Just as an observation of what I saw in an informative video about the analytical continuation of the Riemann Zeta function, is it possible to define the continuation for complex numbers $s\in\mathbb{C}$ with $\Re(s)<1$ as simply $f(\zeta(f(s))$ where $f$ is a function that reflects its argument across the point $1+0i$ or across the line crossing 1 on the real axis?
There may be a way to actually formuate $f$. I would translate, then rotate $\pi$ radians, then translate back.
Ignoring reflections and the function $f$, is there a way to take $s\in\mathbb{C}$ with $\Re(s)<1$, transform it over to the other side where the traditional $\zeta$ function defined, evaluate it there, then transform it back?
There just seems to me to be some symmetry between where it's traditionally defined and where it's continued. I'm guessing the answer to my question is a big fat "No", because the formula for the analytic continuation looks rather complicated, but it does appear to be written in terms of the original $\zeta$ function.
You might want to look at the functional equation.
But before that, you should look at the simplest analytic continuation :
Define inductively $$a_0(n,s) =n^{-s}, \qquad a_{k+1}(n,s) = a_{k}(n,s)-a_k(n+1,s)$$ so that $$a_1(n,s) = n^{-s}-(n+1)^{-s}, \quad a_2(n,s) = n^{-s}-2(n+1)^{-s}+(n+2)^{-s}, \\ a_k(n,s) = \sum_{m=0}^{k} {k \choose m} (-1)^m (n+m)^{-s}$$
Then by summation by parts, using that $b_n = (-1)^{n+1} \implies b_n-b_{n+1} = 2b_n$ and $a_k(n,s) = \mathcal{O}(n^{-s-k})$ : $$\begin{array}{lll}\zeta(s) &=& \sum_{n=1}^\infty a_0(n,s)& \text{converges for } Re(s) > 1\\ \eta(s) &= & \sum_{n=1}^\infty (-1)^{n+1} a_1(n,s) & \text{converges for } Re(s) > 0\\ 2\eta(s) &= &\sum_{n=1}^\infty (-1)^{n+1} a_2(n,s)& \text{converges for } Re(s) > -1\\ & & &\vdots \\ 2^k\eta(s) &=& \sum_{n=1}^\infty (-1)^{n+1} a_{k+1}(n,s) & \text{converges for } Re(s) > -k\end{array}$$