Let \(a\) and \(b\) be coprime natural numbers, and let \(d\) be a divisor of \(ab\). Since \(a\) and \(b\) are coprime, we can use the fact that the divisors of \(ab\) correspond to pairs of divisors of \(a\) and \(b\). Specifically, we can express \(d\) as \(d = d_a \cdot d_b\), where \(d_a\) divides \(a\) and \(d_b\) divides \(b\). The uniqueness of this decomposition follows from the coprimality of \(a\) and \(b\), which ensures that the divisors of \(a\) and \(b\) do not share any common factors. Therefore, there is a one-to-one correspondence between the divisors of \(ab\) and the pairs of divisors of \(a\) and \(b\), which guarantees the uniqueness of the decomposition.

Since \(f\) is multiplicative, we can express the sum over divisors of \(ab\) in terms of the sums over divisors of \(a\) and \(b\). The key idea is to use the fact that the divisors of \(ab\) can be expressed as products of divisors of \(a\) and divisors of \(b\), due to the coprimality condition. Specifically, each divisor \(d\) of \(ab\) can be uniquely written as \(d = d_a * d_b\), where \(d_a\) divides \(a\) and \(d_b\) divides \(b\). Therefore, we can rewrite the sum as a double sum over the divisors of \(a\) and \(b\), which factorizes into the product of the two separate sums.

Multiply out and collect terms.

By definition of \(\zeta \), we have \(\zeta (n) = 1\) for all \(n \geq 1\). Thus, the Dirichlet convolution \((\zeta * \zeta )(n)\) counts the number of ways to write \(n\) as a product of two positive integers, which is exactly the number of divisors of \(n\), i.e., \(\tau (n)\).

From the previous theorem, we have that the Dirichlet convolution of \(\zeta \) with itself is \(\tau \). Taking L-series on both sides, we get \(L(\tau , s) = L(\zeta , s) \cdot L(\zeta , s)\). Since \(L(\zeta , s)\) is the Riemann zeta function \(\zeta (s)\), we conclude that \(L(\tau , s) = \zeta (s)^2\) for \(\Re (s) {\gt} 1\).

By definition, \(d_k\) is the \(k\)-fold Dirichlet convolution of \(\zeta \). When \(k = 0\), this corresponds to the empty convolution, which is defined to be the multiplicative identity in the algebra of arithmetic functions. The multiplicative identity is the function that takes the value \(1\) at \(n=1\) and \(0\) elsewhere, which can be expressed as \(\zeta ^0\).

By definition, \(d_k\) is the \(k\)-fold Dirichlet convolution of \(\zeta \). When \(k = 1\), this means we are taking the convolution of \(\zeta \) with itself once, which simply gives us \(\zeta \). Therefore, \(d_1 = \zeta ^1 = \zeta \).

By definition, \(d_k\) is the \(k\)-fold Dirichlet convolution of \(\zeta \). When \(k = 2\), this means we are taking the convolution of \(\zeta \) with itself twice, which gives us \(\zeta * \zeta \). From the earlier theorem, we know that \(\zeta * \zeta = \tau \), where \(\tau \) is the divisor count function. Therefore, \(d_2 = \zeta ^2 = \tau \).

By definition, \(d_k\) is the \(k\)-fold Dirichlet convolution of \(\zeta \). Therefore, \(d_{k + 1}\) is the \((k + 1)\)-fold convolution of \(\zeta \), which can be expressed as the convolution of \(d_k\) (the \(k\)-fold convolution) with \(\zeta \). Thus, we have \(d_{k + 1} = d_k * \zeta \).

Since \(d_k\) is defined as the \(k\)-fold Dirichlet convolution of \(\zeta \), and we know that the L-series of \(\zeta \) converges for \(\Re (s) {\gt} 1\), it follows that the L-series of \(d_k\) also converges for \(\Re (s) {\gt} 1\). This is because the convolution of functions with convergent L-series will also have a convergent L-series in the same region. Therefore, we can conclude that \(L(d_k, s)\) is summable for \(\Re (s) {\gt} 1\).

