32}=1-\sum _{k=0}^{\infty }{\frac {2k \left( 2\,k+1 \right) \zeta \left( 2\,k+2 \right) }{{4}^{2k+2}}}\tag{5}$$


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

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


First, Mathematica confirms this. For the proof, I would assume that expanding out the zeta and then changing the order of summation will do the trick (you will get sums of the sort $\sum_{k=1}^\infty 2k(2k+1)\frac{1}{{(4 i)}^{2k+2}},$ which are easy to evaluate (by integrating the sum of the geometric series a couple of times):

$$\frac{2 \left(48 i^2+1\right)}{\left(16 i^2-1\right)^3}$$


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