Examples of arithmetic functions
The constant \(\bf 1\) and the identity \(\rm Id\) functions are defined by \[{\bf 1}(n) := 1\qquad \text{ and } \qquad {\rm Id}(n):=n \quad \text{ for all } \quad n \in \mathbb{Z}_+.\]
The Dirac delta function \(\delta_m\) is defined as follows
\[\delta_m(n) = \begin{cases} 1 & \text{if } n = m, \\ 0 & \text{otherwise}. \end{cases}\] We shall abbreviate \(\delta_1\) to \(\delta\).
The divisor function \(\tau(n)\) is the number of positive divisors of \(n\in \mathbb Z_+\), \[\tau(n):=\#\{d\in\mathbb Z_+: d\mid n\}=\sum_{d\mid n}1.\] Some authors also use the notation \(d(n)\) for the divisor function.
More generally, the sum of powers of divisors is defined by \[\sigma_k(n):=\sum_{d\mid n}d^k, \quad \text{ where } \quad k\in \mathbb N.\] Observe that \(\tau(n)=\sigma_0(n)\), and we abbreviate \(\sigma_1\) to \(\sigma\).
The Euler totient function \(\varphi\) is defined by \[\varphi(n):=\#\{m\in[n]: (n, m)=1\}=\sum_{m\in[n]}\delta((n, m)).\] Here and throughout, we use the convention from combinatorics that \([N]:=(0, N]\cap\mathbb Z_+\) for any real number \(N>0\).
The function \(\omega\) is defined as follows: \(\omega(1) = 0\) and \(\omega(n)\) counts the number of distinct prime factors of \(n\) for all \(n \geq 2\).
The function \(\Omega\) is defined as follows: \(\Omega(1) = 0\) and \(\Omega(n)\) counts the number of prime factors of \(n\) with multiplicities for all \(n \geq 2\).
The Liouville function \(\lambda\) is defined as follows \[\lambda(n) = (-1)^{\Omega(n)}.\]
The Möbius function \(\mu(n)\) is defined as follows \[\mu(n) := \begin{cases} 1 & \text{if } n = 1, \\ (-1)^k & \text{if } n \text{ is the product of } k \text{ distinct primes,} \\ 0 & \text{if } n \text{ is divisible by the square of a prime.} \end{cases}\]
The von Mangoldt function \(\Lambda(n)\) is defined as follows \[\Lambda(n) := \begin{cases} \log p & \text{if } n = p^k \text{ is a prime power,} \\ 0 & \text{otherwise.} \end{cases}\]
Sums and products of arithmetic functions are arithmetic functions: \[(f+g)(n):=f(n)+g(n) \quad \text{ and } \quad (f\cdot g)(n):=f(n)\cdot g(n).\]
Proof. Let \(\mathbb D(n):=\{d\in[n]: d\mid n\}\) be the set of all positive divisors of \(n\). \[(\: f \star g)(n)=\sum_{d \mid n} f(d) g\left(\frac{n}{d}\right)=\sum_{d \mid n} f\left(\frac{n}{d}\right) g(d),\] since the mapping \(\mathbb D(n)\ni d\mapsto n/d\in \mathbb D(n)\) is one-to-one.$$\tag*{$\blacksquare$}$$
The set \(\mathbb A:=(\mathbb A, +, \star)\) of all complex-valued arithmetic functions, with addition \(+\) defined by pointwise sum and multiplication \(\star\) defined by Dirichlet convolution, is a commutative ring with additive identity \(0\) and multiplicative identity \(\delta\), which is the Dirac delta at \(1\). Furthermore, if \(f(1) \neq 0\), then \(f\) is invertible.
Proof: Prove it!
The function \(f\) is said to be multiplicative if \(f(1) \neq 0\) and if, for all positive integers \(m, n\in\mathbb Z_+\) such that \((m, n) = 1\), we have \[f(mn) = f(m) f(n).\]
The function \(f\) is completely multiplicative if \(f(1) \neq 0\) and if the condition \[f(mn) = f(m) f(n)\] holds for all positive integers \(m\) and \(n\).
The function \(f\) is strongly multiplicative if \(f\) is multiplicative and if \(f(p^\alpha) = f(p)\) for all prime powers \(p^\alpha\).
Remark.
The condition \(f(1) \neq 0\) is a convention to exclude the zero function from the set of multiplicative functions.
Furthermore, it is easily seen that if \(f\) and \(g\) are multiplicative, then so are \(fg\) and \(f/g\) with \(g \neq 0\) for the quotient.
The function \(f\) is said to be additive if for all positive integers \(m, n\in\mathbb Z_+\) such that \((m, n) = 1\), we have \[f(mn) = f(m) + f(n).\]
The function \(f\) is completely additive if the condition \[f(mn) = f(m) + f(n)\] holds for all positive integers \(m\) and \(n\).
