Elementary bound for the reciprocal of the Riemann zeta function for real $s \ge 2$

Question

I was playing around with $\zeta(s)^{-1}$ and after plugging in some values for $s$, it seems like we have $$\zeta(s)^{-1} = 1 - \epsilon(s)$$ for $\epsilon(s) < 2^{-s}$ where $s \in [2,\infty)$? To rephrase the question: for every real $s \ge 2$, can we show that $\zeta(s)^{-1} > 1 - 2^{-s}$? For now, using the well-known formula: $$\zeta(s)^{-1} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$$ I was able to get a bound like $\zeta(s)^{-1} > 1 - 3\cdot 2^{-s}$ for every $s \in [2,\infty)$ by the following steps: $$\frac{1 - \zeta(s)^{-1}}{2^{-s}} = \sum_{n \ge 2}\frac{-\mu(n)}{(n/2)^{s}} \leq \sum_{n \ge 2}\frac{1}{(n/2)^{2}} = 4(\zeta(2) - 1) < 3.$$ I wonder if anyone could show me a way to get rid of the $3$ or find a $s \in [2,\infty)$ such that $\zeta(s)^{-2} \leq 1 - 2^{-s}$?

Answer 1

The inequality you are aiming for fails for all $s>1$, because $$(1-2^{-s})^{-1}=1+2^{-s}+4^{-s}+8^{-s}+\ldots <1+2^{-s}+3^{-s}+4^{-s}+\ldots =\zeta(s)\,.$$