Analytic Number Theory

Transcription

1 Otto Forster: Analytic Number Theory Lecture notes of a course given in the Winter Semester / at the Department of Mathematics, LMU Munich, Germany

2 Contents. Notations and Conventions. Divisibility. Unique Factorization Theorem 3. Congruences. Chinese Remainder Theorem 7 3. Arithmetical Functions. Möbius Inversion Theorem 4. Riemann Zeta Function. Euler Product 5. The Euler-Maclaurin Summation Formula 7 6. Dirichlet Series Group Characters. Dirichlet L-series Primes in Arithmetic Progressions 5 9. The Gamma Function 57. Functional Equation of the Zeta Function 64. The Chebyshev Functions Theta and Psi 7. Laplace and Mellin Transform Proof of the Prime Number Theorem 84

3 . Notations and Conventions Standard notations for sets Z ring of all integers N set of all integers N set of all integers P set of all primes = {, 3, 5, 7,,...} Q, R, C denote the fields of rational, real and complex numbers respectively A multiplicative group of invertible elements of a ring A [a, b], ]a, b[, [a, b[, ]a, b] denote closed, open and half-open intervals of R R + = [, [ set of non-negative real numbers R + = R + R multiplicative group of positive real numbers x greatest integer x R Landau symbols O, o For two functions f, g : [a, [ C, one writes f(x = O(g(x for x, if there exist constants C > and x a such that f(x C g(x for all x x. Similarily, f(x = o(g(x for x means that for every ε > there exists R a such that f(x ε g(x for all x R. For functions f, g : ]a, b[ C the notions f(x = O(g(x for x a, and f(x = o(g(x for x a, are defined analogously. f(x = f (x + O(g(x

4 . Notations and Conventions is defined as f(x f (x = O(g(x. Asymptotic equality Two functions f, g : [a, [ C are said to be asymptotically equal for x, in symbols f(x g(x for x, if g(x for x x and f(x lim x g(x =. Analogously, for two sequences (a n n n and (b n n n, a n b n a n means lim =. A famous example for asymptotic equality is the Stirling formula n b n n! ( n n, πn e which we will prove in theorem 9.8. Miscellaneous We sometimes write iff as an abbreviation for if and only if.

5 . Divisibility. Unique Factorization Theorem.. Definition. Let x, y Z be two integers. We define x y (read: x divides y, iff there exists an integer q such that y = qx. We write x y, if this is not the case... We list some simple properties of divisibility for numbers x, y, z Z. i (x y y z = x z. ii x for all x Z. iii x = x =. iv x and x for all x Z. v (x y y x = x = ±y..3. Definition. A prime number is an integer p such that there doesn t exist any integer x with < x < p and x p. So the only positive divisors of a prime number p are and p. Note that by definition is not a prime number. Every integer x is either a prime or a product of a finite number of primes. This can be easily proved by induction on x. The assertion is certainly true for x =. Let now x >, and assume that the assertion has already been proved for all integers x < x. If x is a prime, we are done. Otherwise there exists a decomposition x = yz with integers y, z < x. By induction hypothesis, y and z can be written as products of primes y = n m p i, z = q j, i= j= (m, n, p i, q j prime Multiplying these two formulas gives the desired prime factorization of x. Using the convention that an empty product (with zero factors equals, we can state that any positive integer x is a product of primes x = n p i, i= n, p i primes. We can now state and prove Euclid s famous theorem on the infinitude of primes..4. Theorem (Euclid. There exist infinitely many prime numbers. Proof. Assume to the contrary that there are only finitely many primes and that p :=, p := 3, p 3,..., p n Chap. last revised:

6 . Unique factorization theorem is a complete list of all primes. The integer x := p p... p n + must be a product of primes, hence must be divisible by at least one of the p i, i =,..., n. But this is impossible since x p i = (integer + p i is not an integer. Hence the assumption is false and there exist infinitely many primes. Whereas the existence of a prime factorization was easy to prove, the uniqueness is much harder. For this purpose we need some preparations..5. Definition. Two integers x, y Z are called relatively prime or coprime (G. teilerfremd if they are not both equal to and there does not exist an integer d > with d x and d y. This is equivalent to saying that x and y have no common prime factor. In particular, if p is a prime and x an integer with p x, then p and x are relatively prime..6. Theorem. Two integers x, y are coprime iff there exist integers n, m such that nx + my =. Proof. If nx + my =, every common divisor d of x and y is also a divisor of, hence d = ±. So x and y are coprime. Suppose that x, y are coprime. Without loss of generality we may assume x, y. We prove the theorem by induction on max(x, y. The assertion is trivially true for max(x, y =. Let now N := max(x, y > und suppose that the assertion has already been proved for all integers x, y with max(x, y < N. Since x, y are coprime, we have x y, so we may suppose < x < y. Then (x, y x is a pair of coprime numbers with max(x, y x < N. By induction hypothesis there exist integers n, m with nx + m(y x =, which implies (n mx + my =, q.e.d..7. Theorem. Let x, y Z. If a prime p divides the product xy, then p x or p y. Proof. If p x, we are done. Otherwise p and x are coprime, hence there exist integers n, m with np + mx =. Multiplying this equation by y and using xy = kp with an integer k, we obtain y = npy + mxy = npy + mkp = p(ny + mk. 4