The function \(f\) is strongly additive if \(f\) is multiplicative and if \(f(p^\alpha) = f(p)\) for all prime powers \(p^\alpha\).
Let \(f:\mathbb Z_+\to\mathbb C\) be an arithmetic function.
\(f\) is multiplicative if and only if \(f(1) = 1\) and for all \(n = p_1^{\alpha_1} \cdots p_r^{\alpha_r}\), where the \(p_i\) are distinct primes, we have \[f(n) = \prod_{j\in[r]}f\left(p_j^{\alpha_j}\right).\]
\(f\) is additive if and only if \(f(1) = 0\) and for all \(n = p_1^{\alpha_1} \cdots p_r^{\alpha_r}\), where the \(p_i\) are distinct primes, we have \[f(n) = \sum_{j\in[r]}f\left(p_j^{\alpha_j}\right).\]
Proof: Prove it!
If \(f, g:\mathbb Z_+\to \mathbb C\) are multiplicative, then so is \(f \star g\).
Proof: Let \(f\) and \(g\) be two multiplicative functions and let \(m, n\in\mathbb Z_+\) be such that \((m, n)=1\).
Note that each divisor \(d\) of \(mn\) can be written uniquely in the form \(d=a b\) with \(a\mid m\), and \(b\mid n\) and \((a, b)=1\).
Hence, \[(\:f \star g)(m n)=\sum_{d \mid m n} f(d) g\left(\frac{m n}{d}\right)=\sum_{a \mid m} \sum_{b \mid n} f(a b) g\left(\frac{m n}{a b}\right).\]
Since \(f\) and \(g\) are multiplicative and \((a, b)=(m / a, n / b)=1\), then \[(\:f \star g)(m n)=\sum_{a \mid m} \sum_{b \mid n} f(a) f(b) g\left(\frac{m}{a}\right) g\left(\frac{n}{b}\right)=(f \star g)(m)(f \star g)(n)\] as required. $$\tag*{$\blacksquare$}$$
The functions \(\bf 1\), \({\rm Id}\), \(\delta\) are completely multiplicative.
The functions \(\log\) and \(\Omega\) are strongly additive and consequently the function \(\lambda\) is completely multiplicative.
Since \(\tau={\bf 1}\star{\bf 1}\), \(\sigma={\bf 1}\star {\rm Id}\) and \(\sigma_k={\bf 1}\star {\rm Id}^k\) and both \(\bf 1\) and \({\rm Id}\) are completely multiplicative, so are \(d, \sigma\) and \(\sigma_k\).
It is easily seen that, for all \(m, n\in\mathbb Z_+\), we have \[\omega(mn) = \omega(m) + \omega(n) - \omega(m, n),\] since in the sum \(\omega(m) + \omega(n)\), the prime factors of \((m, n)\) have been counted twice. This implies the additivity of \(\omega\).
The Möbius function \(\mu(n)\) is multiplicative. Indeed, \(\mu(1)=1\) and for all prime powers \(p^\alpha\), we also have
\[\mu(p^\alpha) = \begin{cases} -1, & \text{if } \alpha = 1, \\ \ \ \ 0, & \text{otherwise.} \end{cases}\]
So that, if \(n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_r^{\alpha_r}\) where the \(p_i\) are distinct primes, we have \[\mu(p_1^{\alpha_1}) \cdots \mu(p_r^{\alpha_r}) = \begin{cases} (-1)^r, & \text{if } \alpha_1 = \cdots = \alpha_r = 1, \\ \ \ \ \ \ 0, & \text{otherwise.} \end{cases}\] Hence \(\mu(p_1^{\alpha_1}) \cdots \mu(p_r^{\alpha_r}) =\mu(n)\) as desired.
We intend to prove the following identity \(\mu \star \mathbf{1}=\delta\), i.e. \[\begin{aligned} (\mu \star \mathbf{1})(n)=\sum_{d \mid n} \mu(d)=\delta(n)= \begin{cases} 1, & \text { if } n=1,\\ 0, & \text { if } n>1. \end{cases} \end{aligned}\]
Now since \(\mu\) and \(\mathbf{1}\) are multiplicative, so is the function \(\mu \star \mathbf{1}\) by the previous theorem and hence \((\mu \star \mathbf{1})(1)=1=\delta(1)\) is true for \(n=1\).
Besides, it is sufficient to prove \(\mu \star \mathbf{1}=\delta\) for prime powers by the previous lemma. Indeed, \[(\mu \star \mathbf{1})\left(p^{\alpha}\right)=\sum_{j=0}^{\alpha} \mu\left(p^{j}\right)=\mu(1)+\mu(p)=1-1=0=\delta(p^{\alpha})\] as asserted.
