I am looking for series representation of Riemann Zeta function for $\operatorname{Re}(s)<0$ or $\operatorname{Re}(s)>-3$ or $\operatorname{Re}(s)>-4$ whatever is simplier.
Maybe something like from this wikipedia page here or equation 20 from here which works only for $\operatorname{Re}(s)>0$.
$$\zeta(s)=\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$$
UPDATE 1 (04/11/2019)
Finally, I had some time to check it.
I tried this, based on the answer from reuns, when $\Re(s) > -k$: $$\zeta(s) = \frac{\eta(s)}{1-2^{1-s}} = \frac{1}{1-2^{1-s}} \Big( \sum_{j=1}^{k+1} 2^{-j} f_{j-1}(1,s)+ 2^{-k-1} \sum_{n=1}^\infty (-1)^{n+1} f_{k+1} (n,s))\Big)$$
and the numerical calculation does not match with the library function. This is a screenshot from Wolfram Mathematica, which I used for numerical check. I tried to calculate the value of $\zeta(s=0.5+15i)$. I put some comments in order to explain it. Also, instead of infinity, I calculated the sum up to 10000, but if I increase it, it returns more or less the same value.
Let $$f_0(n,s) = n^{-s}, \qquad f_k(n,s) = f_{k-1}(n,s)-f_{k-1}(n+1,s)= \sum_{m=0}^k {k \choose m} (-1)^m (n+m)^{-s}$$ Using the binomial series $(n+a)^{-s} = n^{-s} \sum_{l=0}^\infty {-s \choose l} (a/n)^l$ show that $f_k(n,s) = O(n^{-s-k})$ and $\partial_s f_k(n,s) = O(n^{-s-k}\log n)$
Note $$\sum_{k=1}^n (-1)^{k+1} = \frac{1+(-1)^{n+1}}{2}$$
By induction if for $\Re(s) > 1$ $$\eta(s) = \sum_{j=1}^k 2^{-j}f_{j-1}(1,s) + 2^{-k} \sum_{n=1}^\infty (-1)^{n+1} f_k (n,s)$$ then $$\eta(s) = \sum_{j=1}^k 2^{-j} f_j(1,s) + 2^{-k} \sum_{n=1}^\infty (\sum_{k=1}^n (-1)^{k+1} )( f_k (n,s)- f_k (n+1,s))$$ $$ = \sum_{j=1}^{k+1} 2^{-j} f_{j-1}(1,s)+ 2^{-k-1} \sum_{n=1}^\infty (-1)^{n+1} f_{k+1} (n,s)$$ the latter is absolutely convergent and holomorphic thus analytic for $\Re(s) > -k$ thus it is the analytic continuation of $\eta(s) = (1-2^{1-s}) \zeta(s)$