7 This shows p y, q.e.d..8. Theorem (Unique factorization theorem. Every positive integer can be written as a (finite product of prime numbers. This decomposition is unique up to order. Proof. The existence of a prime factorization has already been proved, so it remains to show uniqueness. Let x = p... p n = q... q m ( be two prime factorizations of a positive integer x. We must show that m = n and after rearrangement p i = q i for all i. We may assume n m. We prove the assertion by induction on n. a If n =, it follows x = and m =, hence the assertion is true in this case. b Induction step n n, (n. We have p q... q m, hence by theorem.7, p must divide one of the factors q i and since q i is prime, we must have p = q i. After reordering we may assume i =. Dividing equation ( by p we get p... p n = q... q m. By induction hypothesis we have n = m and, after reordering, p i = q i for all i, q.e.d. If we collect multiple occurrences of the same prime, we can write every positive integer in a unique way as x = n i= p e i i, p < p <... < p n primes, n, e i >. This is called the canonical prime factorization of x. Sometimes a variant of this representation is useful. For an integer x and a prime p we define ord p (x := sup{e N : p e x}. Then every nonzero integer x can be written as x = sign(x p p ordp(x where the product is extended over all primes. Note that ord p (x = for all but a finite number of primes, so there is no problem with the convergence of the infinite product..9. Definition (Greatest common divisor. Let x, y Z. An integer d is called greatest common divisor of x and y, if the following two conditions are satisfied: i d ist a common divisor of x and y, i.e. d x and d y. ii If d is any common divisor of x and y, then d d. 5

8 . Unique factorization theorem If d and d are two greatest common divisors of x and y, then d d and d d, hence by..v we have d = ±d. Therefore the greatest common divisor is (in case of existence uniquely determined up to sign. The positive one is denoted by gcd(x, y. The existence can be seen using the prime factor decomposition. For x and y, gcd(x, y = p p min(ordp(x,ordp(y and gcd(x, = gcd(, x = x, gcd(, =. Two integers x, y are relatively prime iff gcd(x, y =. The following is a generalization of theorem.6... Theorem. Let x, y Z. An integer d is greatest common divisor of x and y iff i d is a common divisor of x and y, and ii there exist integers n, m such that nx + my = d. Proof. The case when at least one of x, y equals is trivial, so we may suppose x, y. If d is greatest common divisor of x and y, then x/d and y/d are coprime, hence by theorem.6 there exist integers n, m with n x d + my d =, which implies ii. The implication is trivial... Definition (Least common multiple. Let x, y Z. An integer m is called least common multiple of x and y, if the following two conditions are satisfied: i m ist a common multiple of x and y, i.e. x m and y m. ii If m is any common multiple of x and y, then m m. As in the case of the greatest common divisor, the least common multiple of x and y is uniquely determined up to sign. The positive one is denoted by lcm(x, y. For x and y the following equation holds lcm(x, y = p p max(ord(x,ord(y and lcm(x, = lcm(, x = lcm(, =. The definitions of the greatest common divisor and least common multiple can be extended in a straightforward way to more than two arguments. One has gcd(x,..., x n = gcd(gcd(x,..., x n, x n, lcm(x,..., x n = lcm(lcm(x,..., x n, x n. 6

9 . Congruences. Chinese Remainder Theorem.. Definition. Let m Z. Two integers x, y are called congruent modulo m, in symbols x y mod m, if m divides the difference x y, i.e. x y mz. Examples. mod 5, 3 mod 7, 4 mod 7. x mod is equivalent to x is even, x mod is equivalent to x is odd. Remarks. a x, y are congruent modulo m iff they are congruent modulo m. b x y mod iff x = y. c x y mod for all x, y Z. Therefore the only interesting case is m... Proposition. The congruence modulo m is an equivalence relation, i.e. the following properties hold: i (Reflexivity x x mod m for all x Z ii (Symmetry x y mod m = y x mod m. iii (Transitivity (x y mod m (y z mod m = x z mod m..3. Lemma (Division with rest. Let x, m Z, m. Then there exist uniquely determined integers q, r satisfying x = qm + r, r < m. Remark. The equation x = qm + r implies that x r mod m. Therefore every integer x Z is equivalent modulo m to one and only one element of {,,..., m }..4. Definition. Let m be a positive integer. The set of all equivalence classes of Z modulo m is denoted by Z/mZ or briefly by Z/m. From the above remark we see that Z/mZ = {,,..., m }, where x = x mod m is the equivalence class of x modulo m. If there is no danger of confusion, we will often write simply x instead of x. Chap. last revised: -- 7

10 . Congruences Equivalence modulo m is compatible with addition and multiplication, i.e. x x mod m and y y mod m = x + y x + y mod m and xy x y mod m. Therefore addition and multiplication in Z induces an addition and multiplication in Z/m such that Z/m becomes a commutative ring and the canonical surjection Z Z/m, x x mod m, is a ring homomorphism. Example. In Z/7 one has = 7 =, = 8 =, 3 5 = 5 =. The following are the complete addition and multiplication tables of Z/ Theorem. Let m be a positive integer. An element x Z/m is invertible iff gcd(x, m =. Proof. Suppose gcd(x, m =. By theorem.6 there exist integers ξ, µ such that ξx + µm =. This implies ξx mod m, hence ξ is an inverse of x in Z/m. Suppose that x is invertible, i.e. x y = for some y Z/m. Then xy mod m, hence there exists an integer k such that xy = km. Therefore yx km =, which means by theorem.6 that x and m are coprime, q.e.d..6. Corollary. Let m be a positive integer. The ring Z/m is a field iff m is a prime. Notation. If p is a prime, the field Z/p is also denoted by F p. For any ring A with unit element we denote its multiplicative group of invertible elements by A. In particular we use the notations (Z/m and F p. Example. For p = 7 we have the field F 7 = Z/7 with 7 elements. From the above multiplication table we can read off the inverses of the elements of F 7 = F 7 {}. 8

