Section 24.4 Multiplication
At the end of the previous section, you may have noticed something surprising. The Euler products we obtained for the Riemannxxxxxxxxxx
sum([moebius(n)/n^2 for n in [1..10000]]).n()
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1/zeta(2).n()
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1/(pi^2/6).n()
Remark 24.4.1.
Zeta has interesting values at integers, not just for
To compare with the situation for even
Fact 24.4.2.
The arithmetic functions
The Dirichlet series of these functions are also inverses, as ordinary functions:
Alternately,
Theorem 24.4.3.
Use the following notation:
Take
and to be two arithmetic functions.Let
be their Dirichlet product.Let
be the corresponding Dirichlet series (in the variable ).
Then if the series
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
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
Notice that the inner sum is precisely the Dirichlet \(\star\) product (except divided by \(d^s\)). So we may rewrite this as
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.)