Let \(f\) and \(g\) be two arithmetic functions. Then we have \[g = f \star {\bf 1} \quad \iff \quad f = g \star \mu\] Equivalently, by expanding Dirichlet’s convolution, we have \[g(n) = \sum_{d \mid n} f(d) \iff f(n) = \sum_{d \mid n} g(d) \mu\left(\frac{n}{d}\right) \quad \text{ for all } \quad \mathbb Z_+.\]
Proof: Using the identity \(\mu \star \mathbf{1}=\delta\), we deduce \[g = f \star {\bf 1}\quad \iff \quad g \star \mu = f \star ({\bf 1} \star \mu) = f.\] This completes the proof. $$\tag*{$\blacksquare$}$$
Euler’s totient function is multiplicative and \(\varphi=\mu \star \mathrm{Id}\).
Moreover, we have
\[\begin{aligned} \varphi\left(p^{\alpha}\right)=p^{\alpha}-p^{\alpha-1}=p^{\alpha}\left(1-\frac{1}{p}\right), \end{aligned}\]
and by the multiplicativity we obtain
\[\varphi\left(n\right)=n\prod_{\substack{p\mid n\\p\in\mathbb P}}\left(1-\frac{1}{p}\right).\]
If both \(g\) and \(f \star g\) are multiplicative, then \(f\) is also multiplicative.
Proof: Prove it!
If \(g\) is multiplicative, then so is \(g^{-1}\), its Dirichlet inverse. In particular, the set of all multiplicative functions forms a multiplicative group with multiplication defined by the Dirichlet convolution.
Proof: Prove it!
If \(f\) is multiplicative, then we have \[\sum_{d \mid n}\mu(d)f(d) = \prod_{p \mid n} (1 - f(p)).\]
Proof: Prove it!
The von Mangoldt function is an example of a function that is neither multiplicative nor additive.
For every \(n\in\mathbb Z_+\) we have \[( \Lambda\star {\bf 1} )(n)=\sum_{d\mid n}\Lambda(d)=\log n.\]
Proof: The theorem is true if \(n = 1\) since both sides are \(0\).
Therefore, assume that \(n > 1\) and write \(n = \prod_{i=1}^r p_i^{\alpha_i}\). Then \[\log n=\sum_{i=1}^r \alpha_i\log p_i.\]
The only nonzero terms in the sum \(\sum_{d\mid n}\Lambda(d)\) come from those divisors \(d\) of the form \(p_k^m\) for \(m \in [a_k]\) and \(k \in [r]\). Hence, we have \[\sum_{d\mid n}\Lambda(d)=\sum_{i=1}^r\sum_{m=1}^{a_i}\Lambda(p_i^m)=\sum_{i=1}^r\sum_{m=1}^{a_i}\log p_i=\sum_{i=1}^r{a_i}\log p_i=\log n.\]
This completes the proof.$$\tag*{$\blacksquare$}$$
If \(n \geq 2\), we have \[\Lambda(n) = \sum_{d \mid n} \mu(d) \log \frac{n}{d} = -\sum_{d \mid n} \mu(d) \log d.\]
Proof: We know that \[( \Lambda\star {\bf 1} )(n)=\sum_{d\mid n}\Lambda(d)=\log n.\]
So inverting this formula by using the Möbius inversion formula, we obtain \[\Lambda(n) = \sum_{d \mid n} \mu(d) \log \frac{n}{d} = \log n \sum_{d \mid n} \mu(d) - \sum_{d \mid n} \mu(d) \log d,\] which simplifies to \[\Lambda(n) = \delta(n) \log n - \sum_{d \mid n} \mu(d) \log d.\]
Since \(\delta(n) \log n = 0\) for all \(n\in\mathbb Z_+\), the proof is complete.$$\tag*{$\blacksquare$}$$
In view of the multiplicative properties of certain arithmetic functions, we use Dirichlet series rather than power series in analytic number theory.
Here, we ignore convergence problems, and \(D(s, f)\) is the complex number equal to the sum when it converges.
In analytic number theory, it is customary to express a complex number \(s\in\mathbb C\) in the form \[s=\sigma +it\in\mathbb C.\]
Examples
\(D(s, \delta)=1\).
