4
$\begingroup$

We have the following identity: (where $\mu$ is Mobius function) $$\frac{\zeta(s)}{\zeta(2s)} = \sum_{n=1}^{\infty} \frac{|\mu(n)|}{n^s} \quad (\sigma > 1, s = \sigma+it )$$

The proof I read (Titchmarsh p5) is given by the following line:

$$\frac{\zeta (s) }{ \zeta(2s) } \stackrel{*}{=} \prod_p \frac{1-p^{-2s}}{1-p^{-s}} = \prod_p \big( 1 + \frac{1}{p^s} \big) \stackrel{**}{=} \sum_{n=1}^\infty \frac{|\mu(n)|}{n^s}$$


My question is how do we justify $*$ and $**$ rigorously? Below is what I think:


$(*):$ We note that the partial products, $\hat{f_n} = f_1 \cdots f_n$ where $f_i = \big( 1 - \frac{1}{p_i^s} \big)^{-1}$ converges compactly in region $\sigma >1$. $\zeta(s) = \lim_{n \rightarrow \infty} \hat{f}_n(s)$ (by Euler Product formula). Similarly, $\zeta(2s) = \lim_{n \rightarrow \infty} \hat{g}_n(s)$ where $\hat{g}_n = \hat{f}_n (2s)$. We note that $|\zeta(s)| > 0$ for all $\sigma >1 $.

$\frac{1}{\zeta(2s)} = \lim_{n \rightarrow \infty} \frac{1}{\hat{g}_n(s)}$ is well-defined, and so by basic algebra of limits, $$\frac{\zeta(s) }{ \zeta(2s)} = \lim_{n \rightarrow \infty} \frac{\hat{f}_n(s)} { \hat{g}_n(s) } $$

$(**):$ Here we argue by taking a prime $P$, and we have,

$$ \Big| \sum \frac{|\mu(n)| }{n^s} - \prod_{p \le P} (1+ \frac{1}{p^s} ) \Big| \le \sum \frac{1}{(P+n)^s} \rightarrow 0 $$ as $P \rightarrow \infty$.


In general, when do we know we can just expand infinite products like this?

For instance, we also have the identity:

$$ \frac{\zeta^2(s)}{\zeta(2s)} = \sum \frac{2^{v(n)}}{n^s} $$ the proof then utilizes infinite product of infinite sums:

$$ \frac{\zeta^2(s)}{\zeta(2s)} = \prod_p (1 + 2p^{-s} + 2p^{-2s} + \cdots ) = \sum \frac{2^{v(s)}}{n} $$

How do we know the last two equalities coincide?

$\endgroup$
  • 2
    $\begingroup$ Yes, what you propose works. In general, we know we can "just expand" when the products converge absolutely. One must (at least once) prove that this sort of expansion works for absolutely convergent products, of course, before one knows that it works. $\endgroup$ – Daniel Fischer Jul 3 '17 at 12:53
  • $\begingroup$ Thanks, I will do some reading about this, and update my post, hopefully you could check if it is right. $\endgroup$ – CL. Jul 3 '17 at 13:03
  • $\begingroup$ @DanielFischer I have made a post on what I researched. Is it fine? $\endgroup$ – CL. Jul 3 '17 at 14:56
2
$\begingroup$

For the first equality $(*)$ we have the Euler product representation of the zeta function \begin{eqnarray*} \zeta(s)= \prod_{p \in \mathbb{P}} (1-p^{-s})^{-1}. \end{eqnarray*} From your last comment you seem happy now that \begin{eqnarray*} \frac{\zeta(s)}{\zeta(2s)}= \prod_{p \in \mathbb{P}} (1+\frac{1}{p^{s}}). \end{eqnarray*} Now consider the terms created by expanding this product \begin{eqnarray*} \left(1+\frac{1}{2^{s}}\right)\left(1+\frac{1}{3^{s}}\right)\left(1+\frac{1}{5^{s}}\right) \cdots &=& 1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\frac{1}{6^{s}}+\frac{1}{7^{s}} \cdots \\ &=& \sum_{n=1}^{\infty} \frac{\chi(n)}{n^s} \end{eqnarray*} where $\chi(n)=1$ if $n$ is the product of distinct primes (in other words, $n$ is square-free) and $\chi(n)=0$ otherwise. It is easy to see that $\chi$ is essentially the Mobuis $\mu$ function without the powers of $-1$ to count the parity of the number of distinct primes.