11 x x Direct Products. For two rings (resp. groups A and A, the cartesian product A A becomes a ring (resp. a group with component-wise defined operations: (x, x + (y, y := (x + y, x + y (x, x (y, y := (x y, x y. If A, A are two rings with unit element, then (, is the zero element and (, the unit element of A A. For the group of invertible elements the following equation holds: (A A = A A. Note that if A and A are fields, the direct product A A is a ring, but not a field, since there are zero divisors: (, (, = (,..8. Theorem (Chinese remainder theorem. Let m, m be two positive coprime integers. Then the map φ : Z/m m Z/m Z/m, x (x mod m, x mod m is an isomorphism of rings. Proof. It is clear that φ is a ring homomorphism. Since Z/m m and Z/m Z/m have the same number of elements (namely m m, it suffices to prove that φ is injective. Suppose φ(x =. This means that x mod m and x mod m, i.e. m x and m x. Since m and m are coprime, it follows that m m x, hence x = in Z/m m, q.e.d. Remark. The classical formulation of the Chinese remainder theorem is the following (which is contained in theorem.8: Let m, m be two positive coprime integers. Then for every pair a, a of integers there exists an integer a such that a a i mod m i for i =,. This integer a is uniquely determined modulo m m..9. Definition (Euler phi function. Let m be a positive integer. Then ϕ(m is defined as the number of integers k {,,..., m } which are coprime to m. Using theorem.5, this can also be expressed as ϕ(m := #(Z/m, where #S denotes the number of elements of a set S. For small m, the ϕ-function takes the following values 9

12 . Congruences m ϕ(m It is obvious that for a prime p one has ϕ(p = p. More generally, for a prime power p k it is easy to see that ( ϕ(p k = p k p k = p k. p If m and n are coprime, it follows from theorem.8 that (Z/mn = (Z/m (Z/n, hence ϕ(mn = ϕ(nϕ(m. Using this, we can derive.. Theorem. For every positive integer n the following formula holds: ϕ(n = n (, p p n where the product is extended over all prime divisors p of n. Proof. Let n = r i= pe i i be the canonical prime decomposition of n. Then r r ϕ(n = ϕ(p e i i = p e i i ( r = n (, q.e.d. p i p i i= i=.. Theorem (Euler. Let m be an integer and a an integer with gcd(a, m =. Then a ϕ(m mod m. Proof. We use some notions and elementary facts from group theory. Let G be a finite group, written multiplicatively, with unit element e. The order of an element a G is defined as ord(a := min{k N : a k = e}. The order of the group is defined as the number of its elements, ord(g := #G. Then, as a special case of a theorem of Lagrange, one has ord(a ord(g for all a G. We apply this to the group G = (Z/m. By definition ord((z/m = ϕ(m. Let r be the order of a (Z/m. Then ϕ(m = rs with an integer s and we have in (Z/m a ϕ(m = a rs = (a r s = s =, q.e.d... Corollary (Little Theorem of Fermat. Let p be a prime and a an integer with p a. Then a p mod p. i=

13 3. Arithmetical Functions. Möbius Inversion Theorem 3.. Definition. a An arithmetical function is a map f : N C. b The function f is called multiplicative if it is not identically zero and f(nm = f(nf(m for all n, m N with gcd(n, m =. c The function f is called completely multiplicative or strictly multiplicative if it is not identically zero and f(nm = f(nf(m for all n, m N (without restriction. Remark. A multiplicative arithmetical function a : N C satisfies a( =. This can be seen as follows: Since gcd(, n =, we have a(n = a(a(n for all n. Therefore a(, (otherwise a would be identically zero, and a( = a(a( implies a( =. 3.. Examples i The Euler phi function ϕ : N N C, which was defined in (.9, is a multiplicative arithmetical function. It is not completely multiplicative, since for a prime p we have ϕ(p = p p = (p p ϕ(p = (p. ii Let α C be an arbitrary complex number. We define a function p α : N C, n p α (n := n α = e α log(n. Then p α is a completely multiplicative arithmetical function. iii Let f : N Z C be defined by f(p := for primes p and f(n = if n is not prime. This is an example of an arithmetical function which is not multiplicative. Remark. A multiplicative arithmetical function f : N C is completely determined by its values at the prime powers: If n = r i= pe i i is the canonical prime decomposition of n, then f(n = r i= f(p e i i Divisor function τ : N N. This function is defined by τ(n := number of positive divisors of n. Thus τ(p = and τ(p k = + k for primes p. (The divisors of p k are, p, p,..., p k. Chap. 3 last revised: --

14 3. Arithmetical functions The divisor function is multiplicative. This can be seen as follows: Let m, m N be a pair of coprime numbers and m := m m. Looking at the prime decompositions one sees that the product d := d d of divisors d m and d m is a divisor of m and conversely every divisor d m can be uniquely decomposed in this way. This can be also expressed by saying that the map Div(m Div(m Div(m m, (d, d d d is bijective, where Div(n denotes the set of positive divisors of n. This implies immediately the multiplicativity of τ Divisor sum function σ : N N. This function is defined by σ(n := sum of all positive divisors of n. Thus for a prime p we have σ(p = + p and σ(p k = + p + p p k = pk+ p. The divisor sum function is also multiplicative. Proof. Let m, m N be coprime numbers. Then σ(m m = ( d = d d = d m m d m,d m = σ(m σ(m. d m d ( d m d 3.5. Definition. A perfect number (G. vollkommene Zahl is a number n N such that σ(n = n. The condition σ(n = n can also be expressed as d n,d<n d = n, i.e. a number n is perfect if the sum of its proper divisors equals n. The smallest perfect numbers are 6 = + + 3, 8 = The next perfect numbers are 496, 88. The even perfect numbers are characterized by the following theorem. Theorem. a (Euclid If q is a prime such that q is prime, then n := q ( q is a perfect number.

