Need help showing that $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$

Question

This question is I am working on is the extension of the domain of zeta function.
$\zeta(x)$=$1+\frac{1}{2^x}+\frac{1}{3^x}+\frac{1}{4^x}+\frac{1}{5^x}+\frac{1}{6^x}+ \cdots$
$\eta(x)$=$1-\frac{1}{2^x}+\frac{1}{3^x}-\frac{1}{4^x}+\frac{1}{5^x}-\frac{1}{6^x}+ \cdots$
Note: The zeta $\zeta(x)$ function consists of positive terms only and the eta $\eta(x)$ function converges for all positive values for $x$ even when $x$ is less than $0$ where $0\lt x \lt 1$. Also in the equation
$\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$, the left hand side is only defined when $x\gt 1$ because that is the domain of the zeta series. The right hand side can be used to define the zeta function for $0\lt x \lt 1$ and the right hand side totally depends on $\eta(x)$ in the function because it is already defined for all positive values for $x$.

Here is my attempt so far:
$({1\over 1-2^{1-x}})$ $\eta (x)$
$\qquad\qquad\qquad=({1\over 1-2^{1-x}})\cdot 1 -({1\over 1-2^{1-x}})(\frac{1}{2^x})+({1\over 1-2^{1-x}})(\frac{1}{3^x})-({1\over 1-2^{1-x}})(\frac{1}{4^x})+({1\over 1-2^{1-x}})(\frac{1}{5^x})-({1\over 1-2^{1-x}})(\frac{1}{6^x})+\cdots$
$\qquad\qquad\qquad=(\frac{2^x}{2^x-2})-(\frac{1}{2^x-2})$$+$[$({2/3})^{x}\over{2^x-2}$]$-$($2^{-x}\over{2^x-2}$)$+$[$({2/5})^{x}\over{2^x-2}$]$-$($3^{-x}\over{2^x-2}$)$+\cdots$

By adding each pair of terms from above we get(note: I attempted to input the info but its too difficult to do due to the complexity from combining the terms but I have the next step written down and after that I am stuck showing it). Need help showing that $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$. Getting cancellations is hard

Answer 1

An idea for you to develop:

$$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}=\sum_{n=1}^\infty\frac1{(2n)^s}+\sum_{n=1}^\infty\frac1{(2n-1)^s}=\frac1{2^s}\zeta(s)+\sum_{n=1}^\infty\frac1{(2n-1)^s}\implies\ldots$$

Observe you're going to get lots of "poles"...which in fact are removable singularities.

Answer 2

As i said in comment: $\zeta (x) - \eta (x) = \sum\limits_{n=1}^{\infty}\frac{1}{n^x} - \sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}=\sum\limits_{n=1}^{\infty}(\frac{1}{n^x} - \frac{(-1)^{n-1}}{n^x})=2\sum\limits_{n=1}^{\infty}\frac{1}{(2n)^x}=2^{1-x}\sum\limits_{n=1}^{\infty}\frac{1}{n^x}=2^{1-x}\zeta (x) $

Hence: $\zeta (x)(1 - 2^{1-x}) - \eta(x)=0\\ \zeta(x)=\frac{\eta(x)}{1-2^{1-x}}$