11

$$\zeta (2)(1-\frac{1}{4})+\frac{\zeta (4)}{4}(1-\frac{1}{4^2})+\frac{\zeta (6)}{4^2}(1-\frac{1}{4^3})+...=\frac{\pi }{2}$$

The WolframAlph couldn't recognize the closed-form which is $\pi/2$ when I gave the series, enter image description here

so I used the WolframAlph again to compute many terms of infinite series. enter image description here

I think that the WolfarmAlph cannot say the value is $\pi/2$,So we need to prove it.

E.H.E
  • 23,280

3 Answers3

16

Since,

$\begin{align}\sum_{k=1}^\infty\zeta(2k)\,x^{2k} &= \sum_{k=1}^\infty\sum_{n=1}^\infty\frac{x^{2k}}{n^{2k}}\\ &=\sum_{n=1}^\infty\frac{\frac{x^2}{n^2}}{1-\frac{x^2}{n^2}}\\ &=\sum_{n=1}^\infty\frac{x^2}{n^2-x^2}\\ &=-\frac{x}{2}\sum_{n=1}^\infty\left(\frac{1}{x+n} + \frac{1}{x-n}\right)\\ &=-\frac{x}{2}\left(\pi\cot(\pi x)-\frac1x\right)\\ &=\frac{1}{2}\left(1-\pi x\cot(\pi x)\right) \end{align}$

Plug in the particular cases $x = \dfrac{1}{2}$ and $x = \dfrac{1}{4}$ in the series.

r9m
  • 17,938
13

We have: $$\zeta(2n)\left(1-\frac{1}{4^n}\right)=\sum_{k=0}^{+\infty}\frac{1}{(2k+1)^{2n}}\tag{1}$$ hence: $$\sum_{n=1}^{+\infty}\frac{\zeta(2n)}{4^{n-1}}\left(1-\frac{1}{4^n}\right)=4\sum_{n=1}^{+\infty}\sum_{k=0}^{+\infty}\frac{1}{(4k+2)^{2n}}=4\sum_{k=0}^{+\infty}\frac{1}{(4k+2)^2-1}\tag{2}$$ and the result follows from: $$\sum_{k=0}^{+\infty}\frac{1}{(4k+2)^2-1}=\frac{1}{2}\sum_{k=0}^{+\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)=\frac{1}{2}\arctan 1 = \color{red}{\frac{\pi}{8}}.\tag{3}$$

Jack D'Aurizio
  • 353,855
7

$$\zeta(2n) = \frac{1}{\Gamma(2n)}\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\,dx $$ so we have: $$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{n-1}} =\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{2n-2}}= 2\int_{0}^{\infty}\frac{\sinh(x/2)}{e^x-1}\,dx =\int_{0}^{\infty}e^{-x/2}\,dx = 2$$

also we have : $$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{2n-1}} = \int_{0}^{\infty}\frac{\sinh(x/4)}{e^x-1}\,dx =\frac12\int_{0}^{\infty}\frac{1}{e^{\frac{3x}{4}}+e^\frac{x}{4}}\,dx =2-\frac{\pi}{2}$$

last integral we can slove using $u=e^\frac{x}{4}$

putting all together :

$$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{n-1}} -\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{2n-1}} =\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{n-1}}(1-\frac{1}{4^{n}})=\frac{\pi}{2}$$

mnsh
  • 5,875