TITLE: Detailed proof of $\zeta(s)-1/(s-1)$ extends holomorphically to $\Re(s)>0$
QUESTION [10 upvotes]: I'm trying to understand the proof of PNT by Don Zagier. But his proof is too simplified so I can't understand it. I got stumped at step II: $\zeta(s)-1/(s-1)$ extends holomorphically to $\Re(s)>0$.
The proof is available here:
http://mathdl.maa.org/images/upload_library/22/Chauvenet/Zagier.pdf
Can anyone post a detailed proof that $\zeta(s)-1/(s-1)$ extends holomorphically to $\Re(s)>0$？
Or please tell me which book has a detailed proof so I can look it up?
edit:
Zagier's proof:
$\displaystyle \zeta(s)-\frac{1}{s-1} = \sum_{n=1}^{\infty} \int_{n}^{n+1} (\frac{1}{n^s}-\frac{1}{x^s}) dx$
$\displaystyle \left|\int_{n}^{n+1} (\frac{1}{n^s}-\frac{1}{x^s}) dx\right| \leq \frac{|s|}{(n^{\Re(s)+1})}$
therefore $\displaystyle \zeta(s)-\frac{1}{s-1}$ extends holomorphically to $\Re(s)>0$
My question: We have a function
(i) $f(s)=\sum_{1}^{\infty} g_n(s) \quad \forall \Re(s)>0$
(ii) $g_n(s) \leq |s|/(n^{\Re(s)+1}) \quad \forall \Re(s)>0$
Is (i) and (ii) the necessary and sufficient condition that $f(s)$ has holomorphic continuation?
Does there exist functions $f$ and $g_n$ that satisfy (i) and (ii), but does not have holomorphic continuation?
Do we need to prove other conditions, for example $g_n$ must be continuous/holomorphic?

REPLY [3 votes]: Trying to address what I think was troubling the OP.
Fix constants $a,b$ such that $0<a<b$. Consider the set $$L(a,b)=\{x+yi\in\mathbb{C}\mid a<x, |x+yi|<b\}.$$ The functions $g_n(z)$ are holomorphic in the set $L(a,b)$. Item $(ii)$ means that for all $z\in L(a,b)$ and all positive integers $n$ we have
$$
|g_n(z)|\le\frac{b}{n^{1+a}}.
$$
Therefore, by Weierstrass' M-test (note that here the lower bound on the exponent, $1+a$, does not depend on $z$, only on the set $L(a,b)$), the series $f(z)=\sum_{n=1}^\infty g_n(z)$ converges uniformly in the set $L(a,b)$ and therefore gives a holomorphic function on $L(a,b)$.
The right half-plane is the union of the sets $L(a,b)$, so $f(z)$ is holomorphic there.