15 b (Euler Conversely, every even perfect number n may be obtained by the construction in a. The prove is left as an exercise. The above examples correspond to q =, 3, 5, 7. For q =, = 47 = 3 89 is not prime. It is not known whether there exist odd perfect numbers Möbius function µ : N Z. This rather strange looking, but important function is defined by, for n =, µ(n :=, if there exists a prime p with p n, ( r, if n is a product of r different primes. This leads to the following table n µ(n It follows directly from the definition that µ is multiplicative Definition. Let f : N C be an arithmetical function. The summatory function of f is the function F : N C defined by F (n := d n f(d, where the sum is extended over all positive divisors d of n Examples. i The divisor sum function σ(n = d n d is the summatory function of the identity map ι : N N, ι(n := n. ii The divisor function τ : N N can be written as τ(n = d n. Therefore τ is the summatory function of the constant function u : N N, u(n := for all n. 3

16 3. Arithmetical functions 3.9. Theorem (Summatory function of the Euler phi function. For all n N ϕ(d = n. d n This means that the summatory function of the Euler phi function is the identity map ι : N N. Proof. The set M n := {,,..., n} is the disjoint union of the sets A d := {m M n : gcd(m, n = d}, d n. Therefore n = d n #A d. We have gcd(m, n = d iff d m, d n and gcd(m/d, n/d =. It follows that #A d = ϕ(n/d, hence n = d n #A d = d n ϕ(n/d = d n ϕ(d, q.e.d. 3.. Theorem (Summatory function of the Möbius function. µ(d = d n { for n =, for all n >. Therefore the summatory function of the Möbius function is the function δ : N Z, δ (n := { for n =, for all n >. Proof. The case n = is trival. Now suppose n and let n = r j= pe j j be the canonical prime factorization of n. For s r we denote by D s the set of all divisors d n which are the product of s different primes {p,..., p r }, (D = {}. For all d D s we have µ(d = ( s ; but µ(d = for all divisors of n that do not belong to any of the D s. Therefore µ(d = d n r s= d D s µ(d = = ( + ( r =, r ( s #D s = s= r ( r ( s s where we have used the binomial theorem. This proves our theorem. 3.. Definition (Dirichlet product. For two arithmetical functions f, g : N C one defines their Dirichlet product (or Dirichlet convolution f g : N C by s= (f g(n := d n f(dg(n/d. 4

17 This can be written in a symmetric way as (f g(n = kl=n f(kg(l, where the sum extends over all pairs k, l N with kl = n. This shows that f g = g f and (f g(n = d n f(n/dg(d. Example. (f g(6 = f(g(6 + f(g(3 + f(3g( + f(6g(. Remark. Let f be an arbitrary arithmetical function and u the constant function u(n = for all n N. Then (u f(n = d n u(n/df(d = d n f(d. Thus the summatory function of an arithmetical function f is nothing else than the Dirichlet product u f. 3.. Theorem. If the arithmetical functions f, g : N C are multiplicative, their Dirichlet product f g is again multiplicative. Example. Since the constant function u(n = is clearly multiplicative, the summatory function of every multiplicative arithmetical function is multiplicative. Proof. Let m, m N be two coprime numbers. Then (f g(m m = ( m m ( m m f(dg = f(d d g d d d d m m d m,d m = ( m ( m f(d f(d g g d d d m d m = ( m ( m f(d g f(d g d d d m d m = (f g(m (f g(m, q.e.d Theorem. The set F(N, C of all arithmetical functions f : N C is a commutative ring with unit element when addition is defined by (f + g(n := f(n + g(n for all n N and multiplication is the Dirichlet product. The unit element is the function δ : N C defined by δ ( :=, δ (n = for all n >. 5

18 3. Arithmetical functions Remark. The notation δ is motivated by the Kronecker δ-symbol { for i = j, δ ij = otherwise. Using this, one can write δ (n = δ n. Proof. That δ is the unit element is seen as follows (δ f(n = ( n ( n δ (df = δ (f = f(n. d d n All ring axioms with exception of the associative law for multiplication are easily verified. Proof of associativity: ((f g h(n = k,l kl=n = i,j,l ijl=n = i,m im=n (f g(kh(l = k,l kl=n f(ig(jh(l = i,m im=n f(ig(jh(l i,j ij=k f(ig(jh(l j,l jl=m f(i(g h(m = (f (g h(n, q.e.d Theorem (Möbius inversion formula. Let f : N C be an arithmetical function and F : N C its summatory function, F (n = d n f(d for all n N. ( Then f can be reconstructed from F by the formula f(n = ( n µ F (d for all n N. ( d d n Conversely, ( implies (. Proof. The formula ( can be written as F = u f, where u is the constant function u(n = for all n. Theorem 3. says that u is the Dirichlet inverse of the Möbius function: Therefore u µ = µ u = δ. µ F = µ (u f = (µ u f = δ f = f, 6

19 which is formula (. Conversely, from f = µ F one obtains u f = u (µ F = (u µ F = δ F = F, that is formula (, q.e.d Examples. i Applying the Möbius inversion formula to the summatory function of the Euler phi function (theorem 3.9 n = ι(n = d n ϕ(d yields ϕ = µ ι, i.e. ϕ(n = d n This can also be written as ( n µ(d ι = d d n n d µ(d. ϕ(n n = d n µ(d d. ii Example 3.8.i says u ι = σ which implies ι = µ σ, i.e. ( n µ σ(d = n. d d n iii Example 3.8.ii says u u = τ, hence u = µ τ, i.e. ( n µ τ(d = for all n. d d n We now state a second Möbius inversion formula for functions defined on the real interval I := {x R : x } Theorem. For a function f : I C define F : I C by F (x = ( x f for all x, k k x ( where the sum extends over all positive integers k x. Then f(x = ( x µ(kf for all x. k k x ( Conversely, ( implies (. 7

20 3. Arithmetical functions Example. If f is the constant function f(x = for all x, then F (x = x = greatest integer x. The theorem implies the remarkable formula x µ(k = for all x. k k x E.g. for x = 5 this reads 5µ( + µ( + µ(3 + µ(4 + µ(5 =. To prove theorem 3.6, we put it first into an abstract context Let F(I, C denote the vector space of all functions f : I = [, [ C. We define an operation of the ring of all arithmetical functions on this vector space F(N, C F(I, C F(I, C, (α, f α f, where (α f(x := ( x α(kf. k k x 3.8. Theorem. With the above operation, F(I, C becomes a module over the ring F(N, C. Proof. It is clear that F(I, C is an abelian group with respect to pointwise addition (f +g(x = f(x+g(x. So it remains to verify the following laws (for α, β F(N, C and f, g F(I, C. i α (f + g = α f + α g, ii (α + β f = α f + β f, iii α (β f = (α β f, iv δ f = f. The assertions i and ii are trivial. The associative law iii can be seen as follows (α (β f(x = ( x α(k(β f = α(k k k x k x l x/k = ( x α(kβ(lf kl kl x = ( x α(kβ(lf n n x kl=n = ( x β(nf = ((α β f(x. n x(α n ( x β(lf kl 8

