Evaluating $\zeta(s)$ for Re(s)>-3 I found this one here:Evaluating $\zeta'(2)$
$$\zeta(s) =  \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \right) \ , \ \text{Re}(s) >-3.$$
Iam interested in proving this, but i don't know how to approach, Can some one help me. Thank you

For $\Re(s) > 1$ it is obvious that it is true, then you need to prove that the RHS stays analytic (away from $1$) for $\Re(s) > -3$, so that it is the analytic continuation of $\sum_{k\ge 1} k^{-s},\Re(s) > 1$.
I think the easiest way is, with $$f_m(s)=  \frac{m^{1-s}}{1-s} + \frac{m^{-s}}{2} - \frac{sm^{-s-1}}{12} $$
to show that $$
f_m(s)-f_{m-1}(s) = m^{-s}+O((1+|s|)^4 m^{-s-4})\tag{1}$$
which follows from the binomial expansion $$m^z-(m-1)^z = m^z (1-(1-1/m)^z)=-m^z (\sum_{n=1}^N {z\choose n} (-m)^{-n} + O((1+|z|)^{N+1} m^{-N-1}))$$
From $(1)$ you'll get that $$  \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - f_m(s) \right)$$ converges locally uniformly for $\Re(s) > -3,s\ne 1$ so that it is analytic there.
This generalizes to $\Re(s)>-N$ with the Euler Mclaurin summation formula.
