Fact 24.5.1.
Call \(P\) the Dirichlet series for \(\phi\text{;}\) it converges for \(s>2\text{.}\)
Section 24.5 More series and convergence
Subsection 24.5.1 A series for Euler phi and a general theorem
We can now feel confident applying these amazing facts to calculate the Dirichlet series of \(\phi\) in terms of the Riemann \(\zeta\) function. We’ll see a few facts along the way which could serve as templates for many such investigations.
Proof.
From Fact 23.3.2, we recall that \(\phi\star u=N\text{.}\) Also, we know from earlier in this chapter that \(\zeta\) is absolutely convergent for \(s>1\text{.}\)
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 \(s>2\text{.}\)
Fact 24.5.2.
The series for \(N\) may also be written as \(\zeta(s-1)\text{.}\)
Proof.
This follows just from writing it down, as each term in the infinite sum is like that of \(\zeta\) but with a different exponent after cancelling.
We can do even better than this to get a single formula for the series \(P\text{.}\)
Fact 24.5.3.
The series for \(\phi\text{,}\) \(P(s)\text{,}\) evaluates as
\begin{equation*}
P(s)=\sum_{n=1}^\infty \frac{\phi(n)}{n^s}=\frac{\zeta(s-1)}{\zeta(s)}
\end{equation*}
Proof.
Recall that the Riemann zeta function is just the Dirichlet series for \(u\text{;}\) the previous fact is that the series for \(N\) is \(\zeta(s-1)\text{.}\)
Apply Theorem 24.4.3 to the series for \(\phi\) and \(u\text{.}\) When you multiply these two series it should give the series for \(N\text{,}\) and we already showed it all converges. Substitute in the formulas to get \(P(s)\zeta(s)=\zeta(s-1)\) for \(s>2\text{,}\) which suffices to prove the fact.
We can check this with Sage at any particular point if we wish.
It turns out that such Euler products (and hence nice computations) show up quite frequently.
Theorem 24.5.4.
If \(\sum_{n=1}^{\infty}\frac{f(n)}{n^s}\) converges absolutely and \(f\) is multiplicative, then
\begin{equation*}
\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_p\left(1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\cdots\right)\text{.}
\end{equation*}
Proof.
Doing this is Exercise 24.7.2. We have a proof that Moebius \(\mu\)’s Dirichlet series converges to its Euler product in the next subsection; the proof of this is very similar, just more general.
Subsection 24.5.2 A missing step: Convergence of Dirichlet series
Before we start using all these facts in the next section, we have to acknowledge there is a missing step thus far. Namely, we haven’t demonstrated much about convergence of these series or products, much less that they converge to each other. Although it is fun to play around, and numerical experimentation will convince you they are very likely, we need more to really use these tools with abandon.
Our goal in this subsection is to prove for the Moebius \(\mu\) function that its Dirichlet series converges to the Euler product. Proofs for most other such functions (such as the Riemann zeta function) are similar enough to leave more general proofs to a graduate course.
Fact 24.5.5.
For \(s> 1\) we have
\begin{equation*}
\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)=\prod_p (1-p^{-s})
\end{equation*}
Proof.
This proof follows the outline of [E.2.1, Theorem 9.3a] closely; see also [E.2.1, Theorem 9.2]. 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})\text{.}
\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_{\substack{n\text{ squarefree}\\\text{ only }p_i\mid n,\; 1\leq i\leq k}}\frac{\mu(n)}{n}\text{.}
\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_{\substack{n\text{ squarefree}\\\text{ only }p_i\mid n,\; 1\leq i\leq k}}\frac{\mu(n)}{n}=\sum_{n\in A_k}\frac{\mu(n)}{n}\text{.}
\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}\text{.}
\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 \(\sum_{n\notin A_k}\frac{\mu(n)}{n^s}\text{.}\) 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 \(s\text{,}\) because it’s bounded by \(\pm\zeta\text{,}\) 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 \(n>p_k\text{,}\) 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 \(k\to \infty\text{,}\) 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!
Claim 24.5.6.
With all notation as in Fact 24.5.5, we have
\begin{equation*}
\left|\sum_{n\notin A_k}\frac{\mu(n)}{n^s}\right|\leq \sum_{n\notin A_k}\left|\frac{\mu(n)}{n^s}\right|\leq \sum_{n>p_k}\left|\frac{\mu(n)}{n^s}\right|\leq \sum_{n>p_k}\frac{1}{n^s}
\end{equation*}
Proof.
The first inequality follows if we can put the absolute value inside the summation, which is an extended triangle inequality. To be careful, we should wish the series involved to actually converge, but 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 \(p_k\text{,}\) 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.