21 Proof of iv: (δ f(x = k x ( x ( x δ (kf = δ (f = f(x, q.e.d. k 3.9. Now we take up the proof of theorem 3.6. Equation ( can be written as F = u f with the constant function u(n =. Multiplying this equation by the Möbius function yields µ F = µ (u F = (µ u f = δ f = f, which is equation (. Conversly, from f = µ F it follows u f = u (µ F = (u µ F = δ F = F, which is equation (, q.e.d. 9

22 4. Zeta function 4. Riemann Zeta Function. Euler Product 4.. Definition. For a complex s C with Re(s >, the Riemann zeta function is defined by the series ζ(s := n= n s. Let us first study the convergence of this infinite series. Following an old tradition, we denote the real and imaginary part of s by σ resp. t, i.e. We have s = σ + it, σ, t R. n s = n s = e s log n = e σ log(n it log n = n σ e it log n, therefore = n s n. σ Since converges for all real σ >, we see that the zeta series converges absolutely n= nσ and uniformly in every halfplane H(σ, σ >, where H(σ := {s C : Re(s > σ }. It follows by a theorem of Weierstrass that ζ is a holomorphic (= regular analytic function in the halfplane H( = {s C : Re(s > }. We will see later that ζ can be continued analytically to a meromorphic function in the whole complex plane C, which is holomorphic in C {} and has a pole of first order at s =. A weaker statement is 4.. Proposition. lim σ ζ(σ =. Proof. Let R > be any given bound. Since n= n that N n= n R +. =, there exists an N > such Chap. 4 last revised: 3-4-4

23 The function σ N n= N n= n σ is continuous on R, hence there exists an ε > such that R for all σ with σ < + ε. nσ A fortiori we have n= n σ R for all < σ < + ε. This proves the proposition Theorem (Euler product. For all s C with Re(s > one has ζ(s = p P p s, where the product is extended over the set P of all primes. Proof. Since p s < /p /, we can use the geometric series p = s p, ks k= which converges absolutely. If P P is any finite set of primes, the product ( = ( + p ks p + s p + s p s p P p P k= can be calculated by termwise multiplication and we obtain ( p P k= = p ks n N(P n s, where N(P is the set of all positive integers n whose prime decomposition contains only primes from the set P. (Here the unique prime factorization is used. Letting P = P m be set of all primes m and passing to the limit m, we obtain the assertion of the theorem. Remark. The Euler product can be used to give another proof of the infinitude of primes. If the set P of all primes were finite, the Euler product p P ( p s would be continuous at s =, which contradicts the fact that lim σ ζ(σ = We recall some facts from the theory of analytic functions of a complex variable about infinite products. Let G C be an open set. For a continuous function f : G C and a compact subset K G we define the maximum norm f K := sup{ f(z : z K} R +.

24 4. Zeta function (The supremum is < since f is continous. Let now f ν sequence of holomorphic functions. The infinite product F (z := ( + f ν (z ν= : G C, ν, be a is said to be normally convergent on a compact subset K G, if f ν K <. ν= In this case, the product converges absolutely and uniformly on K. (The converse is not true, as can be seen by taking the constant functions f ν = for all ν. The product is said to be normally convergent in G if it converges normally on any compact subset of K G. The limit F of a normally convergent infinite product of holomorphic functions + f ν is again holomorphic and F (z = for a particular point z G if and only if one of the factors vanishes in z Theorem. The Riemann zeta function has no zeroes in the half plane H( = {s C : Re(s > }. For its inverse one has ζ(s = p P ( = p s where µ is the Möbius function. n= µ(n n s, Proof. The first assertion follows from the fact that the Euler product for the zeta function converges normally in H( and all factors ( p s have no zeroes in H(. Inverting the product representation for /ζ(s yields /ζ(s = ( p s. To prove the last equation, let P a finite set of primes and N (P the set of all positive integers n that can be written as a product n = p p... p r of distinct primes p j P, (r. Then, since ( r = µ(p... p r, ( = p s p P n N (P µ(n n s. Letting P = P m be set of all primes m and passing to the limit m, we obtain the assertion of the theorem. Note that µ(n = for all n N m N (P m We recall now some facts about the logarithm function. (By logarithm we always mean the natural logarithm with basis e = We have the Taylor expansion log( + z = ( n= n zn n for all z C with z <.

25 From this follows ( log = z n= z n n for all z C with z <. (Of course here the principal branch of the logarithm with log( = is understood. If f : G C is a holomorphic function without zeroes in a simply connected domain G C, then there exists a holomorphic branch of the logarithm of f, i.e. a holomorphic function log f : G C with e (log f(z = f(z for all z G. This function log f is uniquely determined up to an additive constant πin, n Z. Since the zeta function has no zeroes in the simply connected halfplane H(, we can form the logarithm of the zeta function, where we select the branch of log ζ that takes real values on the real half line ], [ Theorem. For the logarithm of the zeta function in the halfplane H(, the following equation holds: The function log ζ(s = p P p + s k= k p P p ks. F (s := k= k p P p ks is bounded in H(. Remark. If one defines the prime zeta function by P (s := p P p s for s H(, the formula of the theorem may be written as log ζ(s = k= P (ks k Proof. Using the Euler product we obtain log ζ(s = ( log = p s p P p P = p + s k p. ks p P p P = P (s + F (s, where F (s = k= 3 k= kp = ks k= p P k= kp ks P (ks. k

26 4. Zeta function To prove the boundedness of F (s = k= k p P p = ks k= P (ks k in H(, we use the estimate (with σ = Re(s > P (ks P (kσ P (k = p P n= n n and obtain for all s H( p k n k n= dx x = dx k x = k k F (s k= k(k =, q.e.d Corollary (Euler. p P p = =. Proof. Since the difference P (s log ζ(s is bounded for Re(s > we get, using proposition 4., ( lim P (σ = lim =. σ σ p σ This implies the assertion. p P Remark. The corollary gives another proof that there are infinitely many primes, but says more. Comparing with n= n <, we can conclude that the density of primes is in some sense greater than the density of square numbers. The following theorem is a variant of theorem 4.7 and gives an interesting formula for the difference between P (s and log ζ(s Theorem. We have the following representation of the prime zeta function for Re(s > P (s = p P p = log ζ(s + s k= µ(k k 4 log ζ(ks.

