asymptotic growth of zeta function on the real line

Question

let $r$ be real $r> 1$ then $$\zeta(r) = O\left(1+\frac{1}{r-1}\right)$$

Can you tell me how to prove this formula?

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I found this after posting $$\sum_{n \ge 1} \frac{1}{n^r} = 1 + \sum_{n \ge 2} \frac{1}{n^r} \le 1 + \int_2^\infty \frac{dx}{x^r} = 1 - \frac{1}{r-1} \frac{1}{2}$$ and it seems a right idea but the sign is wrong.

Answer 1

Sure, it comes from Laurent's formula $$ \zeta(r) = \frac{r}{r-1} - r\int_1^\infty \frac{\{x\}}{x^{1+r}}dx$$