Section24.5Multiplication and Inverses

Subsection24.5.1A series for Euler phi

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 use three facts here:

\(\phi\star u=N\)
\(\zeta\) is absolutely convergent for \(s>1\)
The Dirichlet series of \(\phi\) is absolutely convergent as well, as \[0<\sum_{n=1}^\infty \frac{\phi(n)}{n^s}\leq \sum_{n=1}^\infty \frac{n}{n^s}=\sum_{n=1}^\infty\frac{1}{n^{s-1}}\] which converges by the integral test if \(s>2\) (it's also \(\zeta(s-1)\)).

So the series for \(\phi\) (call it \(P\)) and \(u\) multiply to the series for \(N\).

But now we have three other facts.

The series for \(u\) is \(\zeta(s)\).
The series for \(N\) (from last time, or above) is \(\zeta(s-1)\).
So \(P(s)\zeta(s)=\zeta(s-1)\) for \(s>2\).

That means \[\sum_{n=1}^\infty \frac{\phi(n)}{n^s}=\frac{\zeta(s-1)}{\zeta(s)}\; .\] We can check this with Sage too.

Subsection24.5.2A general theorem

It turns out that such Euler products (and hence nice computations like this) show up quite frequently.

Theorem24.5.1

If \(\sum_{n=1}^{\infty}\frac{f(n)}{n^s}\) converges absolutely and \(f\) is multiplicative, then \[\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_p\left(1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\cdots\right)\, .\]

Proof

Subsection24.5.3A Missing Step: Convergence of Dirichlet Series

Before we start using this in the next section, we have to acknowledge there is a missing step thus far.

We don't actually know 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 and its series that it 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.

Fact24.5.2

For \(s> 1\) we have \[\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)=\prod_p (1-p^{-s})\]

Proof

This proof follows closely with that of Jones and Jones.

We start with the identity we already know: \[\sum_{d\mid n}\frac{\mu(d)}{d}=\prod_{p\mid n}(1-p^{-1})\, .\] Assuming we multiply these products out through the \(k\)th prime, we get \[\prod_{i=1}^k \left(1-\frac{1}{p_i}\right)=\] \[1-\frac{1}{p_1}-\frac{1}{p_2}-\cdots +\frac{1}{p_1 p_2}+\frac{1}{p_1 p_3}+\cdots -\frac{1}{p_1 p_2 p_3}-\frac{1}{p_1 p_2 p_4}-\cdots =\] \[ \sum_{n\text{ squarefree and only }p_i\mid n,1\leq i\leq k}\frac{\mu(n)}{n}\, .\]

Next, let's introduce the set \[A_k=\{n \mid n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k},e_i\geq 0\}\] 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, the big right hand side sum is equal to \[\sum_{n\in A_k}\frac{\mu(n)}{n}\; .\] I can replace \(p_i\) with \(p_i^s\) with no harm (the FTA gives the same relations), so I can now write \[\prod_{i=1}^k (1-p_i^{-s})=\sum_{n\in A_k}\frac{\mu(n)}{n^s}\, .\]

Our next step is to get a bound on the difference between the infinite product and infinite series. So let's take a closer look at \(n\notin A_k\).

All \(n\notin A_k\) must be \(n>p_k\) since \(n\) cannot have any of the first \(k\) primes as factors.
So at the very least we know that \[\left| \prod_{i=1}^k (1-p_i^{-s})-\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}\right|=\left|\sum_{n\notin A_k}\frac{\mu(n)}{n^s}\right|\; .\]
What this means is we have a specific formula for an error up through the \(k\)th prime.
Note that this only makes sense because the infinite sum involving \(\mu\) converges for the same \(s\) as \(\zeta(s)\) does. This is because it's bounded by \(\pm\zeta\), and \(\zeta\) converged absolutely for \(s>1\) (recalling our calculus tests).

Next I can do a few extra things to bound this error even better. \[\left|\sum_{n\notin A_k}\frac{\mu(n)}{n^s}\right|\leq \sum_{n\notin A_k}\left|\frac{\mu(n)}{n^s}\right|\leq \sum_{n>p_k}\left|\frac{\mu(n)}{n^s}\right|\leq \sum_{n>p_k}\frac{1}{n^s}\, .\] There are a few things to explain here.

We can put the absolute value inside the summation by an extended triangle inequality; this is only legitimate if the final thing converges, but we are about to show that.
The second inequality is again 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.
The last one is using the boundedness by \(\zeta\) we used a second ago.

Finally, this 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 it all converges and we have our Euler product for this Dirichlet series!