27 Proof. We start from the formula of theorem 4.7 log ζ(s = k= P (ks. k We have as in the proof of theorem 4.7 the estimate P (ks P (kσ kσ kσ, (where σ = Re(s, which implies log ζ(s k= k σ = σ k= k = ζ( σ =: c σ with the constant c = ζ(. Therefore the series k= (µ(k/k log ζ(ks converges absolutely: µ(k k log ζ(ks k= k= k c kσ = cζ( σ Substituting log ζ(ks = l= P (kls/l we get k= µ(k k log ζ(ks = = k,l= µ(kp (kls kl = µ(k P (ns n n= k n = P (s, q.e.d. <. n= kl=n = n= µ(k P (kls kl δ (n P (ns n We conclude this chapter with an interesting application of therem Theorem. The probability that two random numbers m, n N are coprime is 6/π 6%, more precisely: For real x let Then Copr(x := {(m, n N N : m, n x and m, n coprime}. #Copr(x lim = x x ζ( = 6 π. Proof. Let A(x be the set of all pairs m, n of integers with m, n x and A k (x := {(n, m A(x : gcd(m, n = k}. 5

28 4. Zeta function Then A(x is the disjoint union of all A k (x, k =,,..., x, and for every k we have a bijection ( x Copr A k (x, (m, n (km, kn. k Therefore ( x #Copr = x. k k x Now we can apply the inversion formula of theorem 3.6 and obtain #Copr(x = x. µ(k k k x Since (x/k x/k <, it follows that (x/k x/k < x/k, hence #Copr(x x µ(k( x x( + log x = O(x log x, k k k x k x so we can write #Copr(x x = k x µ(k k ( log x + O x. On the other hand k= µ(k/k = /ζ( by theorem 4.5, hence µ(k k ζ( ( k = O. x k x k>x Combining this with the previous estimate yields #Copr(x = ( log x x ζ( + O, x which implies the assertion of the theorem. Remark. The fact ζ( = π 6 will be proven in the next chapter. 6

29 5. The Euler-Maclaurin Summation Formula 5.. We define a periodic function saw : R R with period by saw(x := x x This is a kind of sawtooth function, see figure. 3 With this function, we can state a first form of the Euler-Maclaurin summation formula. This formula shows how a sum can be approximated by an integral and gives an exact error term. 5.. Theorem (Euler-Maclaurin I. Let x be a real number and f : [x, [ C a continuously differentiable function. Then we have for all integers n m x n n f(k = (f(m + f(n + f(xdx + k=m Proof. We have n k=m m n f(k (f(m + f(n = n (f(k + f(k +. k=m On the other hand we get by partial integration k+ k saw(xf (xdx = k+ k m (x k f (xdx = (x k f(x k+ k saw(xf (xdx. k+ k f(xdx k+ = (f(k + + f(k f(xdx. Summing up from k = m to n yields the assertion of the theorem. k Chap. 5 last revised:

30 5. Euler-Maclaurin summation Using this theorem, we can construct an analytic continuation of the zeta function Theorem. The Riemann zeta function can be analytically continued to a meromorphic function in the halfplane H( = {s C : Re(s > } with a single pole of order at s =. The continued function can be represented in H( as ζ(s = + s s saw(x x dx. s+ Proof. Applying theorem 5. to the function f(x = /x s we get N n= ( n = + N + s N s dx N x s s For Re(s > we have lim N /N s = and N lim N dx x s = lim N saw(x x dx. s+ ( s N = s s. Therefore we can pass to the limit N in the formula above and get for Re(s > ζ(s = + s s We will now show that the integral F (s := saw(x x s+ dx saw(x dx. ( xs+ exists for all s C with σ := Re(s > and represents a holomorphic function in the halfplane H(. This will then complete the proof of the theorem, since the right hand side of the formula ( defines a meromorphic continuation of the zeta function to H( with a single pole at s =. The existence of the integral follows from the estimate saw(x x s+ x, σ+ since (/xσ+ dx < for σ >. To prove the holomorphy of F it suffices by the theorem of Morera to show that for all compact rectangles R H( F (sds =. R This can be seen as follows: Since R H( is compact, there exist a σ > such that Re(s σ for all s R. Therefore we have on R [, [ the majorization saw(x x s+ x σ + 8

31 and we can apply the theorem of Fubini saw(x F (s ds = dx ds R = R x s+ ( saw(x ds dx =, R xs+ }{{} = q.e.d. There exists also a proof of the holomorphy of F without recourse to Lebesgue integration theory: We write with F (s = f n (s = n+ n saw(x x s+ dx = n= n+ n saw(x x s+ dx = saw(x n+ x n dx = dx. x s+ n x s+ f n (s The function f n is holomorphic in C (it is easily checked directly that g(z = b a tz dt is holomorphic in the whole z-plane and satisfies an estimate f n (s n σ + for all s H(σ Since n= /nσ + < for all σ >, the series F = n= f n converges uniformly on every compact subset of H(. By a theorem of Weierstraß, the limit function F is holomorphic in H( Definition. The Euler-Mascheroni constant is defined as the limit ( N C := lim N n log N. n= The existence of this limit can be proved using the Euler-Maclaurin summation formula (5.. This is left to the reader as an exercise Theorem. There exist uniquely determined functions β k : R R, k N, with the following properties: i All functions β k are periodic with period, i.e. β k (x + n = β k (x for all n Z, and the functions β k with k are continuous. ii β = saw. 9 n=