$\endgroup$
2
$\begingroup$

This is my attempt to answer my question rigorously, please tell me if it is inappropriate to do so ( I did not want to post edit the OP as it would be too long.)


Definitions: An infinite product $\prod a_n$ is convergent if exists an index $m$ such that $p_{m,n} = a_m \cdots a_n $ converges to a limit $\hat{p}_m \not=0 $ as $n \rightarrow \infty$. We define $\prod a_n := a_0 \ldots a_{m-1} \hat{p}_m$.

An infinite product $\prod b_n$ where $b_n:= 1+ a_n$, is said to be an absolutely convergent infinite product if $\prod ( 1 + |a_n| )$ converges.


Observation: Suppose $\prod a_n$ is conergent. We have $\hat{p}_m $ exists for all $m\in \mathbb{N}$ by definition. Also, $\lim \hat{p}_m = \frac{\prod a_n}{p_{0,m-1}} \rightarrow 1 $ as $m \rightarrow \infty$ (wlog, product is not equal to $0$). Further $\lim p_m = \lim \hat{p}_m/ \hat{p}_{m+1} \rightarrow 1 $ as $m \rightarrow \infty$.

Proposition 1: $\prod b_n $, $b_n:= 1 + a_n$, converges iff $\sum_{n \ge m} \log (1+a_n) $ converges for some $m \in \mathbb{N}$.

Proof: Suppose $\prod b_n$ converges. From observation $b_n:= 1 + a_n \rightarrow 1$ as $n \rightarrow \infty$ and $\lim \hat{p}_n \rightarrow 1$ as $n \rightarrow \infty$. By continuity, and noting that the argument lies in $\mathbb{C}_{-}$, we have a well defined equation: $$ \lim_{n\rightarrow \infty} \log p_{m,n } = \log \hat{p}_n \Rightarrow \sum_{n \ge m} \log (1+a_n) \text{ converges to some } \alpha $$ Conversely, by taking $e$ to the power of the given result, yields convergene of $\prod b_n$.

Proposition 2: $\prod b_n$, $b_n:=1 + a_n$ converges absolutely iff $\sum |a_n| $ converges.

Proof: Suppose $\prod b_n$ converges absolutely, then $\prod (1 + |a_n|) $ converges. Hence $\sum \log ( 1 + |a_n| ) $ converges. Now note the inequality, $$ \frac{1}{2} |x| \le | \log (1+ x ) | \le 2 |x| $$ for $|x| \le \frac{1}{2}$. By applying first proposition, we deduce the result. As a corollary if $\prod b_n$ converges absolutely then $\sum |a_n| $ converges, $\sum \log (1+ a_n) $ converges absolutely, and $\prod b_n$ converges.


Example: In the case of proving identity $$ \frac{\zeta^2(s)}{\zeta(2s)} = \sum_{n=1}^\infty \frac{2^{v(n)}}{n^s}, \, ( \sigma > 1 )$$ we have the infinite product, $$ \prod _p (1 + a_p) = \prod_p ( 1 + 2p^{-s} + 2 p^{-2s} + \cdots ) $$ Where $a_p = 2( \sum_{n =1 }^\infty p^{-ns} ) $. Now for some constant $C \in \mathbb{R}_{>0}$ $$\sum_p |a_p| \le \sum_p ( \sum_n \frac{1}{p^{ n \sigma}} ) \le C \sum_p \frac{1}{p^\sigma} < \infty $$ Hence, the infinite product converges, and we may apply term by term multiplication, i.e. $$ \prod_{p \le P } (1 + a_p) \rightarrow \prod_p (1 + a_p) $$ as $P \rightarrow \infty$.


References: Trench, W. (1999). Conditional Convergence of Infinite Products. The American Mathematical Monthly, 106(7), p.646.

$\endgroup$
  • $\begingroup$ For $\Re(s) > 0 $ : $\prod_{p \le k} (1+p^{-s}+p^{-2s}+\ldots) = \sum_{n \in A_k} n^{-s}$ where $A_k$ are the natural numbers whose largest prime factor is $\le k$. With multiplicative functions it works the same way : $$\prod_{p \le k} (1+f(p)p^{-s}+f(p^2) p^{-2s})+\ldots) = \sum_{n \in A_k} f(n) n^{-s}$$ (assuming $1+f(p)p^{-s}+f(p^2) p^{-2s}+\ldots$ converges absolutely) $\endgroup$ – reuns Jul 3 '17 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.