# Is $\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta (2)$?

$$\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta(2)$$

By using numerical calculation, I found this relationship between the values of zeta function at even integers and $\zeta(2)$, but this needs proving, any help?

Since: $$\frac{1-\pi x\cot(\pi x)}{2}=\sum_{n\geq 1}\zeta(2n)\,x^{2n} \tag{1}$$ we have, by differentiation: $$\sum_{n\geq 1}2n\zeta(2n) x^{2n-1} = \frac{\pi}{4\sin^2(\pi x)}\left(2\pi x-\sin(2\pi x)\right) \tag{2}$$ and by evaluating both sides at $x=\frac{1}{2}$ the claim follows.

• Faster then me as always! (+1) Commented Sep 29, 2015 at 18:40
• @tired: that's just because you're tired :D Commented Sep 29, 2015 at 18:42
• I overcame my tiredness and give another proof :) Commented Sep 29, 2015 at 20:11

Taking the derivative of $$\sum_{n=0}^\infty x^n=\frac1{1-x}\tag{1}$$ and multiplying by $x$ yields $$\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}\tag{2}$$ Therefore, \hspace{-1cm}\begin{align} \frac83\sum_{n=1}^\infty\frac{n}{2^{2n}}\zeta(2n) &=\frac83\sum_{n=1}^\infty\frac{n}{4^n}\sum_{k=1}^\infty\frac1{k^{2n}}\tag{3a}\\ &=\frac83\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{n}{\left(4k^2\right)^n}\tag{3b}\\ &=\frac83\sum_{k=1}^\infty\frac{4k^2}{\left(4k^2-1\right)^2}\tag{3c}\\ &=\frac23\sum_{k=1}^\infty\left[\frac1{(2k-1)^2}+\frac1{(2k+1)^2}+\frac1{2k-1}-\frac1{2k+1}\right]\tag{3d}\\ &=\frac23\sum_{k=1}^\infty\left[\frac2{(2k-1)^2}-\left(\frac1{(2k-1)^2}-\frac1{(2k+1)^2}\right)+\left(\frac1{2k-1}-\frac1{2k+1}\right)\right]\tag{3e}\\ &=\frac23\sum_{k=1}^\infty\frac2{(2k-1)^2}\tag{3f}\\[6pt] &=\zeta(2)\tag{3g} \end{align} Explanation:
$\text{(3a)}$: expand $\zeta(2n)$
$\text{(3b)}$: change the order of summation
$\text{(3c)}$: apply $(2)$
$\text{(3d)}$: partial fractions
$\text{(3e)}$: unify the sum of the odd squares by introducing a telescoping series
$\text{(3f)}$: the telescoping series cancel: $-1+1=0$
$\text{(3g)}$: the sum of the reciprocals of the odd squares is $\frac34$ of the sum of the reciprocals of all the squares

• Nice (+1), i just saw that our approaches have a lot in common! Commented Sep 29, 2015 at 20:12

Ok, let give me a slightly different answer the Jack,

Our sum is

$$S=\frac{8}{3}\sum_{n=1}^{\infty}\frac{n}{2^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}=\frac{8}{3}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{n}{(2^2 k^2)^n}\underbrace{=}_{(1)}\frac{8}{3}\sum_{k=1}^{\infty}\frac{4 k^2}{\left(4 k^2-1\right)^2}\underbrace{=}_{(2)}\\\frac{2}{3}\left(\partial_x\cot(1/x)-1+\underbrace{2\sum_{k=1}^{\infty}\frac{1}{k^2\pi^2 x^2-1}}_{-Q(x)}\right)\big|_{x=\frac{2}{\pi}}\underbrace{=}_{(3)}\frac{2}{3}\left(\partial_x\cot(1/x)\right)\big|_{x=\frac{2}{\pi}}\underbrace{=}_{(4)}\frac{\pi^2}{6}$$

in (1) we just differentiated a geometric series. In (2) we used the Mittag-Leffler expansion of $\cot(1/z)$ and differentiated it $$\partial_z\cot(1/z)=\partial_z\left(z+2z\sum_{k=1}\frac{1}{1-\pi^2 k^2 z^2}\right)=1+\underbrace{2\sum_{k=1}\frac{1}{1-\pi^2 k^2 z^2}}_{Q(z)}+\underbrace{4\sum_{k=1}\frac{\pi^2 k^2 z^2}{(1-\pi^2 k^2 z^2)^2}}_{\frac{3}{2}S \quad\text{if}\quad z=2/\pi}$$

In (3) we again used the Mittag Leffler expansion to show that the sum on the left is $Q(2/\pi)=-1+\cot(1/x)/x\big|_{\frac{2}{\pi}}=-1$

In (4) we used $\partial_z \cot(1/z)=\frac{z^2}{\sin^2(1/z)}$ and $\sin(\pi/2)=1$