From the definition of \(d_k\) as the \(k\)-fold Dirichlet convolution of \(\zeta \), we can express \(d_k\) as \(\zeta ^k\). The L-series of a Dirichlet convolution corresponds to the product of the L-series of the individual functions. Since \(L(\zeta , s)\) is the Riemann zeta function \(\zeta (s)\), it follows that \(L(d_k, s) = L(\zeta ^k, s) = (L(\zeta , s))^k = \zeta (s)^k\) for \(\Re (s) {\gt} 1\) where the series converges.

The function \(d_k\) is defined as the \(k\)-fold Dirichlet convolution of \(\zeta \). Since \(\zeta \) is a multiplicative function, and the Dirichlet convolution of multiplicative functions is also multiplicative, it follows that \(d_k\) is multiplicative for all \(k\). This can be shown by induction on \(k\), using the fact that the convolution of a multiplicative function with another multiplicative function remains multiplicative.

The function \(d_k\) counts the number of ways to write a natural number as an ordered product of \(k\) natural numbers. For a prime power \(p^a\), the number of ways to factor it into \(k\) factors corresponds to the number of non-negative integer solutions to the equation \(x_1 + x_2 + ... + x_k = a\), where each \(x_i\) represents the exponent of \(p\) in the factorization of the corresponding factor. This is a classic combinatorial problem, and the number of solutions is given by the formula \((a + k - 1).choose (k - 1)\), which counts the ways to distribute \(a\) indistinguishable items (the prime factors) into \(k\) distinguishable boxes (the factors).

The function \(d_k\) is multiplicative, so to compute \(d_k(n)\) for a general natural number \(n\), we can factor \(n\) into its prime power decomposition: \(n = p_1^{a_1} p_2^{a_2} ... p_m^{a_m}\). Since \(d_k\) is multiplicative, we have:

Using the explicit formula for prime powers from the previous theorem, we can substitute to get:

This gives us a complete formula for \(d_k(n)\) in terms of the prime factorization of \(n\).

The function \(\sigma ^R\) is defined as the sum of the \(s\)-th powers of the divisors of \(n\). When \(s\) is a natural number \(k\), this definition coincides with the classical divisor power sum function \(\sigma _k(n)\), which also sums the \(k\)-th powers of the divisors of \(n\). Therefore, for natural exponents, we have \(\sigma ^R_k(n) = \sigma _k(n)\) when we view \(\sigma _k(n)\) as a complex number. This can be shown by directly comparing the definitions and noting that both functions sum over the same set of divisors with the same exponentiation.

This follows immediately from the definition.

Note that \(d\mid p^i\) implies that \(d=p^j\) with \(0\leq j\leq i\). Thus,

Note that \(d\ mapsto n/d\) forms a one-to-one mapping between the divisors of \(n\). Using this in combination with the definiton we have that

By definition we have that

This immediately follows from the fact that exponentiation with a fixed power is a homomorphism.

The function \(\sigma ^R(\nu )\) is defined as the sum of the \(\nu \)-th powers of the divisors of \(n\). The function \(\text{pow}^R(\nu )\) is defined as \(n \mapsto n^\nu \) for \(n \neq 0\) and \(0\) for \(n = 0\). The Dirichlet convolution of \(\zeta \) (the constant function \(1\)) and \(\text{pow}^R(\nu )\) is exactly \(\sigma ^R(\nu )\), since for each divisor \(d\) of \(n\), we have \((\zeta * \text{pow}^R(\nu ))(n) = \sum _{d|n} 1 \cdot d^\nu = \sigma ^R(\nu )(n)\). Thus, we have \(\sigma ^R(\nu ) = \zeta * \text{pow}^R(\nu )\).

Recall from Lemma ?? that \(\sigma ^R\) is \(\zeta \) convolved with Definition 12.1.5. Since both of these are multiplicative functions, their convolution is also multiplicative.

Since \(\sigma ^R_s\) is multiplicative, it suffices to understand it at primes powers.

Applying Lemma ??.

