# Serre's proof that zeta function is meromorphic

I try to understand the proof of Chap. VI, n° 3.1, Prop. 10 in Serre's "A course in arithmetic" (page 70). The goal is to prove that zeta-function can be written as \begin{align*} \zeta(s)=\frac{1}{s-1}+\phi(s) \end{align*} with an holomorphic $\phi$ for any $s\in\mathbb{C}$ with $\mathfrak{Re}(s)>0$.

I understand the proof except for one detail: $\phi$ is given explicitly by $\phi=\sum_{n=1}^\infty \phi_n$ with \begin{align*} \phi_n(s)=\int_n^{n+1} \left(n^{-s}-t^{-s}\right) dt. \end{align*} To prove the convergence of $\phi$, we want to show that \begin{align*} \left|\phi_n(s)\right|\leq\frac{|s|}{n^{x+1}} \quad \text{with}\quad x=\mathfrak{Re}(s)\quad(*) \end{align*} For this Serre first notes that \begin{align*} \left|\phi_n(s)\right| \leq \sup_{n\leq t\leq n+1}\left| n^{-s}-t^{-s}\right| \end{align*} Which is clear since the range of integration is $(n+1)-n=1$. Then he sais that the derivative of $n^{-s}-t^{-s}$ is equal to $\frac{s}{t^{s+1}}$. And from this somehow follows (*). My question is: How exactly does this follow?

I noted the following statements for $s=x+iy$ where $x=\mathfrak{Re}(s)$ and $y=\mathfrak{Im}(s)$ and $f(t)=n^{-s}-t^{-s}$

1. Because $|t^{iy}|=|e^{iy\ln(t)}|=1$ (since it is on the unit circle) and $|t^{x+1}|=t^{x+1}$ (since $t$ and $x$ are real) we see that \begin{align*} \left|f'(t)\right|=\left|\frac{s}{t^{x+1+iy}}\right|=\frac{|s|}{|t^{x+1}| |t^{iy}|}=\frac{|s|}{t^{x+1}} \end{align*}
2. Because $f(n)=0$ we can calculate \begin{align*} |f(n+1)|=|f(n+1)-f(n)|=\left|\int_n^{n+1}f'(t)dt\right|\leq\sup_{n\leq t\leq n+1}\left|f'(t)\right|=\sup_{n\leq t\leq n+1}\frac{|s|}{t^{x+1}}=\frac{|s|}{n^{x+1}} \end{align*}

Am I close? :-)

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Looks exactly right, you should have more confidence :). The key idea that $|\int f| \le \int|f|$ is pervasive and often considered a form of the triangle inequality. Great choice of book, by the way. – Erick Wong Feb 21 '13 at 15:09
I don't see how this implies $|\phi_n(s)|\leq\frac{|s|}{n^{x+1}}$ and what it has to do with the derivative. – born Feb 21 '13 at 15:38
Oh, I must have taken this for granted. In the first equation after (*) you cite that $|\phi_n(s)|$ is bounded by $\sup |f(t)|$. Although point $2$ only refers to $f(n+1)$, use the same argument to get a bound for $f(t)$. – Erick Wong Feb 21 '13 at 15:48
The key idea is that the derivative controls how quickly $f$ can vary. Since $f'$ is small, $f(n+\epsilon)$ cannot stray too far from $f(n)$, which is $0$. – Erick Wong Feb 21 '13 at 15:49

By the definition of $\phi_n$ and the Fundamental theorem of calculus we have
$$|\phi_n(s)| = \left| \int_n^{n+1} f(t) \, dt \right| \stackrel{f(n)=0}{=} \left| \int_n^{n+1} \left(\int_n^t f'(y) \, dy \right) \, dt \right| \leq \int_n^{n+1} |f'(y)| \, dy$$
Use your calculation from 1. (and note that $y \in [n,n+1]$, hence $y\geq n$).