<s>[INST] Product of two series to get a series decomposition of zeta in the critical strip $\def\sfrac#1#2{%
    \small#1%
    \kern-.05em\lower0.1ex/\kern-.025em%
    \lower0.4ex\small#2}$I've been working on gaining an intuitive understanding of the analytic continuation of the zeta function, but I've gotten stuck at this part where I have to multiply two very strange series together.
The approach is quite simple: first start with the Dirichlet eta function, defined by $η(s) = \frac{1}{1^s} + \frac{-1}{2^s} + \frac{1}{3^s} + \frac{-1}{4^s} + ... $. Then the following relationship holds:
$ζ(s) = η(s) · \frac{1}{1-\frac{2}{2^s}}$
If $\Re\{s\} > 1$, then the right factor can be expanded into a Taylor series, yielding
$ζ(s) = \left(\frac{1}{1^s} + \frac{-1}{2^s} + \frac{1}{3^s} + \frac{-1}{4^s} + ... \right) \cdot \left(\frac{1}{1^s} + \frac{2}{2^s} + \frac{4}{4^s} + \frac{8}{8^s} + ... \right) \\
= \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + ...$
where the product can be evaluated as a Dirichlet convolution.
This is because the image of $\frac{2}{2^s}$ lies within the radius of convergence of the $1 + x + x^2 + x^3 + ...$ series expansion of $\frac{1}{1-x}$, so the Taylor expansion is valid.
However, if $\Re\{s\} < 1$, then the image of $\frac{2}{2^s}$ lies outside the ROC, so that expansion doesn't work. Instead, we can take advantage of the functional equation
$f(x) + f(\frac{1}{x}) = 1$
where $f(x) = \frac{1}{1-x}$.
Substituting in $x=\frac{2}{2^s}$, and allowing that $\frac{1}{\left(\frac{2}{2^s}\right)} = \frac{\sfrac{1}{2}}{{\sfrac{1}{2}}^s}$, you get
$\frac{1}{1-\frac{2}{2^s}} = 1 - \frac{1}{1-\frac{\sfrac{1}{2}}{{\sfrac{1}{2}}^s}}$
where $\frac{\sfrac{1}{2}}{{\sfrac{1}{2}}^s}$ now has an image lying within the ROC of the aforementioned series expansion for $\Re\{s\} < 1$. So we can expand the whole right side as 
$\frac{1}{1-\frac{2}{2^s}} = \frac{-\sfrac{1}{2}}{{\sfrac{1}{2}}^s} + \frac{-\sfrac{1}{4}}{{\sfrac{1}{4}}^s} + \frac{-\sfrac{1}{8}}{{\sfrac{1}{8}}^s} + \frac{-\sfrac{1}{16}}{{\sfrac{1}{16}}^s} + ...$
Finally, putting the whole thing together, we get
$ζ(s) = \left(\frac{1}{1^s} + \frac{-1}{2^s} + \frac{1}{3^s} + \frac{-1}{4^s} + ... \right) \cdot \left(\frac{-\sfrac{1}{2}}{{\sfrac{1}{2}}^s} + \frac{-\sfrac{1}{4}}{{\sfrac{1}{4}}^s} + \frac{-\sfrac{1}{8}}{{\sfrac{1}{8}}^s} + \frac{-\sfrac{1}{16}}{{\sfrac{1}{16}}^s} + ...\right)$
Which, for $0 < \Re\{s\} < 1$, is the product of a conditionally convergent Dirichlet series, and an absolutely convergent "fractional Dirichlet series."
And here I'm stumped. How can I expand this product? I understand the result should be some kind of "fractional Dirichlet series" where the denominators are dyadic rationals raised to the power of s, and I understand I basically want to perform some kind of Dirichlet convolution type thing here.
But how do I actually do it? What does the resulting expression look like?
 [/INST] the series for $\eta(s)$ is absolutely convergent for $\Re(s) > 0$ if you group the terms by two :
$\eta(s) = \displaystyle\sum_{n=1}^\infty (2n-1)^{-s} - (2n)^{-s} = \sum_{n=1}^\infty \mathcal{O}(s (2n)^{-s-1})$ ( from the Taylor expansion of order 1 of $(1-x)^{-s}$ when $x \to 0$)
$\eta(s) = (1-2^{1-s}) \ \zeta(s)$ and for $\displaystyle \Re(s) < 1 : \ \ \frac{1}{1-2^{1-s}} = -\frac{2^{s-1}}{1-2^{s-1}} = - \sum_{k=1}^\infty 2^{k(s-1)}$.
you get :
$$\zeta(s) = - \sum_{k=1}^\infty 2^{k(s-1)} \eta(s) = - \sum_{n,k} \left( (2n-1)^{-s} - (2n)^{-s} \right) 2^{k(s-1)}$$ 
which is an absolutely convergent double sum for $\Re(s) \in ]0;1[$. 
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