Exercise 2.25

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
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ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Consider the function $\zeta(s) = \sum_{n=1}^\infty 1/n^s$. $\zeta$ is called the Riemann zeta function. It converges for $s > 1$. Prove the formal identity (Euler's identity) $$\zeta(s) = \prod_p (1-1/p^s)^{-1}.$$

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
We prove this equality, not only formally, but for all complex value $s$ such that $\mathrm{Re}(s)>1$.

Let $s \in \C$ and $f(n) = \frac{1}{n^s},\ n\in \N^*$.

$f$ is completely multiplicative : $f(mn) = f(m) f(n)$ for $m,n \in \N^*$.

Moreover  $\sum_{n=1}^\infty f(n)$ is absolutely convergent for $\mathrm{Re}(s) > 1$. Indeed, if $s = u+iv, u,v \in \R$, $\vert f(n) \vert = \vert n^{-s}\vert = \vert e^{-s \log(n)} \vert = \vert e^{-u \log(n)} e^{-iv \log(n)} \vert = e^{-u \log(n)} = \frac{1}{n^u}$, so $\sum\limits_{n=1}^\infty \vert f(n) \vert=\sum\limits_{n=1}^\infty 1/n^u$ converges if $u=\mathrm{Re}(s) >1$.

With these properties of $f$ ($f$ multiplicative and $\sum_{n=1}^\infty f(n)$  absolutely convergent), we will show that 
$$\sum_{n=1}^\infty f(n) = \prod_p (1 +f(p)+f(p^2)+\cdots).$$
Let $S^* = \sum\limits_{n = 1}^\infty \vert f(n) \vert < \infty$, and $S = \sum\limits_{n = 1}^\infty  f(n) \in \C$. For each prime number $p$, $\sum\limits_{k=1}^\infty \vert f(p^k) \vert$ converges (this sum is less than  $S^*$), so $\sum\limits_{k=0}^\infty f(p^k)$ converges absolutely. Thus, for $x \in \R$, the two finite products
$$P(x) = \prod_{p\leq x }\sum_{k=0}^\infty f(p^k), \qquad P^*(x) =  \prod_{p\leq x} \sum_{k=0}^\infty \vert f(p^k) \vert$$
are well defined.

If $p, q$ are two prime numbers, as $\sum_{i=0}^\infty f(p^i), \sum_{j=0}^\infty f(q^j)$ are absolutely convergent, $(f(p^i) f(q^j))_{(i,j) \in \N^2}$ is summable, so the sum of these elements can be arranged in any order :
$$ \sum_{i=0}^\infty f(p^i)  \sum_{k=0}^\infty f(q^k) = \sum_{(i,j) \in \N^2} f(p^i) f(q^j) =  \sum_{(i,j) \in \N^2} f(p^iq^j).$$
 If $p_1,\cdots,p_t$ are all the prime $p\leq x$, repeating $t$ times these products, we obtain
\begin{align*}
P(x) & = \prod_{p\leq x }\sum_{k=0}^\infty f(p^k)\
&= \sum_{i_1=0}^\infty f(p_1^{i_1})\cdots  \sum_{i_t=0}^\infty f(p_t^{i_t})\
&= \sum_{(i_1,\ldots,i_k)\in \N^k} f(p_1^{i_1} \cdots p_t^{i_t})\
&=\sum_{n\in \Delta} f(n),
\end{align*}
where $\Delta$ is the set of integers $n \in \N^*$ whose prime factors are not greater than $x$. Let $\overline{\Delta} = \N^* \setminus \Delta$ : this is the set of numbers $n \in \N^*$ such that at least a prime factor is greater than $x$. So
$$P(x) = \sum_{n\in \Delta} f(n)  =  S - \sum_{n \in \overline{\Delta}} f(n).$$
Then
$$\vert P(x) - S \vert \leq \sum_{n\in \overline{\Delta}} \vert f(n) \vert \leq \sum_{n\geq x} \vert f(n) \vert.$$
So $\lim\limits_{x 	o + \infty} P(x) = S$, that is
$$\prod_p \sum_{k=0}^\infty f(p^k) = \sum_{n=1}^\infty f(n).$$
Finally, 
\begin{align*}
\sum_{n=1}^\infty \frac{1}{n^s} &=  \prod_p \left ( 1 + \frac{1}{p^s}+ \cdots + \frac{1}{p^{ks}}+ \cdotsight )\
&=  \prod_p (1-1/p^s)^{-1}
\end{align*}
\end{proof}

Exercise 2.25

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Exercise 2.25

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