I'm try to know if the precison of Euler-McLaurin summation depends by the index N inside the formula as follow:
$\zeta(s)_N = \sum_{k=1}^{N} k^{-s} + \frac{N^{1-s}}{1-s} + \int_{N}^{\infty} \frac{x-[x]}{x^{s+1}} dx$
where $[x]$ is the integer part of x and N arbitrary integer $\geq 1$.
In particolar given two $N_1 < N_2$ I'm asking if $|\zeta(s)_{N_1} -\zeta(s)| > |\zeta(s)_{N_2}-\zeta(s)|$ and if there is some advantage choose bigger N or at least is indipendent?
Thanks for cooperation
GM
It seems to be more than intuitive if you remember thay, if $$t = [t] + \{t \}$$ $$\int_0^x \{t \}^n\,dt=\frac {x- \{x \}+\{x \}^{n+1}}{n+1 }=\frac {[x]+\{x \}^{n+1}}{n+1 }$$
Just try for $s=\pi$ and let $$A_N=\sum_{k=1}^{N} k^{-\pi} - \frac{N^{1-\pi}}{\pi-1}\quad \quad \text{and}\quad \quad B_N=\int_{N}^{\infty} \frac{x-[x]}{x^{\pi+1}} dx$$
$$\left( \begin{array}{cccc} N & A_N & B_N & \zeta(\pi)_N=A_N+B_N \\ 5 & 1.14935 & 9.086\times 10^{-4} & 1.15026 \\ 10 & 1.16984 & 1.087\times 10^{-4} & 1.16995 \\ 15 & 1.17351 & 3.087\times 10^{-5} & 1.17354 \\ 20 & 1.17475 & 1.253\times 10^{-5} & 1.17477 \\ 25 & 1.17531 & 6.291\times 10^{-6} & 1.17532 \\ \end{array} \right)$$