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

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?

Solution 1:

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

Solution 2:

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

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

Solution 3:

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

Solution 4:

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

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