Zeta function's asymptotic $\zeta(\sigma+it) = \mathcal{O}(|t|^{1-\sigma+\epsilon})$ I have one question about some asymptotic property of Zeta functions.
(Proposition 2.7-Ch 6) If $\sigma,t\in \mathbb R$, $|t|\ge1$, $0\le\sigma_0\le1$ and $\sigma_0\le\sigma$, then for every $\varepsilon>0$, there exists a constant $c_\varepsilon$ such that $|\zeta(s)|\le c_\varepsilon|t|^{1-\sigma_{0}+\varepsilon}$ where $s=\sigma+it$. 
I tried the proof of this theorem,but I couldn't have more steps from below thing. 
$|\zeta(s)|\le |\frac{1}{s-1}|+2|s|^{1-\sigma_0+\varepsilon} \sum_{n=1}^{\infty}{\frac{1}{n^{1+\varepsilon}}}$ .
The first part is easily transformed to the form of $c_{\varepsilon}|t|^{1-\sigma_{0}+\epsilon}$. However, the second summation is difficult.
How can I change the second part to the form of $c_{\varepsilon}|t|^{1-\sigma_{0}+\epsilon}$ ? I have a trouble because of the term of $|s|$.  
 A: Steins's method (p.174) is to use that

with $\displaystyle\delta_n(s) = \int_n^{n+1}(n^{-s}-x^{-s})dx = \int_n^{n+1}\int_n^x s t^{-s-1}dtdx\tag{1}$ for $Re(s) > 1$ and by analytic continuation for $Re(s) > 0$ :
  $\displaystyle\zeta(s)=\frac{1}{s-1}+\sum_{n=1}^\infty \delta_n(s)\tag{2}$

Now $(1)$ gives two inequalities : $|\delta_n(s)| \le |s| n^{-Re(s)-1}$ and  $|\delta_n(s)| \le 2n^{-Re(s)}$ so that for any $a \in(0,1)$ : $$|\delta_n(s)| \le (2n^{-Re(s)})^{1-a}(|s| n^{-Re(s)-1})^a=|s|^a 2^{1-a}n^{-Re(s)-a}\tag{3}$$
and summing for $Re(s) > 1-a>0$ :
$$\left|\zeta(s)-\frac{1}{s-1}\right| \le \sum_{n=1}^\infty |\delta_n(s)| \le|s|^a 2^{1-a} \sum_{n=1}^\infty n^{-Re(s)-a}=|s|^a 2^{1-a}\zeta(a+Re(s))\tag{4}$$
i.e. for $|t| > 1$ and $1 > Re(s) =\sigma > 1-a>0$ :
$$|\zeta(\sigma+it)| \le \frac{1}{\sqrt{(a-1)^2+1}}+(t^2+\sigma^2)^{a/2} 2^{1-a}\zeta(a+\sigma)< 4|t[^a \zeta(a+\sigma)\tag{5}$$
A: @user1952009, I attatched the full answer your questions and my questions. 
Could you review this?
Given conditions are $\sigma,t\in \mathbb R$, $|t|\ge1$, $0\le\sigma_0\le1$ and $\sigma_0\le\sigma$. 
Now, I will prove this. 
First, $\frac{1}{|s-1|} \le 1/|t| = \frac{|t|^{1-\sigma_0+\varepsilon}}{|t|^{2-\sigma_0+\varepsilon}} \le |t|^{1-\sigma_0+\varepsilon}$ 
as $|t|\ge1$. 
Now, we have to prove the second part. From your hint, there is a positive number $\delta \ge 1$ such that if $|t|>\delta$, then 
$1/2<\frac{|s|}{|t|}<3/2$ $\Rightarrow$  $|s|^{1-\sigma_0+\varepsilon} \le $ $(3/2)^{1-\sigma_0+\varepsilon}$ $|t|^{{1-\sigma_0+\varepsilon}} $. 
Our final obstacles are the case $|t| \le \delta$. 
$|s|^{1-\sigma_0+\varepsilon}\le|\delta s|^{1-\sigma_0+\varepsilon} \le  (3/2)^{1-\sigma_0+\varepsilon}|\delta t|^{1-\sigma_0+\varepsilon} \le (3\delta/2 )^{1-\sigma_0+\varepsilon} |t|^{1-\sigma_0+\varepsilon} $ because of the part (i).
Since $\delta$ is a constance dominated by 1/2, we are done. 
