Let $$\zeta(s) = \sum_{n = 1}^\infty \frac{1}{n^s}$$ be the Riemann zeta function (here we just consider real $s$). We do have a description given by $$\zeta(s) = \frac{s}{s-1}-s\int_{1}^\infty \frac{u-[u]}{u^{s+1}} du$$ Can we deduce by this formula or another one some fairly "precise" estimates for giving upper and lower bounds for $\zeta(s)$ for all natural numbers $s$ (or just an infinite subset of the natural numbers).
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$\begingroup$ I assume that the Euler-Maclaurin summation formula would do the trick. I would consult Montgomery & Vaughan's Multiplicative Number Theory, Appendix B, for details. $\endgroup$– Craig FranzeCommented Sep 16, 2018 at 20:55
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$\begingroup$ $\zeta(s) \in ( 2^{-s}+\frac{3^{1-s}}{s-1}, 2^{-s}+3^{-s}+\frac{3^{1-s}}{s-1})$ $\endgroup$– reunsCommented Sep 17, 2018 at 3:11
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Using the Euler-Maclaurin summation formula gives \begin{multline} \zeta(s)=\sum_{n=1}^{N}n^{-s}+\frac{N^{1-s}}{s-1}+N^{-s}\sum_{k=1}^{K}\binom{s+k-2}{k-1}B_k N^{-k+1}/k\\ -\binom{s+K-1}{K}\int_{N}^{\infty}B_K(\{x\})x^{-s-K}dx, \end{multline} where $B_K$ refers to the Bernoulli polynomial. Details can be found in Montgomery and Vaughan's Multiplicative Number Theory book (Appendix B; this is established in equation (B.24)).
answered Sep 16, 2018 at 21:20
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$\begingroup$ Thanks Craig, that is very helpful. But right now I am looking for some very simple, yet precise upper and lower bound estimates (so, for instance, without any Bernoulli terms). Do you know some results pointing into that direction? $\endgroup$– tobiasCommented Sep 16, 2018 at 22:46
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$\begingroup$ For instance, it would be great to have some estimate of the form $1+a(s) \leq \zeta(s) \leq 1+b(s)$ for all natural numbers $s$, where $a(s)$ and $b(s)$ are asymptotically equivalent. $\endgroup$– tobiasCommented Sep 16, 2018 at 23:22
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$\begingroup$ You could just use the special case where $K=1$; note that $B_1=-1/2$, and $|B_1(\{x\})|\le 1/2$, and so $$\sum_{n<N}n^{-s}+\frac{N^{1-s}}{s-1}<\zeta(s)<\sum_{n\le N}n^{-s}+\frac{N^{1-s}}{s-1}$$ $\endgroup$ Commented Sep 17, 2018 at 4:49