Presumably, the most important example of a Dirichlet series is the Riemann zeta function \[D(s, {\bf 1})=\zeta(s):=\sum_{n=1}^{\infty} \frac{1}{n^{s}}.\]
Let \(f, g\) and \(h\) be three arithmetic functions. Then \[h=f \star g \quad \Longleftrightarrow \quad D(s, h)=D(s, f) \cdot D(s, g).\]
Proof: We have \[D(s, f)\cdot D(s, g)=\sum_{k, m=1}^{\infty} \frac{f(k) g(m)}{(k m)^{s}}=\sum_{n=1}^{\infty} \frac{1}{n^{s}} \sum_{d \mid n} f(d) g\left(\frac{n}{d}\right)=\sum_{n=1}^{\infty} \frac{(f \star g)(n)}{n^{s}},\] which completes the proof. $$\tag*{$\blacksquare$}$$
Remark. The set \(\mathbb D:=(\mathbb D, +, \cdot)\) of formal Dirichlet series with addition \(+\) and multiplication \(\cdot\) defined respectively by \[D(s, f) + D(s, g)=D(s, f+g), \quad \text{ and } \quad D(s, f) \cdot D(s, g)=D(s, f\star g),\] forms a commutative ring with additive identity \(0\) and multiplicative identity \(1\). Moreover, \(\mathbb D:=(\mathbb D, +, \cdot)\) is isomorphic to the ring of arithmetic functions \(\mathbb A=(\mathbb A, +, \star)\) via the mapping \(\mathbb A\ni f\mapsto D(s, f)\in \mathbb D\).
Let \(f\) be an arithmetic function. Then \(f\) is multiplicative if and only if \[D(s, f)=\prod_{p\in\mathbb P}\left(1+\sum_{k=1}^{\infty} \frac{f\left(p^{k}\right)}{p^{s k}}\right).\] The above product is called the Euler product of \(D(s, f)\).
Proof: Expanding the product we obtain a formal sum of all products of the form \(\frac{f(p_1^{a_1}) \cdots f(p_r^{a_r})}{(p_1^{a_1} \cdots p_r^{a_r})^s},\) where \(p_1, \ldots, p_r\) are distinct prime numbers, \(a_1, \ldots, a_r \in \mathbb{Z}_{+}\), and \(r \in \mathbb{N}\).
By multiplicativity, the numerator can be written as \(f(p_1^{a_1} \cdots p_r^{a_r})\).
The Fundamental Theorem of Arithmetic implies that the products \(p_1^{a_1} \cdots p_r^{a_r}\) are in one-to-one correspondence with all natural numbers.
This gives a formal proof of the desired identity. $$\tag*{$\blacksquare$}$$
By the previous lemma \(\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=\prod_{p\in\mathbb P}\big(1-\frac{1}{p^s}\big)^{-1}\), hence \[\frac{1}{\zeta(s)}=\prod_{p\in\mathbb P}\bigg(1-\frac{1}{p^s}\bigg)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s},\] which implies \(\mu\star{\bf 1}=\delta\).
Taking logarithm we obtain \[\log \zeta(s)=\sum_{p\in\mathbb P}\log\bigg(1-\frac{1}{p^s}\bigg)^{-1}=\sum_{p\in\mathbb P}\sum_{k=1}^{\infty}\frac{1}{kp^{ks}}.\]
By formal differentiation, we have \[-\zeta'(s)=\sum_{n=1}^{\infty}\frac{\log n}{n^s}, \quad\text{ and } \quad -\frac{\zeta'(s)}{\zeta(s)}=\sum_{p\in\mathbb P}\sum_{k=1}^{\infty}\frac{\log p}{p^{ks}}=\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}.\]
Thus, \(\Lambda=\mu\star \log\) and consequently \(\Lambda\star {\bf 1}=\log\).
For each Dirichlet series \(D(s, f)\) with \(s=\sigma +it\in\mathbb C\), there exists \(\sigma_{a} \in \mathbb{R} \cup\{ \pm \infty\}\), called the abscissa of absolute convergence, such that
\(D(s, f)\) converges absolutely in the half-plane \(\sigma>\sigma_{a}\);
\(D(s, f)\) does not converge absolutely in the half-plane \(\sigma<\sigma_{a}\).
Remarks.
In particular, the series \(D(s, f)\) defines an analytic function in the halfplane \(\sigma>\sigma_{a}\). By abuse of notation, this function will be still denoted by \(D(s, f)\).
If \(|\: f(n)| \leqslant \log n\), then the series \(D(s, f)\) is absolutely convergent in the half-plane \(\sigma>1\), and hence \(\sigma_{a} \leqslant 1\).
Remarks.
At \(\sigma=\sigma_{a}\), the series may or may not converge absolutely. For instance, \(\zeta(s)\) converges absolutely in the half-plane \(\sigma>\sigma_{a}=1\), but does not converge on the line \(\sigma=1\).
On the other hand, the Dirichlet series associated to the function \(f(n)=\) \(1 /(\log (en))^{2}\) has also \(\sigma_{a}=1\) for the abscissa of absolute convergence, but converges absolutely at \(\sigma=1\).
Proof: Let \(S:=\{s\in \mathbb C: D(s, f) \ \text{converges absolutely}\}\).