The function \(\text{pow}^R(\nu )\) is defined as \(n \mapsto n^\nu \) for \(n \neq 0\) and \(0\) for \(n = 0\). The L-series of \(\text{pow}^R(\nu )\) at \(s\) is given by the sum \(\sum _{n=1}^{\infty } n^{\nu - s}\). This series converges to the Riemann zeta function \(\zeta (s - \nu )\) for \(\Re (s - \nu ) {\gt} 1\), since the zeta function is defined as \(\zeta (s) = \sum _{n=1}^{\infty } n^{-s}\) for \(\Re (s) {\gt} 1\). Therefore, we have \(L(\text{pow}^R(\nu ), s) = \zeta (s - \nu )\) under the condition that \(\Re (s - \nu ) {\gt} 1\).

We apply ?? which states that if there exists a constant \(C\) such that \(\| f(n)\| \leq C \cdot n^r\) for all \(n\) sufficiently large, then the abscissa of absolute convergence of \(L(f, s)\) is at most \(r + 1\). In our case, we can take \(f(n) = n^\nu \) and observe that \(\| n^\nu \| = n^{\Re (\nu )}\). Thus, we can choose \(C = 1\) and \(r = \Re (\nu )\), which gives us the desired result that the abscissa of absolute convergence of \(L(\text{pow}^R(\nu ), s)\) is at most \(\Re (\nu ) + 1\).

The divisor power sum function \(\sigma _\nu \) is the Dirichlet convolution of the constant function \(1\) (i.e., \(\zeta \)) and the power function \(n \mapsto n^\nu \). The L-series of a Dirichlet convolution is the product of the L-series of the individual functions. Since \(L(1, s) = \zeta (s)\) and \(L(n \mapsto n^\nu , s) = \zeta (s - \nu )\), we have \(L(\sigma _\nu , s) = \zeta (s) \cdot \zeta (s - \nu )\) for \(\Re (s) {\gt} 1\) and \(\Re (s - \nu ) {\gt} 1\).

This is a direct consequence of the multiplicative properties of the functions involved and the definition of the L-series. The left-hand side can be expressed as a product of L-series corresponding to the functions \(\zeta \), \(n \mapsto n^{-\alpha }\), and \(n \mapsto n^{-\beta }\). The right-hand side involves the L-series of the convolution of \(\sigma _\alpha \) and \(\sigma _\beta \), which can be expressed in terms of the L-series of \(\zeta \) and the power functions. By carefully applying the properties of Dirichlet convolutions and the definitions of the L-series, we can derive the stated formula.

This is a special case of the previous theorem where we set \(\alpha = \beta = 0\).

This follows from the multiplicative properties of the divisor function \(\tau \) and the definition of the L-series. The left-hand side can be expressed as a product of L-series corresponding to \(\zeta \) and the function \(n \mapsto \tau (n^2)\). The right-hand side is the L-series of the function \(n \mapsto \tau (n)^2\). By analyzing the Euler products and using the fact that \(\tau (n)\) counts divisors, we can derive the stated equality.

This follows from the previous two theorems. From the corollary of Ramanujan’s formula, we have \(\zeta (s)^4 = \zeta (2s) \sum _{n=1}^{\infty } \tau (n)^2 n^{-s}\). From the Baby Rankin-Selberg result, we have \(\zeta (s) \sum _{n=1}^{\infty } \tau (n^2) n^{-s} = \sum _{n=1}^{\infty } \tau (n)^2 n^{-s}\). Combining these two results, we can express \(\zeta (s)^4\) in terms of \(\zeta (s)\) and \(\sum _{n=1}^{\infty } \tau (n^2) n^{-s}\), which leads to the conclusion that \(\zeta (s)^3 = \zeta (2s) \sum _{n=1}^{\infty } \tau (n^2) n^{-s}\).

This is an alternative expression for \(\zeta (s)^3\) that can be derived from the previous results. By expressing \(\zeta (s)^3\) in terms of the L-series of \(\tau (n^2)\) and using the properties of Dirichlet convolutions, we can rewrite the sum in a way that involves summing over divisors \(d\) and corresponding \(m\) such that \(d^2 m = n\). This rearrangement of the series allows us to express \(\zeta (s)^3\) in the desired form.

