Boundedness of Riemann zeta function: for every $a>1$ there is $\frac{1}{a-1} \le \zeta(a) \le \frac{a}{a-1}$

Question

I've got a question. How to prove that for every $a>1$ there is $\frac{1}{a-1} \le \zeta(a) \le \frac{a}{a-1}$?

Thanks!

Answer 1

For every $a>1$ we have: $$ \zeta(a)=\frac{1}{\Gamma(a)}\int_{0}^{+\infty}\frac{x^{a-1}}{e^{x}-1}\,dx$$ but if $x\in\mathbb{R}^+$, $\frac{1}{e^x-1}\geq\frac{1}{x e^{x}}$, from which the inequality $\zeta(a)\geq\frac{1}{a-1}$ readily follows.

On the other hand, we also have the inequality $\frac{1}{e^{x}-1}\leq \frac{1}{x e^{x/2}}$, from which it follows that: $$ \zeta(a)\leq \frac{2^{a-1}}{a-1},$$ that is stronger than the given inequality for $a\in (1,2)$.

As an alternative, we also have: $$\zeta(a)= \frac{1}{1^a}+\frac{1}{2^a}+\ldots \geq \int_{1}^{+\infty}\frac{dx}{x^a} = \frac{1}{a-1}$$ and: $$\zeta(a)-1= \frac{1}{2^a}+\frac{1}{3^a}+\ldots\leq \int_{1}^{+\infty}\frac{dx}{x^a} = \frac{1}{a-1}$$ since $\frac{1}{x^a}$ is a decreasing and convex function on $\mathbb{R}^+$.