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The problem I am trying to solve is: I need to prove the relation between the Dirichlet eta function and the Riemann zeta function $\eta(s) = \left(1-2^{1-s}\right) \zeta(s)$.

But I have no clue where to start or how to approach this problem. Any hints are welcome.

Bernard
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Jutrop
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1 Answers1

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Assignment:

Prove the following fact:

$$\eta\left(\text{s}\right):=\sum_{\text{n}\space\ge\space1}\frac{\left(-1\right)^{\text{n}-1}}{\text{n}^\text{s}}=\zeta\left(\text{s}\right)\left(1-2^{1-\text{s}}\right)$$

Where $\eta\left(\text{s}\right)$ is the Dirichlet eta function and $\zeta\left(\text{s}\right)$ is the Riemann zeta function.

Solution:

First, let's add the sum over all odd and even numbers:

$$\eta\left(\text{s}\right)=\underbrace{\sum_{\text{n}\space\ge\space1}\frac{\left(-1\right)^{2\text{n}-1-1}}{\left(2\text{n}-1\right)^\text{s}}}_\text{odd part}+\underbrace{\sum_{\text{n}\space\ge\space1}\frac{\left(-1\right)^{2\text{n}-1}}{\left(2\text{n}\right)^\text{s}}}_\text{even part}=\sum_{\text{n}\space\ge\space1}\frac{\left(-1\right)^{2\left(\text{n}-1\right)}}{\left(2\text{n}-1\right)^\text{s}}+\sum_{\text{n}\space\ge\space1}\frac{\left(-1\right)^{2\text{n}-1}}{\left(2\text{n}\right)^\text{s}}\tag1$$

Now, we know that $\left(-1\right)^{2\left(\text{n}-1\right)}=1\space\forall\space\text{n}\in\mathbb{N}$ and $\left(-1\right)^{2\text{n}-1}=-1\space\forall\space\text{n}\in\mathbb{N}$, so we can write:

$$\eta\left(\text{s}\right)=\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}-1\right)^\text{s}}-\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}\right)^\text{s}}\tag2$$

Now, Let's add and subtract the same thing on the RHS:

$$ \begin{alignat*}{1} \eta\left(\text{s}\right)&=\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}-1\right)^\text{s}}-\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}\right)^\text{s}}+\underbrace{\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}\right)^\text{s}}-\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}\right)^\text{s}}}_{=\space0}\\ \\ &=\underbrace{\underbrace{\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}-1\right)^\text{s}}}_\text{odd part}+\underbrace{\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}\right)^\text{s}}}_\text{even part}}_{\text{odd part}\space+\space\text{even part}}-\left\{\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}\right)^\text{s}}+\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}\right)^\text{s}}\right\}\\ \\ &=\sum_{\text{n}\space\ge\space1}\frac{1}{\text{n}^\text{s}}-2\sum_{\text{n}\space\ge\space1}\frac{1}{\left(2\text{n}\right)^\text{s}} \end{alignat*} \tag3 $$

Now, by definition, the Riemann zeta function is given by:

$$\zeta\left(\text{s}\right):=\sum_{\text{n}\space\ge\space1}\frac{1}{\text{n}^\text{s}}\tag4$$

And we can use the fact that:

$$\frac{1}{\left(2\text{n}\right)^\text{s}}=\frac{1}{2^\text{s}\cdot\text{n}^\text{s}}=\frac{1}{2^\text{s}}\cdot\frac{1}{\text{n}^\text{s}}=\frac{2^{-\text{s}}}{\text{n}^\text{s}}\tag5$$

So:

$$ \begin{alignat*}{1} \eta\left(\text{s}\right)&=\zeta\left(\text{s}\right)-2\sum_{\text{n}\space\ge\space1}\frac{2^{-\text{s}}}{\text{n}^\text{s}}\\ \\ &=\zeta\left(\text{s}\right)-2\cdot2^{-\text{s}}\sum_{\text{n}\space\ge\space1}\frac{1}{\text{n}^\text{s}}\\ \\ &=\zeta\left(\text{s}\right)-2^{1-\text{s}}\sum_{\text{n}\space\ge\space1}\frac{1}{\text{n}^\text{s}}\\ \\ &=\zeta\left(\text{s}\right)-2^{1-\text{s}}\zeta\left(\text{s}\right)\\ \\ &=\zeta\left(\text{s}\right)\left(1-2^{1-\text{s}}\right) \end{alignat*} \tag6 $$

Jan Eerland
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    Thank you for your answer – Jutrop May 07 '20 at 12:19
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    However, this relation also holds for $0<\Re(s)<1$, but the equation of $\zeta$ you used only works for $\Re(s)>1$. I'm pretty sure there's a way to fix that, but still. – DUO Labs May 08 '20 at 20:29
  • For completion: $$\zeta(s)=\frac{1}{1-2^{1-s}} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^s} \operatorname{Re}(s)>0 \wedge s \neq 1$$ – alex.peter Mar 09 '24 at 18:42