32 5. Euler-Maclaurin summation iii β k is differentiable in ], [ and β k(x = β k (x for all < x < and k. iv β k (xdx = for all k Proof. By condition iii, the function β k is uniquely determined in the intervall ], [ by β k up to an additive constant. This constant is uniquely determined by condition iv. Thus by ii-iv, all β k are uniquely determined in ], [, and by periodicity even in R Z. It remains to be shown that the definition of β k, k can be extended continuously across the integer points. This is equivalent with lim β k(ε = lim β k ( ε. ε ε For k one has hence by iv β k ( ε β k (ε = ε ε β k (xdx, lim (β k( ε β k (ε = ε β k (xdx =, q.e.d. Example. Let us calculate β. The condition β (x = β (x = x for < x < leads to β (x = x x + c with an integration constant c. Since we have c =, i.e. ( x xdx = 6 4 =, β (x = x x + = x(x + for x. / 3 Graph of β 3

33 5.6. Theorem. The functions β n have the following Fourier expansions β k (x = ( k β k+ (x = ( k n= n= which converge uniformly on R. Formula ( is also valid for k = and x R Z. cos(πnx, k, ( (πn k sin(πnx, k, ( (πn k+ Proof. a We first calculate the Fourier series n Z c ne inx of β. The coefficients c n are given by the integral c n = β (xe πinx dx. By theorem 5.5.iv we have c =. Let now n. Using partial integration we get and hence c n = xe πinx dx = πin xe πinx + πin x e πinx dx = πin x e πinx + πin ( x x + e πinx dx = Thus we have the Fourier series β (x = n Z e πin (πn = n= (πn. e πinx + e πinx (πn = e πinx dx = i πn xe πinx dx = n= cos(πnx (πn. i πn + (πn, By the general theory of Fourier series, the convergence is with respect to the L -norm f L = ( f(x dx /, but since n= /n < and β is continuous, we have even uniform convergence. b Since the right hand sides of the formulae of the theorem satisfy the same recursion and normalization relations (5.5.iii-iv as the functions β k, it follows that the given Fourier expansions are valid for all β k, k. To prove the formula for β (x = saw(x = n= sin(πnx, x R Z, πn 3

34 5. Euler-Maclaurin summation it suffices to show that the series sin(πnx n= converges uniformly on every interval πn [δ, δ], < δ <, since then termwise differentiation of the Fourier series of β is allowed. To simplify the notation we will prove the equivalent statement sin nx converges uniformly on [δ, π δ], ( < δ < π. n Define n= S m (x := m ( m sin nx = Im e. inx n= n= For δ x π δ we have m S m (x e inx = eimx e ix n= It follows for m k > m sin nx n = m n=k n=k m n=k sin δ S n (x S n (x n ( S n (x n e ix/ e ix/ = sin x + S m(x n + m + S k (x k, k sin δ ( k m + + m + + k. sin δ hence also n=k sin nx n k sin δ for all x [δ, π δ], which proves the asserted uniform convergence and thereby completes the proof of the theorem Definition. It follows immediately from (5.5.iii-iv that β n, restricted to the open interval ], [, is a polynomial of degree n with rational coefficients. The n-th Bernoulli polynomial B n (X Q[X] is defined by B n (x n! = β n (x for < x <, n and B (X =. The Bernoulli numbers B k are defined by B n := B n (, n. Strictly speaking, it is not correct to use the same symbol B k for the Bernoulli polynomials and the Bernoulli numbers. However this notation is the usual one. To avoid confusion, we will always indicate the variable when we are dealing with Bernoulli polynomials. 3

35 We know already the first Bernoulli polynomials B (X = X and B (X = X(X +, 6 hence B =, B =, B =. 6 An easy consequence of theorem 5.6 is 5.8. Theorem. For the Bernoulli numbers the following relations hold: i B k+ = for all k. ii k (k! B k = ( (π k n, k n= hence ζ(k = (πk (k! B k for all k. iii sign(b k = ( k for all k. Remarks. a Formula ii of the theorem says in particular ζ( = n= n = π 6, which was already used in the previous chapter. b Since lim σ ζ(σ =, formula ii shows the asymptotic growth of the Bernoulli numbers B k B k (k! (π k for k Theorem (Generating function for the Bernoulli polynomials. For fixed x R, text the function is a complex analytic function of t with a removable singularity at e t t =. The Taylor expansion at t = of this function has the form te xt e t = n= B n (x n! t n. In particular, for x = one has t e t = n= B n n! tn. Proof. Define B n (x by the above Taylor expansions. We will show that 33

36 5. Euler-Maclaurin summation (a B (x =, B (x = x, (b B n(x = nb n (x, (n, (c B n(xdx =, (n. Then theorem 5.5 implies B n(x n! Proof of (a = β n (x for < x < and all n. te xt e t = t( + xt + O(t t + t + O(t 3 = + xt + O(t + t + O(t = ( + xt( t + O(t = + (x t + O(t, which shows B (x = and B (x = x. Proof of (b We calculate x x te xt e t = n= B n(x n! te xt e t t n in two ways and x te xt e t = t e xt e t = n= B n (x n! t n+ = n= B n (x (n! tn Comparing coefficients we get B n(x = nb n (x. Proof of (c te xt e t dx = On the other hand ext e t te xt e t dx = n= ( x= x= = et e t e t =. t n B n (xdx n!. Comparing coefficients, we get B n(xdx = for all n, q.e.d. 5.. Recursion formula. Theorem 5.9 can be used to derive a recursion formula for the Bernoulli numbers. Since (e t /t = t n /n!, we have ( k= n= ( B k k! tk (l +! tl =. l= 34

