Speed of decay of $\zeta(x)-1$ as $x \to \infty$

Question

I am trying to find some numerical bound on the Riemann zeta function $$ \zeta(x) = \sum_{k\ge 1} 1/n^x. $$ I am only interested in the case which $x > 1$, so the above expression is valid. More precisely, what is the decay of the function $$ \zeta(x)-1 \text{ as } x\to \infty? $$ I suppose that the next term is of order $1/2^x$ (due to the definition of the zeta function), but can I also control the constant? That is, do I have the existence of a positive constant $C>0$ such that $$ |\zeta(x)-1| \le C e^{- (\ln 2) \cdot x} $$ and if so, is the value of $C$ known?

Answer 1

The first term is $1$, the next is $2^{-x}$. When $x > 2$ (say), if $n \ge 3$ $$n^{-x} = 3^{-x} (n/3)^{-x} < 3^{-x} (n/3)^{-2}$$ so since $\sum_n n^{-2}$ converges $$ \sum_{n=3}^\infty n^{-x} < c 3^{-x}$$ for some positive constant $c$. Thus $\zeta(x) = 1 + 2^{-x} + O(3^{-x})$.