Specific theorem used to derrive $\frac{1}{\zeta(s)}=\sum_{n=0}^\infty\frac{\mu(n)}{n^s}$

Question

In the book Prime Obsession by John Derbyshire he used the following reasoning to derive that identity $$(a+b)(c+d) = ac+ad+bc+bd$$ $$(a+b)(c+d)(e+f)=ace+acf+ade+adf+bce+bcf+bde+bdf$$ The resulting sum of the product involves adding up all possible products where a term is grabbed from each parenthesis. He then applies that reasoning to the euler product: $$\frac{1}{\zeta(s)}=\prod_{p}({1-p^{-s}})$$ When the product rule above is applied, only prime power terms remain in the sum, and each one is multiplied by either $1$ or $-1$ depending on the power. This gives: $$\frac{1}{\zeta(s)}=\sum_{n=0}^\infty\frac{\mu(n)}{n^s}$$ The identity used looks like it can be generalized to the following formula: $$\prod_{n=1}^k({a_n+b_n})=\sum{?}$$ Though I'm not sure what I should put into the sigma to represent the multiple combinations of the different terms that should end up in the sum (it might require nested sums?). Is there a specific theorem that generalizes the multiplication rule used to derive the identity? How exactly does such a theorem still work as we let $k->\infty$ (i.e under what conditions does it work for infinite products)? Is there a clearer way to see how the mobius function pops out of this?

Answer 1

Let $p_n$ denote $n^{th}$ prime number.

Take $$P_k = \prod_{n=1}^k (1-a_n) $$

It is easy to see by induction that $$P_k = 1 - \sum_{1 \le i_1 \le k}a_{i_1} +\sum_{1 \le i_1 < i_2 \le k}{a_{i_1}a_{i_2}} - \ldots+ (-1)^k a_1a_2\ldots a_k $$ where the general terms is $$(-1)^r\sum_{1\le i_1 < i_2 < \cdots<i_r\le k}a_{i_1}a_{i_2}\cdots {a_{i_r}} $$

Now take $a_n = p_n^{-s}$, then $$(-1)^r\sum_{1\le i_1 < i_2 < \cdots<i_r\le k}a_{i_1}a_{i_2}\cdots {a_{i_r}} = (-1)^r \sum_{1\le i_1 < i_2 < \cdots<i_r\le k}p_{i_1}^{-s}p_{i_2}^{-s}\cdots {p_{i_r}^{-s}}$$

If you consider $m = p_{i_1}^{-s}p_{i_2}^{-s}\cdots {p_{i_r}^{-s}}$ then $\mu(m) = (-1)^r$, also see that there in the expansion of $P_k$ only squarefree terms appear and since $\mu(m) =0$ when $m$ is not squarefree, we can easily see that the coefficient of $m^{-s}$ in the expansion of $$P_k = \prod_{n=1}^k (1-a_n) $$ is $\mu(m)$.

By taking $\lim_{k\to \infty}$ we get the desired result.

Update

We can write $$\prod_{n=1}^{k} (a_n+b_n) = \prod_{n=1}^{k} a_n \left(1+\frac{b_n}{a_n} \right) = (a_1 a_2 \ldots a_n) \prod_{n=1}^{k} (1 + c_n) $$ where $c_n = b_n / a_n$ and we have already worked out $\prod_{n=1}^{k} (1 - (-c_n))$.

I am not sure if this identity has a particular name but a very similar expression appears in principle of inclusion and exclusion

Answer 2

The most broad-scope generalization of this is the Euler product formula for multiplicative functions.

Theorem (Euler product): Let $f(n)$ be a multiplicative arithmetical function such that $\sum_n|f(n)|$ converges. Then we have

$$ \prod_p\sum_{k\ge1}f(p^k) $$

where $p$ runs over all positive prime numbers.

Proof. By fundamental theorem of arithmetic, we know that every positive integer $n$ can be written as a product of prime powers $n=\prod_pp^{r_p}$ where $r_p>0$ when $p|n$ and $r_p=0$ otherwise. By iterating through all $n\ge1$, we can just iterate through all possible combinations:

\begin{aligned} \sum_{n\ge1}f(n) &=\sum_{r_2,r_3,\cdots\ge0}f(2^{r_2}3^{r_3}\cdots)=\sum_{r_2\ge0}f(2^{r_2})\sum_{r_3\ge0}f(2^{r_3})\cdots \\ &=\prod_p\sum_{r_p\ge0}f(p^{r_p})=\prod_p\sum_{r\ge0}f(p^r) \end{aligned}

Plugging $f(n)=n^{-s}$ in, we see that for $\Re(s)>1$ there is

$$ \zeta(s)=\sum_{n\ge1}{1\over n^s}=\prod_p\sum_{k\ge1}{1\over p^{ks}}=\prod_p\left(1-{1\over p^s}\right)^{-1} $$

Moreover, by plugging in $f(n)=\mu(n)n^{-s}$, we observe that for $\Re(s)>1$ there is

$$ \sum_{n\ge1}{\mu(n)\over n^s}=\prod_p\left(1-{1\over p^s}\right) $$

Combining these two results, we see that whenever $\Re(s)>1$ there is

$$ \sum_{n\ge1}{\mu(n)\over n^s}={1\over\zeta(s)}\tag{$*$} $$

NOTE: The prime number theorem is equivalent to make ($*$) hold for $\Re(s)\ge1$, and the Riemann hypothesis is equivalent to make ($*$) hold for $\Re(s)>1/2$.