If \(S=\varnothing\), then put \(\sigma_{a}=+\infty\). Otherwise define \[\sigma_{a}:=\inf \{\sigma: s=\sigma+i t \in S\}.\]
\(D(s, f)\) does not converge absolutely if \(\sigma<\sigma_{a}\) by the definition of \(\sigma_{a}\).
On the other hand, suppose that \(D(s, f)\) is absolutely convergent for some \(s_{0}=\sigma_{0}+i t_{0} \in \mathbb{C}\) and let \(s=\sigma+i t\) be such that \(\sigma \geqslant \sigma_{0}\). Since \[\left|\frac{f(n)}{n^{s}}\right|=\left|\frac{f(n)}{n^{s_{0}}}\right| \times \frac{1}{n^{\sigma-\sigma_{0}}} \leqslant\left|\frac{f(n)}{n^{s_{0}}}\right|\] we infer that \(D(s, f)\) converges absolutely at any point \(s\) with \(\sigma \geqslant \sigma_{0}\).
Now by the definition of \(\sigma_{a}\), there exist points arbitrarily close to \(\sigma_{a}\) at which \(D(s, f)\) converges absolutely, and therefore by above \(D(s, f)\) converges absolutely at each point \(s\) such that \(\sigma>\sigma_{a}\). $$\tag*{$\blacksquare$}$$
Let \(D(s, f)=\sum_{n=1}^{\infty} f(n) n^{-s}\) be a Dirichlet series. Assume that \[|\:f(n)| \leqslant M n^{\alpha} \quad \text{ for all } \quad n\in\mathbb Z_+,\] for some \(\alpha \geqslant 0\) and \(M>0\) independent of \(n\).
Then \(D(s, f)\) converges absolutely in the half-plane \(\sigma>\alpha+1\).
In particular, \(\sigma_a\le \alpha+1\).
Proof: Indeed, observe that \[\begin{aligned} |D(s, f)|=\sum_{n=1}^{\infty}|\: f(n) n^{-s}|\le M\sum_{n=1}^{\infty}n^{-(\sigma-\alpha)}<\infty, \end{aligned}\] whenever \(\sigma>\alpha+1\), as desired.$$\tag*{$\blacksquare$}$$
Let \(f, g\) be two arithmetic functions. If the Dirichlet series \(D(s, f)\) and \(D(s, g)\) are absolutely convergent at a point \(s_{0}\), then \(D(s, f\star g)\) converges absolutely at \(s_{0}\) and we have \(D(s_0, f\star g)=D(s_0, f)D(s_0, g)\).
Proof: We have \[D(s_0, f)D(s_0, g)=\sum_{n=1}^{\infty} \frac{f\star g(n)}{n^{s_{0}}}=D(s, f\star g),\] where the rearrangement of the terms in the double sums is justified by the absolute convergence of the two series \(D(s, f)\) and \(D(s, g)\) at \(s=s_{0}\).
Then the absolute convergence of \(D(s_0, f\star g)\) follows, as we have \[\sum_{n=1}^{\infty}\left|\frac{f\star g(n)}{n^{s_{0}}}\right| \leqslant\left(\sum_{n=1}^{\infty}\left|\frac{f(n)}{n^{s_{0}}}\right|\right)\left(\sum_{n=1}^{\infty}\left|\frac{g(n)}{n^{s_{0}}}\right|\right).\qquad \tag*{$\blacksquare$}\]
Let \(f\) be an arithmetic function such that \(f(1) \neq 0\). Let \(f^{-1}\) be the convolution inverse of the fuction \(f\), i.e. \(f \star f^{-1}=\delta\). Then \[D(s, f^{-1})=\frac{1}{D(s, f)}\] at every point \(s\) where \(D(s, f)\) and \(D(s, f^{-1})\) converge absolutely.
Example
The Möbius function satisfies \(\mu^{-1}=\mathbf{1}\), hence for \(\sigma>1\), we have \[\sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}}=\frac{1}{\zeta(s)}.\]
In particular, \(\sum_{n=1}^{\infty} \frac{\mu(n)}{n^{2}}=\frac{6}{\pi^2}\), since \(\zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^2}{6}\).
Let \(f, g:\mathbb Z_+\to \mathbb C\) be arithmetic functions. Let \(F(x):=\sum_{1\le n \leq x} f(n)\).
