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Let $s = \sigma+ i t$ with $0\leq \sigma\leq 1$, $|t|\leq X$, where $1\leq X<2$.

It is easy to use the Euler-Maclaurin formula to prove a result of the form $$\left|\zeta(s) - 1 - \frac{X^{1-s}}{s-1}\right| \leq \frac{c}{X^\sigma},$$ where $c$ is a constant.

The exact value of $c$ depends on the order up to which one takes Euler-Maclaurin and several choices one can make in estimating the terms. Amusingly, my collaborators and I can get values for $c$ very close but slightly greater than $2/3$.

What is the best $c$? Is it perhaps $2/3$? The issue could in principle be solved computationally (at least if the answer is "no"), though the fact that there are three parameters ($\sigma$, $t$ and $X$) makes that approach cumbersome. It would be nicer to have a non-brute-force proof.

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One can in fact get a $c$ better than $2/3$. Sketch:

  • for $X\leq 11/10$, do an Euler-Maclaurin expansion up to order $3$, and use the fact that $B_3(x)$ is small for $x$ small ($0\leq x\leq 1/10$;

  • for $X>11/10$, do an Euler-Maclaurin expansion up to order $4$, and use the lower bound on $X$ to better bound the factors $|s/X|$, $|s (s+1)/X^2|$ and $|s (s+1) (s+2)/X^3|$.

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