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Let $B \ge 2$ be integer and $[x]$ denote the nearest integer to real $x$.

For $2 \le B \le 10^5$ computations with mpmath suggest:

$$ [\zeta(1-1/B)]=-B+1 \qquad (1)$$

Is (1) true for all $B \ge 2$?

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  • $\begingroup$ $\zeta(1+x)=\frac{1}{x}+\gamma+o(1)$. $\endgroup$
    – Wojowu
    Commented Sep 25, 2021 at 13:04
  • $\begingroup$ @Wojowu Many thanks, I suppose this would require some additional effort to get rid of $+\gamma$. $\endgroup$
    – joro
    Commented Sep 25, 2021 at 13:13
  • $\begingroup$ You don't want to get rid of it - when the $o(1)$ term gets small enough, this implies $\frac{1}{x}<\zeta(1+x)<\frac{1}{x}+1$. To see that it is good enough for $B\geq 2$ would require an explicit estimate on $o(1)$ which I will leave to someone else. $\endgroup$
    – Wojowu
    Commented Sep 25, 2021 at 13:43
  • $\begingroup$ @Wojowu I suspect the nearest integer might be outside the bounds. $\endgroup$
    – joro
    Commented Sep 25, 2021 at 17:03

1 Answer 1

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We can make the error mentioned by Wojowu in his comment explicit by using some results on the Laurent coefficients of the zeta function. There are a few results on this, but I'll just use Theorem 2 of this paper by Berndt (which gives a more general result for the Hurwitz zeta function). The relevant result is

Theorem. (Berndt) Write $$ \zeta(s) = \frac{1}{s-1} + \sum_{n=0}^\infty \gamma_n (s-1)^n $$ for the Laurent series expansion of $\zeta$ around $s=1$. Then for $n\geq 1$, we have $$ \left|\gamma_n\right| \leq \frac{4}{n\pi^n}. $$ For $B\geq 2$ an integer, write $$ \zeta\left(1-\frac{1}{B}\right) = -B + \gamma + R(B), $$ where $\gamma=\gamma_0\approx .577...$ is the usual Euler-Mascheroni constant and $$ R(B) = \sum_{n=1}^\infty \gamma_n \left(-\frac{1}{B}\right)^n. $$ Applying the triangle inequality and Berndt's result, we have $$ \left|R(B)\right| \leq \sum_{n=1}^\infty \frac{4}{n(B\pi)^n} \leq \frac{4}{B}\sum_{n=1}^{\infty} \frac{1}{n\pi^n}. $$ The sum on the right converges rapidly and is numerically $\approx 0.383...$. Thus $$ \left| R(B) \right| \leq \frac{2}{B}. $$ (By a more careful analysis, the constant 2 can be replaced with 1.) For $B \geq 26$, it follows that $$ \frac{1}{2} < \gamma+ R(B) < 1. $$ Therefore $$ \left[ \zeta\left(1-\frac{1}{B} \right) \right] = -B+1 $$ for $B \geq 26$. The remaining cases can be verified via computer.

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  • $\begingroup$ Fractional parts seem to decrease, this might suggest a proof working for these cases too...$$\begin{array}{rl}1 & -0.5 \\ 2 & -1.46035 \\ 3 & -2.44758 \\ 4 & -3.44129 \\ 5 & -4.43754 \\ 6 & -5.43505 \\ 7 & -6.43328 \\ 8 & -7.43196 \\ 9 & -8.43093 \\ 10 & -9.43011 \\ 11 & -10.4294 \\ 12 & -11.4289 \\ 13 & -12.4284 \\ 14 & -13.428 \\ 15 & -14.4277 \\ 16 & -15.4274 \\ 17 & -16.4271 \\ 18 & -17.4268 \\ 19 & -18.4266 \\ 20 & -19.4264 \\ 21 & -20.4263 \\ 22 & -21.4261 \\ 23 & -22.426 \\ 24 & -23.4258 \\ 25 & -24.4257 \\ 26 & -25.4256\end{array}$$ $\endgroup$ Commented Sep 26, 2021 at 5:37
  • $\begingroup$ (In fact these fractional parts seem to monotonously converge to $1-\gamma$) $\endgroup$ Commented Sep 26, 2021 at 7:50
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    $\begingroup$ @მამუკაჯიბლაძე In problems like these where explicit estimates are needed, there's always a balance to strike between technical simplicity and range of applicability. I'm sure there's a proof that works for these small values, but it's (likely) not going to be as neat as the method I used. Such technical scruples would probably only obscure or complicate the analysis, rather than simplifying it. Since there are so few exceptional cases to check, I think technical simplicity and aesthetically pleasing inequalities are preferable to having a wider range of applicability. $\endgroup$ Commented Sep 26, 2021 at 22:30

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