-2
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Sep 16, 2018 at 20:55
  • $\begingroup$ $\zeta(s) \in ( 2^{-s}+\frac{3^{1-s}}{s-1}, 2^{-s}+3^{-s}+\frac{3^{1-s}}{s-1})$ $\endgroup$
    – reuns
    Sep 17, 2018 at 3:11

1 Answer 1

0
$\begingroup$

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)).

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – tobias
    Sep 16, 2018 at 22:46
  • $\begingroup$ 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. $\endgroup$
    – tobias
    Sep 16, 2018 at 23:22
  • $\begingroup$ 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 $$\sum_{n<N}n^{-s}+\frac{N^{1-s}}{s-1}<\zeta(s)<\sum_{n\le N}n^{-s}+\frac{N^{1-s}}{s-1}$$ $\endgroup$ Sep 17, 2018 at 4:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.