Then for any \(a, b\in\mathbb N\) with \(a<b\), we have \[\begin{aligned} \sum_{n=a+1}^{b} f(n) g(n)= F(b) g(b)-F(a) g(a+1) -\sum_{n=a+1}^{b-1} F(n)(g(n+1)-g(n)). \end{aligned}\]
Let \(x, y\in \mathbb R_+\) with \(\lfloor y\rfloor<\lfloor x\rfloor\), and let \(g\in C^1([y, x])\). Then \[\sum_{y<n \leq x} f(n) g(n)=F(x) g(x)-F(y) g(y)-\int_{y}^{x} F(t) g^{\prime}(t) d t\]
In particular, if \(x \geq 2\) and \(g\in C^1([1, x])\), then \[\sum_{n \leq x} f(n) g(n)=F(x) g(x)-\int_{1}^{x} F(t) g^{\prime}(t) d t.\]
Let \(D(s, f)=\sum_{n=1}^{\infty} f(n) n^{-s}\) be a Dirichlet series. Assume that \[\Big|\sum_{x<n \leqslant y} f(n)\Big| \leqslant M y^{\alpha} \quad \text{ for all } \quad 0<x<y,\] for some \(\alpha \geqslant 0\) and \(M>0\) independent of \(x\) and \(y\).
Then \(D(s, f)\) converges in the half-plane \(\sigma>\alpha\).
Furthermore, we have in this half-plane \[|D(s, f)| \leqslant \frac{M|s|}{\sigma-\alpha}, \quad \text { and } \quad \bigg|\sum_{x<n \leqslant y} \frac{f(n)}{n^{s}}\bigg| \leqslant \frac{M}{x^{\sigma-\alpha}}\left(\frac{|s|}{\sigma-\alpha}+1\right).\]
The latter statement ensures that \(D(s, f)\) converges uniformly in any compact subset of the half plane \(\sigma>\alpha\).
Proof: Set \(A(x)=\sum_{1\le n \leqslant x} f(n)\) and \(S(x, y)=A(y)-A(x)\).
By partial summation we have \[\sum_{x<n \leqslant y} \frac{f(n)}{n^{s}}=\frac{S(x, y)}{y^{s}}+s \int_{x}^{y} \frac{S(x, u)}{u^{s+1}} d u.\]
By hypothesis we have \(\left|S(x, y) / y^{s}\right| \leqslant M y^{\alpha-\sigma}\), so that \(S(x, y) / y^{s}\) tends to 0 as \(y \to \infty\) in the half-plane \(\sigma>\alpha\).
Therefore if one of \[D(s, f)=\sum_{n \in\mathbb N} \frac{f(n)}{n^{s}}, \quad \text { or } \quad s \int_{1}^{\infty} \frac{A(u)}{u^{s+1}} d u.\] converges, then so does the other, and the two quantities converge to the same limit.
But since \[\left|\frac{A(u)}{u^{s+1}}\right| \leqslant \frac{M}{u^{\sigma-\alpha+1}},\] we infer that the integral converges absolutely for \(\sigma>\alpha\), and hence \(D(s, f)\) is convergent in this half-plane.
Therefore for all \(\sigma>\alpha\), we obtain \[\sum_{n=1}^{\infty} \frac{f(n)}{n^{s}}=s \int_{1}^{\infty} \frac{A(u)}{u^{s+1}} d u,\] and hence \[|D(s, f)| \leqslant M|s| \int_{1}^{\infty} \frac{du}{u^{\sigma-\alpha+1}}=\frac{M|s|}{\sigma-\alpha}.\]
Similarly \[\Big|\sum_{x<n \leqslant y} \frac{f(n)}{n^{s}}\Big| \leqslant \frac{M}{y^{\sigma-\alpha}} +M|s| \int_{x}^{\infty} \frac{d u}{u^{\sigma-\alpha+1}} \leqslant \frac{M}{x^{\sigma-\alpha}}\left(\frac{|s|}{\sigma-\alpha}+1\right)\] as required.$$\tag*{$\blacksquare$}$$
For each Dirichlet series \(D(s, f)\), there exists \(\sigma_{c} \in \mathbb{R} \cup\{ \pm \infty\}\), called the abscissa of convergence, such that \(D(s, f)\) converges in the half-plane \(\sigma>\sigma_{c}\) and does not converge in the half-plane \(\sigma<\sigma_{c}\). Furthermore, \[\sigma_{c} \leqslant \sigma_{a} \leqslant \sigma_{c}+1.\]
Proof: Suppose first that \(D(s, f)\) converges at a point \(s_{0}=\sigma_{0}+i t_{0}\) and fix a small real number \(\varepsilon>0\). By Cauchy’s theorem, there exists \(x_{\varepsilon} \geqslant 1\) such that, for all \(y>x \geqslant x_{\varepsilon}\), we have \[\bigg|\sum_{x<n \leqslant y} \frac{f(n)}{n^{s_{0}}}\bigg| \leqslant \varepsilon.\]
Let \(s=\sigma+i t \in \mathbb{C}\) such that \(\sigma>\sigma_{0}\). Using the previous lemma with \(s\) replaced by \(s-s_{0}\) and \(\alpha=0\), we obtain \[\bigg|\sum_{x<n \leqslant y} \frac{f(n)}{n^{s}}\bigg| \leqslant \varepsilon\left(\frac{\left|s-s_{0}\right|}{\sigma-\sigma_{0}}+1\right).\] so that \(D(s, f)\) converges by Cauchy’s theorem.