Follows from previous arguments.

The series \(\sum _{n=1}^{\infty } |\mu (n)| n^{-s}\) has Euler product \(\prod _{p} (1 + p^{-s})\). On the other hand, \(\zeta (2s)=\prod _p (1 - p^{-2s})^{-1}\). The product of these two Euler products is \(\prod _p (1 - p^{-s})^{-1} = \zeta (s)\), which gives the desired formula.

The function \(\mu ^2(n)\) is the indicator function for squarefree numbers, meaning it is \(1\) if \(n\) is squarefree and \(0\) otherwise. The sum \(\sum _{d^2|n} \mu (d)\) counts the contributions from divisors \(d\) such that \(d^2\) divides \(n\). If \(n\) is squarefree, then the only divisor \(d\) such that \(d^2 | n\) is \(d=1\), which contributes \(\mu (1) = 1\). If \(n\) is not squarefree, then there exists a prime \(p\) such that \(p^2 | n\), and the corresponding divisor \(d=p\) will contribute \(\mu (p) = -1\), which will cancel out the contribution from \(d=1\). Therefore, we have \(\mu ^2(n) = \sum _{d^2|n} \mu (d)\).

The Liouville function \(\lambda (n)\) is defined as \((-1)^{\Omega (n)}\), where \(\Omega (n)\) is the total number of prime factors of \(n\) counted with multiplicity. This means that for each prime factor of \(n\), we contribute a factor of \(-1\) to the product, and the overall sign of \(\lambda (n)\) depends on whether the total number of prime factors is even or odd. Thus, we have \(\lambda (n) = (-1)^{\Omega (n)}\) by definition.

Let \(f\) and \(g\) be completely multiplicative functions. We want to show that their Dirichlet convolution \(h = f * g\) is also completely multiplicative.

Let \(f\) be a completely multiplicative function. To show that \(f\) is multiplicative, we need to verify that \(f(1) = 1\) and that \(f(ab) = f(a)f(b)\) for all coprime natural numbers \(a\) and \(b\). Since \(f\) is completely multiplicative, we have \(f(1) = 1\) by definition. For coprime \(a\) and \(b\), we can write \(ab\) as a product of prime factors, and since \(f\) is completely multiplicative, it will factor as the product of the values of \(f\) at those prime factors. This means that \(f(ab) = f(a)f(b)\) for coprime \(a\) and \(b\), which shows that \(f\) is multiplicative.

The Liouville function \(\lambda (n)\) is defined as \((-1)^{\Omega (n)}\), where \(\Omega (n)\) counts the total number of prime factors of \(n\) with multiplicity. To show that \(\lambda \) is completely multiplicative, we need to verify that \(\lambda (1) = 1\) and that \(\lambda (ab) = \lambda (a)\lambda (b)\) for all natural numbers \(a\) and \(b\).

The Liouville function \(\lambda (n)\) is multiplicative, and its value at prime powers is given by \(\lambda (p^k) = (-1)^k\). The Dirichlet series of \(\lambda \) can be expressed as an Euler product over primes:

The Liouville function \(\lambda (n)\) is defined as \((-1)^{\Omega (n)}\), where \(\Omega (n)\) counts the total number of prime factors of \(n\) with multiplicity. The Möbius function \(\mu (n)\) is defined as \(0\) if \(n\) has a squared prime factor, and otherwise it is \((-1)^{\omega (n)}\), where \(\omega (n)\) counts the number of distinct prime factors of \(n\). For square-free numbers, we have \(\Omega (n) = \omega (n)\), since there are no repeated prime factors. Therefore, for square-free numbers, we have \(\lambda (n) = (-1)^{\omega (n)} = \mu (n)\), which shows that the Liouville function agrees with the Möbius function on square-free numbers.

The Euler totient function \(\varphi (n)\) counts the positive integers up to \(n\) that are relatively prime to \(n\). It is a multiplicative function, and its value at prime powers is given by \(\varphi (p^k) = p^k - p^{k-1}\). The Dirichlet series of \(\varphi \) can be expressed as an Euler product over primes: