Analytic number theory

2 Dirichlet series and properties

A very powerful tool in mathematics for studying sequences of numbers is to study their generating functions. If \(a_n\) is a sequence, one way to turn \(a_n\) into a generating function is to consider \(F(z) = \sum _0^\infty a_nz^n\). Properties of the sequence then relate to properties of the function, and vice versa. For example, we might expect the behavior of \(F(z)\) as \(z \to 1^{-}\) to be related to the asymptotics of the sequence. Also if \(F\) is analytic near \(1\), then the coefficients \(a_n\) can be obtained via Cauchy’s integral formula. These generating functions are good at capturing additive properties of sequences, as \(x^nx^m = x^{n+m}\).

Another type of generating function that can be made from sequences is an \(L\)-function. These capture more of the multiplicative structure of the sequence. These look like something of the form \(L(s) =\sum _0^\infty \frac {a_n}{n^s}\), called a Dirichlet series. Both of these constructions are specializations of the more general construction ca where given two sequences \(\lambda _n,a_n\), we can consider \(\sum a_ne^{-\lambda _n s}\). The basic example is \(\zeta (s) = \sum \frac {1}{n^s}\), the Riemann zeta function. Note the series converges absolutely and uniformly on compact sets for \(\real (s)>1\).

As any good generating function should, \(\zeta (s)\) tells us a great deal about the distribution of the primes. Below is a simple of example of how it can be used.

Theorem 2.1. The series \(\frac {1}{p}\) diverges, where \(p\) ranges over the positive primes in \(\ZZ \).

Proof. \(\sum _1^\infty \frac {1}{n^s}= \prod _p(\frac {1}{1-\frac {1}{p^s}})\), which diverges as \(s \to 1^+\). Taking the log of the right hand side we get \(\sum _p -\log (1-\frac {1}{p^s}) = \sum _p \frac {1}{p} + \sum _{m \geq 2, p} \frac {1}{mp^{ms}}\). The second sum is \(\leq \sum _{m \geq 2, p} \frac {1}{p^{ms}} = \sum _{p} \frac {1}{p^{2s}-p^s} \leq 2\sum _{p}\frac {1}{p^{2s}}\), which is finite as \(s \to 1\). Thus \(\sum _p \frac {1}{p}\) diverges. □

First we will prove some lemmas about Dirichlet series. Note that a necessary condition for convergence at some point is that \(|a_n|\) should be bounded by some polynomial. Here is a partial converse:

Lemma 2.2. If \(|\sum _1^n a_n| = O(n^r)\), \(r>0\), then the Dirichlet series for \(a_n\) converges uniformly for \(\real (s)\geq r+\ee \) for any \(\ee >0\) to a holomorphic function. If \(|a_n| = O(n^r)\), then it converges absolutely and uniformly for \(\real (s)>r+1+\ee \).

Proof. Let \(f_n = \sum _1^n a_n\). We will show that the partial sums are uniformly Cauchy by using summation by parts. \(\sum _k^m \frac {a_n}{n^s} = \sum _k^m\frac {f_n-f_{n-1}}{n^s}\) = \(\frac {f_m}{m^s} - \frac {f_{k-1}}{(k-1)^s} + \sum _k^{m-1}f_n (\frac {1}{n^s}-\frac {1}{(n+1)^s})\). The first two terms go to \(0\) uniformly as \(k \to \infty \), and the difference in the sum is equal to \(\int _{n}^{n+1}(-s)x^{-(s+1)}\), which in absolute value is at most \(|s|n^{-s+1}\). But then since \(f_n = O(n^r)\), the last term in absolute value is at most \(\sum _k^{m-1}\frac {cn^r}{n^{-(r+\ee +1)}}\), which is uniformly bounded in \(m\) and goes to \(0\) as \(k \to \infty \) since \(\ee +1 > 1\). The last statement is clear. □

Lemma 2.3. If \(F,G\) are Dirichlet series for \(a_n,b_n\) that converge, and \(F,G\) agree where they converge, then \(a_n = b_n\).

Proof. Let \(a_i\) be the first nonzero term of \(a_n\). Then \(F \sim \frac {a_i}{i^s}\) as \(s \to \infty \), so this behavior determines \(i\) and \(a_i\). By subtracting this term off, we can recover the rest of the sequence. We have used properties of \(F\) that agree with \(G\), so \(a_n = b_n\). □

Note by analyticity, if they agree on a small set, they agree.

Lemma 2.4. Suppose that \(F,G\) are convergent Dirichlet series for \(a_n,b_n\). Then \(FG\) is the Dirichlet series for \(a_n\star b_n\), where \(\star \) denotes the Dirichlet convolution.

Proof. For sufficiently large \(s\), \(F,G\) converge absolutely and uniformly. Then when we take their product, we can change the order of summation to get the result. □