# How to prove that $\frac{\zeta(2) }{2}+\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}+\frac{\zeta (8)}{2^7}+\cdots=1$?

How can one prove this identity?

$$\frac{\zeta(2) }{2}+\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}+\frac{\zeta (8)}{2^7}+\cdots=1$$

There is a formula for $\zeta$ values at even integers, but it involves Bernoulli numbers; simply plugging it in does not appear to be an efficient approach.

• I think this is a normal question. I don't know why "on hold"? – E.H.E Dec 24 '14 at 19:28
• There is a recursive formula for $\zeta(2n)$ derived in this answer. However, in that answer it is shown that $$\sum_{k=1}^\infty\zeta(2k)x^{2k}=\frac12(1-\pi x\cot(\pi x))$$ With $x=\frac12$, this immediately gives $$\sum_{k=1}^\infty\zeta(2k)\left(\frac12\right)^{2k-1} =1-\frac\pi2\cot\left(\frac\pi2\right)=1$$ This is what is used in Random Variable's answer. – robjohn Dec 31 '14 at 18:13

Since $$\zeta(2n) = \frac{1}{(2n-1)!}\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\,dx$$ we have: $$\sum_{n=1}^{+\infty}\frac{\zeta(2n)}{2^{2n-1}} = \int_{0}^{+\infty}\frac{\sinh(x/2)}{e^x-1}\,dx =\frac{1}{2}\int_{0}^{+\infty}e^{-x/2}\,dx = \color{red}{1}.$$

• @mathworker21: I can mention that the dominated (or just monotone) convergence theorem clearly applies here. – Jack D'Aurizio Jul 14 '17 at 18:41

\begin{align} \sum_{n=1}^\infty\frac{\zeta(2n)}{2^{2n-1}} &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac2{k^{2n}2^{2n}}\tag{1}\\ &=2\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{(4k^2)^n}\tag{2}\\ &=2\sum_{k=1}^\infty\frac1{4k^2-1}\tag{3}\\ &=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k+1}\right)\tag{4}\\[6pt] &=1\tag{5} \end{align} Explanation:
$(1)$: expand $\zeta(2n)=\sum\limits_{k=1}^\infty\frac1{k^{2n}}$
$(2)$: change the order of summation
$(3)$: sum of a geometric series
$(4)$: partial fractions
$(5)$: telescoping sum

• @robjohn, in $(3)$ the numerator should have a $4k^2$. $$\sum_{n=1}^{\infty} \frac{1}{(4k^2)^n} = \frac{1}{1- \frac{1}{4k^2}} = \frac{4k^2}{4k^2 - 1}$$ – Amad27 Mar 20 '15 at 16:45
• @Amad27: Notice that the sum starts from $n=1$. This means that the sum is $\dfrac{\frac1{4k^2}}{1-\frac1{4k^2}}$. – robjohn Mar 20 '15 at 17:10
• Ah! I couldn't catch the index! Sorry robjohn! – Amad27 Mar 20 '15 at 18:17
• This one appears to be simplest possible answer. +1 It is as if you see the result in question at the level of $1 + r + r^{2} + \cdots = 1/(1 - r)$. – Paramanand Singh Jun 20 '16 at 10:29
• Plain and simple - as always @robjohn :) – AD. May 23 at 7:43

The Laurent expansion of $\cot (z)$ at the origin in terms of the Riemann zeta function is $$\cot (z) = \frac{1}{z} - 2 \sum_{k=1}^{\infty}\zeta(2k) \frac{z^{2k-1}}{\pi^{2k}} \ , \ 0 < |z| < \pi.$$

Letting $\displaystyle z= \frac{\pi}{2}$, $$\cot \left(\frac{\pi}{2} \right) = \frac{2}{\pi} - \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{\zeta(2k)}{2^{2k-1}}.$$

But $\cot \left(\frac{\pi}{2} \right)=0$.

Therefore,

$$\sum_{k=1}^{\infty} \frac{\zeta(2k)}{2^{2k-1}} = 1.$$


\begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\zeta\pars{2n} \over 2^{2n - 1}}} = \sum_{n = 2}^{\infty}{\zeta\pars{n} \over 2^{n - 1}} - \sum_{n = 1}^{\infty}{\zeta\pars{2n + 1} \over 2^{2n}} \\[3mm] = &\ -\sum_{n = 2}^{\infty}\pars{-1}^{n}\zeta\pars{n}\pars{-\,\half}^{n - 1} - \sum_{n = 1}^{\infty}\bracks{\zeta\pars{2n + 1} - 1}\pars{\half}^{2n}\ -\ \underbrace{\sum_{n = 1}^{\infty}\pars{\half}^{2n}}_{\ds{1 \over 3}} \\ = &\ -\bracks{\Psi\pars{1 + z} + \gamma}_{\ z\ =\ -1/2} \\[3mm] & - \bracks{% {1 \over 2z} - \half\,\pi\cot\pars{\pi z} - {1 \over 1 - z^{2}} + 1 - \gamma - \Psi\pars{1 + z}}_{\ z\ =\ 1/2} - {1 \over 3} \\[8mm] = &\ -\Psi\pars{\half} - {2 \over 3}\ +\ \underbrace{\Psi\pars{3 \over 2}}_{\ds{\Psi\pars{1/2} + 1/\pars{1/2}}} - {1 \over 3} = \color{#f00}{1} \end{align}

$$\Psi$$ and $$\gamma$$ are the Digamma function and the Euler-Mascheroni constant, respectively. We used the Digamma Recurrence Formula $$\ds{\Psi\pars{z + 1} = \Psi\pars{z} + 1/z}$$ and the identities $$\mathbf{6.3.14}$$ and $$\mathbf{6.3.15}$$ in this link.