##
Section 24.4 Multiplication

At the end of the previous section, you may have noticed something surprising. The Euler products we obtained for the Riemann \(\zeta\) function and the Dirichlet series of the Möbius function are multiplicative inverses of each other:

\begin{equation*}
\prod_p \frac{1}{1-p^{-s}}=1/\left(\prod_p 1-p^{-s}\right)\text{.}
\end{equation*}

We can check this numerically as well; in the following examples, we use \(s=2\text{.}\)

They agree up to quite a few digits when we approximate both representations of the number, so that is a start at reasonability!

Finally, recall from our exploration of the average value of

\(\sigma\) in

Section 20.4 that

\(\zeta(2)=\frac{\pi^2}{6}\) (though there we just used this as a sum, and didn’t call it

\(\zeta(2)\)). Compare this computation with the ones above.

Let’s reinterpret this connection between the Euler products of the \(\zeta\) function and the Möbius series just a little bit. Assuming we can prove that all this makes sense (which we haven’t, yet), we have the following two analogous facts.

###
Fact 24.4.2.

The arithmetic functions \(u\) and \(\mu\) are inverses as arithmetic functions; that is, \(u\star \mu=I\text{.}\)

The Dirichlet series of these functions are also inverses, as ordinary functions:

\begin{equation*}
\prod_p \frac{1}{1-p^{-s}}=1/\left(\prod_p 1-p^{-s}\right)
\end{equation*}

Alternately, \(\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=1/\zeta(s)\)

This analogy is not a coincidence.

###
Theorem 24.4.3.

Use the following notation:

Take \(f(n)\) and \(g(n)\) to be two arithmetic functions.

Let \(h=f\star g\) be their Dirichlet product.

Let \(F,G,H\) be the corresponding Dirichlet series (in the variable \(s\)).

Then if the series \(F\) and \(G\) converge *absolutely* for any particular \(s\text{,}\) then \(H\) converges and \(H=FG\) for that \(s\) as well.

### Proof.

First, we need there is a key fact you may or may not have seen in calculus, related to absolute convergence (see for example

Active Calculus). Roughly speaking, when series converge

*absolutely*, you can mess around with them with a lot with impunity. See, for instance, Mertens’ Theorem on convergence of Cauchy products. Interestingly, neither

[E.4.6] nor

[E.2.1, Theorem 9.6] say much more about this in their presentation of this standard proof. See

Exercise 24.7.3 if you have

*not* encountered this!

In any case, since \(F\) and \(G\) *do* converge absolutely, we can and will mess around a lot with the product

\begin{equation*}
F(s)G(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}\sum_{m=1}^\infty\frac{g(m)}{m^s}\text{.}
\end{equation*}

In particular, we can group the products by the terms

\(\frac{f(n)g(m)}{n^sm^s}\) (the same way we did in proving things about

\(\star\) in

Subsection 23.4.3), without loss of equality.

We can further group by when \(n\) and \(m\) are complementary divisors of the same number (I suggest using specific numbers to try this out). This gives

\begin{equation*}
F(s)G(s)=\sum_{d=1}^\infty\sum_{nm=d}\frac{f(m)g(n)}{d^s}\text{.}
\end{equation*}

Notice that the inner sum is precisely the Dirichlet \(\star\) product (except divided by \(d^s\)). So we may rewrite this as

\begin{equation*}
F(s)G(s)=\sum_{d=1}^\infty \frac{(f\star g)(d)}{d^s}\text{.}
\end{equation*}

The numerators are the definition of

\(h\text{,}\) so this is just

\(H(s)\text{,}\) as desired. (In

[E.4.6, Theorem 11.5] the additional detail that

*any* Dirichlet series with these values must be the one for

\(f\star g\) is proved, which requires a uniqueness result for the series we will omit.)

This is a quite remarkable and deep connection between the discrete/algebraic point of view and the analytic/calculus point of view. It is a shame that this is not exploited more in the standard calculus curriculum, though see

[E.6.8] for a very good resource for those who wish to do so.