From Fact 23.3.2, we recall that \(\phi\star u=N\). Also, we know from earlier in this chapter that \(\zeta\) is absolutely convergent for \(s>1\).

Then the Dirichlet series of \(\phi\) is absolutely convergent as well, as \begin{equation*}0<\sum_{n=1}^\infty \frac{\phi(n)}{n^s}\leq \sum_{n=1}^\infty \frac{n}{n^s}=\sum_{n=1}^\infty\frac{1}{n^{s-1}}\end{equation*} which converges by the integral test if \(s>2\).

Apply Theorem 24.4.3 to the series for \(\phi\) and \(u\); then they multiply to the series for \(N\) and it all converges.

Recall that the Riemann zeta function is just the Dirichlet series for \(u\), and that the series for \(N\) (see the previous remark) is \(\zeta(s-1)\).

Then the previous proof can be expanded to state \(P(s)\zeta(s)=\zeta(s-1)\) for \(s>2\), which suffices to prove the fact.

This proof follows [C.1.1] closely. First we will come up with a way to write a partial product as a specific sum. Then we will use this to get a precise error between partial products and the infinite sum, and finally bound said error by something going to zero, the final step of which we separate out as an independent claim.

We will begin with the identity we already know as defining \(\mu\) in Definition 23.1.1: \begin{equation*}\sum_{d\mid n}\frac{\mu(d)}{d}=\prod_{p\mid n}(1-p^{-1})\, .\end{equation*} Assuming we multiply these products out through the \(k\)th prime, we get \begin{equation*}\prod_{i=1}^k \left(1-\frac{1}{p_i}\right)=\end{equation*} \begin{equation*}1-\frac{1}{p_1}-\frac{1}{p_2}-\cdots +\frac{1}{p_1 p_2}+\frac{1}{p_1 p_3}+\cdots -\frac{1}{p_1 p_2 p_3}-\frac{1}{p_1 p_2 p_4}-\cdots =\end{equation*} \begin{equation*} \sum_{n\text{ squarefree and only }p_i\mid n,1\leq i\leq k}\frac{\mu(n)}{n}\, .\end{equation*} This certainly suggests the entire fact is true.

Next, let's introduce the set \begin{equation*}A_k=\{n \mid n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k},e_i\geq 0\}\end{equation*} This is the set of all integers built out of the first \(k\) primes. Since \(\mu(n)=0\) unless it has no higher prime powers, then in this notation the big right hand side sum is equal to \begin{equation*}\prod_{i=1}^k \left(1-\frac{1}{p_i}\right)=\sum_{n\text{ squarefree and only }p_i\mid n,1\leq i\leq k}\frac{\mu(n)}{n}=\sum_{n\in A_k}\frac{\mu(n)}{n}\; .\end{equation*} Since the Fundamental Theorem of Arithmetic gives all these relations, I can replace \(p_i\) with \(p_i^s\) with no harm and write \begin{equation*}\prod_{i=1}^k (1-p_i^{-s})=\sum_{n\in A_k}\frac{\mu(n)}{n^s}\, .\end{equation*}

Our next step is to get a bound on the difference between the infinite product and infinite series, \begin{equation*}\prod_{i=1}^k (1-p_i^{-s})-\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}\end{equation*} By the work we just did, this is \(\sum_{n\notin A_k}\frac{\mu(n)}{n^s}\). This is the difference between the infinite sum and the partial product through the \(k\)th prime. Further, we know this error is finite for any given allowable \(s\), because it's bounded by \(\pm\zeta\), and \(\zeta\) converges absolutely for \(s>1\) (recall the comparison test for infinite series).

Let's put absolute values on this error bound: \begin{equation*}\left| \prod_{i=1}^k (1-p_i^{-s})-\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}\right|=\left|\sum_{n\notin A_k}\frac{\mu(n)}{n^s}\right|\end{equation*} To get a more explicit bound, we now deduce that any \(n\notin A_k\) must be \(n>p_k\), since \(n\) cannot have any of the first \(k\) primes as factors. Armed with this, the following Claim 24.5.6 will finish the proof: \begin{equation*}\left|\sum_{n\notin A_k}\frac{\mu(n)}{n^s}\right|\leq \sum_{n>p_k}\frac{1}{n^s}\end{equation*} The latter error \(\sum_{n>p_k}\frac{1}{n^s}\) must go to zero as \(k\to \infty\), since this is the tail of a convergent infinite series. That means that the partial products converge to the series; we know that is finite, so everything converges and we have our Euler product for this Dirichlet series!

The first inequality follows if we can put the absolute value inside the summation. This is an extended triangle inequality, which is only legitimate if the final thing converges; however, we already showed this at the end of the proof of the main fact.

The second inequality is due to the fact that any \(n\notin A_k\) must be bigger than \(p_k\), so the set of all integers above \(p_k\) would just yield a bigger sum (since all terms are now positive after the first step).

The final inequality uses that \(\mu=0,1,-1\) always.