Proving the relation between the Dirichlet eta function and the Riemann zeta function 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.
 A: 
Assignment:
Prove the following fact:
$$\eta\left(\text{s}\right):=\sum_{\text{n}=1}^\infty\frac{\left(-1\right)^{\text{n}-1}}{\text{n}^\text{s}}=\zeta\left(\text{s}\right)\left(1-2^{1-\text{s}}\right)$$
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:
$$\eta\left(\text{s}\right)=\underbrace{\sum_{\text{n}=1}^\infty\frac{\left(-1\right)^{2\text{n}-1-1}}{\left(2\text{n}-1\right)^\text{s}}}_\text{odd part}+\underbrace{\sum_{\text{n}=1}^\infty\frac{\left(-1\right)^{2\text{n}-1}}{\left(2\text{n}\right)^\text{s}}}_\text{even part}=\sum_{\text{n}=1}^\infty\frac{\left(-1\right)^{2\left(\text{n}-1\right)}}{\left(2\text{n}-1\right)^\text{s}}+\sum_{\text{n}=1}^\infty\frac{\left(-1\right)^{2\text{n}-1}}{\left(2\text{n}\right)^\text{s}}\tag1$$
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:
$$\eta\left(\text{s}\right)=\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}-1\right)^\text{s}}-\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}\right)^\text{s}}\tag2$$
Now, Let's add and subtract the same thing on the RHS:
$$\eta\left(\text{s}\right)=\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}-1\right)^\text{s}}-\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}\right)^\text{s}}+\underbrace{\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}\right)^\text{s}}-\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}\right)^\text{s}}}_{=\space0}=$$
$$\underbrace{\underbrace{\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}-1\right)^\text{s}}}_\text{odd part}+\underbrace{\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}\right)^\text{s}}}_\text{even part}}_{\text{odd part}\space+\space\text{even part}}-\left\{\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}\right)^\text{s}}+\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}\right)^\text{s}}\right\}=\sum_{\text{n}=1}^\infty\frac{1}{\text{n}^\text{s}}-2\sum_{\text{n}=1}^\infty\frac{1}{\left(2\text{n}\right)^\text{s}}\tag3$$
Now, by definition, the Riemann zeta function is given by:
$$\zeta\left(\text{s}\right):=\sum_{\text{n}=1}^\infty\frac{1}{\text{n}^\text{s}}\tag4$$
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:
$$\eta\left(\text{s}\right)=\zeta\left(\text{s}\right)-2\sum_{\text{n}=1}^\infty\frac{2^{-\text{s}}}{\text{n}^\text{s}}=\zeta\left(\text{s}\right)-2\cdot2^{-\text{s}}\sum_{\text{n}=1}^\infty\frac{1}{\text{n}^\text{s}}=\zeta\left(\text{s}\right)-2^{1-\text{s}}\sum_{\text{n}=1}^\infty\frac{1}{\text{n}^\text{s}}=$$
$$\zeta\left(\text{s}\right)-2^{1-\text{s}}\zeta\left(\text{s}\right)=\zeta\left(\text{s}\right)\left(1-2^{1-\text{s}}\right)\tag6$$