Now we may proceed as before. Let \[S:=\{s\in \mathbb C: D(s, f) \ \text{ converges}\}.\]
If \(S=\varnothing\), then we put \(\sigma_{c}=+\infty\). Otherwise define \[\sigma_{c}:=\inf \{\sigma: s=\sigma+i t \in S\}.\]
\(D(s, f)\) does not converge if \(\sigma<\sigma_{c}\) by the definition of \(\sigma_{c}\).
On the other hand, there exist points \(s_{0}\) with \(\sigma_{0}\) being arbitrarily close to \(\sigma_{c}\) at which \(D(s, f)\) converges.
By above, \(D(s, f)\) converges at any point \(s\) such that \(\sigma>\sigma_{0}\). Since \(\sigma_{0}\) may be chosen as close to \(\sigma_{c}\) as we want, it follows that \(D(s, f)\) converges at any point \(s\) such that \(\sigma>\sigma_{c}\).
The inequality \(\sigma_{c} \leqslant \sigma_{a} \leqslant \sigma_{c}+1\) remains to be shown.
The lower bound is obvious. For the upper bound, it suffices to show that if \(D(s_{0}, f)\) converges for some \(s_{0}\), then it converges absolutely for all \(s\) such that \(\sigma>\sigma_{0}+1\). Now if \(D(s, f)\) converges at some point \(s_{0}\), then \[\lim_{n\to \infty}f(n)n^{-s_{0}}=0.\]
Thus there exists a positive integer \(n_{0}\) such that, for all \(n \geqslant n_{0}\), we have \(\left|\: f(n)\right| \le n^{\sigma_{0}}\), hence \(D(s, f)\) is absolutely convergent in the half-plane \(\sigma>\sigma_{0}+1\) as required. $$\tag*{$\blacksquare$}$$
A Dirichlet series \(D(s, f)=\sum_{n=1}^{\infty} f(n) n^{-s}\) defines a holomorphic function of the variable \(s\) in the half-plane \(\sigma>\sigma_{c}\), in which \(D(s, f)\) can be differentiated term by term so that, for all \(s=\sigma+it\) with \(\sigma>\sigma_{c}\), we have \[\partial^k_sD(s, f)=\sum_{n=1}^{\infty} \frac{(-1)^{k}(\log n)^{k} f(n)}{n^{s}}, \quad \text{ for } \quad k\in\mathbb Z_+.\]
Proof: The partial sums of a Dirichlet series is a holomorphic function of the variable \(s\) that converges uniformly on any compact subset of the half-plane \(\sigma>\sigma_{c}\). Hence, \(D(s, f)\) must be holomorphic in that region.
Consequently, term-by-term differentiation is allowed, and since \(n^{-s}=e^{-s\log n}\), then \[\partial^k_sn^{-s}=(-1)^{k}(\log n)^{k}n^{-s}.\quad \tag*{$\blacksquare$}\]
Let \(D(s, f)=\sum_{n=1}^{\infty} f(n) n^{-s}\) be a Dirichlet series with abscissa of convergence \(\sigma_{c}\).
If \(D(s, f)=0\) for all \(s\) such that \(\sigma>\sigma_{c}\), then \(f(n)=0\) for all \(n \in \mathbb{Z}_+\).
In particular, if \(D(s, f)=D(s, f)\) for all \(s\) such that \(\sigma>\sigma_{c}\), then \(f(n)=g(n)\) for all \(n \in \mathbb{Z}_+\).
Proof: Suppose the contrary and let \(k\in \mathbb Z_+\) be the smallest integer such that \(f(k) \neq 0\). Then \[D(s, f)=\sum_{n=k}^{\infty} f(n) n^{-s}=0\] for all \(s\) such that \(\sigma>\sigma_{c}\). Then we have \[G(s)=k^{s}D(s, f)=k^{s} \sum_{n=k}^{\infty} \frac{f(n)}{n^{s}}=0.\]
Therefore, for all \(s\) such that \(\sigma>\sigma_{c}\) we have \[G(s)=f(k)+\sum_{n=k+1}^{\infty} f(n)\left(\frac{k}{n}\right)^{s}=0.\]
Hence \[0=\lim _{\sigma \rightarrow \infty} G(\sigma)=f(k)\neq 0,\] which is impossible.$$\tag*{$\blacksquare$}$$
If \(f(n) \geqslant 0\) for all \(n\in\mathbb Z_+\), and \(D(s,f)=\sum_{n=1}^{\infty} f(n) n^{-s}\) is a Dirichlet series with abscissa of convergence \(\sigma_{c} \in \mathbb{R}\), then \(D(s, f)\) has a singularity at \(s=\sigma_{c}\).
Proof: We may assume that \(\sigma_{c}=0\) and \(|D(0, f)|<\infty\).
By the Taylor expansion of \(D(s, f)\) about \(a>0\) we have
\[D(s, f)=\sum_{k=0}^{\infty} \frac{(s-a)^{k}}{k!} \partial^k_sD(a, f)=\sum_{k=0}^{\infty} \frac{(s-a)^{k}}{k!} \sum_{n=1}^{\infty} \frac{(-1)^{k}(\log n)^{k} f(n)}{n^{a}},\] which must converge at some \(s=b<0\). Hence
\[0\le \sum_{k=0}^{\infty} \sum_{n=1}^{\infty} \frac{((a-b) \log n)^{k} f(n)}{n^{a} k!}<\infty.\]
Each term is nonnegative, so the order of summation may be changed \[\sum_{n=1}^{\infty} \frac{f(n)}{n^{a}} \sum_{k=0}^{\infty} \frac{((a-b) \log n)^{k}}{k!}=\sum_{n=1}^{\infty} \frac{f(n)}{n^{b}}=\infty,\] since \(b<0=\sigma_{c}\), giving a contradiction. $$\tag*{$\blacksquare$}$$
Let \(f\) be a multiplicative function satisfying \[\sum_{p\in \mathbb P} \sum_{k=1}^{\infty}|\: f(p^{k})|<\infty.\] Then the series \(\sum_{n \geqslant 1} f(n)\) is absolutely convergent and we have \[\sum_{n=1}^{\infty} f(n)=\prod_{p\in \mathbb P}\left(1+\sum_{k=1}^{\infty} f\left(p^{k}\right)\right).\]
Proof: Let us first notice that the inequality \(\sum_{p\in \mathbb P} \sum_{k=1}^{\infty}|\: f(p^{k})|<\infty\) implies the convergence of the product \[\prod_{p\in\mathbb P}\left(1+\sum_{k=1}^{\infty}\left|\: f\left(p^{k}\right)\right|\right)\le \exp\bigg(\sum_{p\in \mathbb P} \sum_{k=1}^{\infty}|\: f(p^{k})|\bigg)<\infty.\]
Now let \(x \geqslant 2\) and set \(P(x)=\prod_{p \in \mathbb P_{\le x}}\big(1+\sum_{k=1}^{\infty}\big|\: f(p^{k})\big|\big)\).
The convergence of the series \(\sum_{k=1}^{\infty}\left|\: f\left(p^{k}\right)\right|\) enables us to rearrange the terms when we expand \(P(x)\), hence \[P(x)=\sum_{{\rm gpf}(n) \le x}|\: f(n)|,\] where \({\rm gpf}(1)=1\) and \({\rm gpf}(n)\) is the greatest prime factor of \(n\ge2\).
Since each integer \(n \leqslant x\) satisfies the condition \({\rm gpf}(n) \leqslant x\), we have \[\sum_{1\le n \leqslant x}|\: f(n)| \leqslant P(x)\]
Since \(P(x)\) has a finite limit as \(x \to \infty\), the above inequality implies that \(\sum_{n \geqslant 1} |\: f(n)|<\infty\). The second part of the theorem follows from \[\bigg|\sum_{n=1}^{\infty} f(n)-\prod_{p \in\mathbb P_{\le x}}\Big(1+\sum_{k=1}^{\infty} f(p^{k})\Big)\bigg| \leqslant \sum_{n>x}|\: f(n)|,\] and the fact that the right-hand side tends to \(0\) as \(x \to \infty\). $$\tag*{$\blacksquare$}$$
Let \(f\) be a multiplicative function and let \(s_{0}\in \mathbb C\). Then the three following assertions are equivalent.
One has \[\sum_{p\in\mathbb P} \sum_{k=1}^{\infty} \frac{|\: f(p^{k})|}{p^{s_{0} k}}<\infty.\]
The series \(D(s, f)\) is absolutely convergent in the half-plane \(\sigma>\sigma_{0}\).
The product \[\prod_{p\in \mathbb P}\left(1+\sum_{k=1}^{\infty} \frac{f(p^{k})}{p^{s k}}\right)\] is absolutely convergent in the half-plane \(\sigma>\sigma_{0}\). If one of these conditions holds, then we have for all \(\sigma>\sigma_{0}\) (in particular for all \(\sigma>\sigma_{a}\)) that \[D(s, f)=\prod_{p\in\mathbb P}\left(1+\sum_{k=1}^{\infty} \frac{f(p^{k})}{p^{s k}}\right).\]