37 The Cauchy product n= c nt n of the two series has coefficients c n = n k= B k k!(n k +! = (n +! n ( n + B k. k k= Hence comparing coefficients we get B = and n ( n + B k = for all n. k k= With this formula one can recursively calculate all B n. The first non zero coefficients are k B k Theorem (Euler-Maclaurin II. Let x be a real number and f : [x, [ C a r-times continuously differentiable function. Then we have for all integers n m x and all r n n f(k = (f(m + f(n + f(xdx k=m + r k= m B k ( f (k (n f (k (m n (k! m B r (x (r! f (r (xdx Here B r (x is the periodic function defined by B r (x := B r (x x = (r!β r (x. Proof. We start with theorem 5. n n f(k = (f(m + f(n + f(xdx + k=m and evaluate the last integral by partial integration. m n m saw(xf (xdx. Since β (x = saw(x for k < x < k + and β is continuous and periodic, we get n m saw(xf (xdx = = = n k=m n k+ k β (xf (x k=m n saw(xf (xdx k+ k n k=m k+ k β (xf (xdx (β (k+f (k+ β (kf (k k=m = B! (f (n f (m 35 n m β (xf (xdx. n m β (xf (xdx

38 5. Euler-Maclaurin summation This proves the case r = of the theorem. The general case is proved by induction. Induction step r r +. n β r (xf (r (xdx = β r+ (xf (r (x n n + β r+ (xf (r+ (xdx m m n This proves the assertion for r +. = β r+ (xf (r+ (xdx [since β r+ (k = B r+ m (r +! = ] = β r+ (xf (r+ (x n n β r+ (xf (r+ (xdx m = B r+ (r +! (f (r+ (n f (r+ (m m m n m β r+ (xf (r+ (xdx. Remark. If f is infinitely often differentiable and we pass to the limit r, the error term n B r (x (r! f (r (xdx m will in general not converge to. In case f is real and f (r does not change sign in the interval [m, n], one has the following estimate n B r (x (r! f (r (xdx B n r f (r (xdx = B r (r! (r! f (r (n f (r (m, m which means that the error of the approximation n n f(k (f(m + f(n + f(xdx + k=m m m r k= B k ( f (k (n f (k (m (k! is by absolute value not larger than the last term of the sum. Hence by increasing r one gets better approximations as long as the absolute values of the added terms decrease. 5.. Theorem. The Riemann zeta function can be analytically continued to a meromorphic function in the whole plane C with a single pole of order at s =. For Re(s > r, the continued function can be represented as ζ(s = + s + r k= s(s +... (s + r B k s(s +... (s + k (k! B r (x (r! x s+r dx. Proof. This is proved by applying theorem 5. to the sum n k= /ks and passing to the limit n. That the last integral defines a holomorphic function for Re(s > r, follows from the fact that the function B r (x is bounded and for all s C with Re(s r + δ. x s+r x +δ 36

39 6. Dirichlet Series 6.. Definition. A Dirichlet series is a series of the form f(s = n= a n n s, (s C, where (a n n is an arbitrary sequence of complex numbers. The abscissa of absolute convergence of this series is defined as σ a := σ a (f := inf{σ R : n= a n n σ < } R {± }. If n= ( a n /n σ does not converge for any σ R, then σ a = +, if it converges for all σ R, then σ a =. An analogous argument as in the case of the zeta function shows that a Dirichlet series with abscissa of absolute convergence σ a converges absolutely and uniformly in every halfplane H(σ, σ > σ a. Example. The Dirichlet series g(s := ( n n= n s has σ a (g =. We will see however that the series converges for every s H(. Of course the convergence is only conditional and not absolute if < Re(s. We need some preparations. 6.. Lemma (Abel summation. Let (a n n and (b n n be two sequences of complex numbers and set A n := n a k, k= A = (empty sum. Then we have for all n m n a k b k = A n b n A m b m k=m n k=m A k (b k+ b k. Remark. This can be viewed as an analogon of the formula for partial integration b a F (xg(xdx = F (bg(b F (ag(a b a F (xg (xdx. Chap. 6 last revised: -- 37

40 6. Dirichlet series Proof. n a k b k = k=m n (A k A k b k = k=m = A n b n + n k=m n A k b k k=m n k=m n A k b k A k b k+ A m b m k=m n = A n b n A m b m A k (b k+ b k, k=m A k b k+ q.e.d Lemma. Let s C with σ := Re(s >. Then we have for all m, n n s s m s σ n. σ m σ Proof. We may assume n m. Since d ( = s dx x s x, s+ n dx s m x = s+ n s m. s Taking the absolute values, we get the estimate n s m s s n m dx s = xσ+ σ n σ m σ, q.e.d. Remark. For s C and an angle α with < α < π/, we define the angular region Ang(s, α := {s + re iφ : r and φ α}. For any s Ang(s, α {s } we have s s Re(s s = cos φ cos α, hence the estimate in lemma 6.3 can be rewritten as n s m s cos α n for all s Ang(, α. σ m σ 6.4. Theorem. Let f(s = n= a n n s 38

41 be a Dirichlet series such that for some s C the partial sums N a n n= n s for N. Then the Dirichlet series converges for every s C with are bounded Re(s > σ := Re(s. The convergence is uniform on every compact subset K H(σ = {s C : Re(s > σ }. Hence f is a holomorphic function in H(σ. Proof. Since f(s = n= n s a n n s s = ã n n s s n= where ã n := a n n s, we may suppose without loss of generality that s =. By hypothesis there exists a constant C > such that N a n C for all N N. n= The compact set K is contained in some angular region Ang(, α with < α < π/. We define C α := cos α and σ := inf{re(s : s K} >. Now we apply the Abel summation lemma 6. to the sum a n (/n s, s K. Setting A N := N n= a n, we get for N M N n=m a n n s = A N N N A s M M + s n=m This leads to the estimate (with σ = Re(s N n=m a n n s C M s + C C M σ + C C α A n ( n s (n + s. N n s (n + s n=m N n=m ( n σ (n + σ = C M σ + C C α ( M σ N σ C M σ ( + C α C ( + C α M σ. 39

42 6. Dirichlet series This becomes arbitrarily small if M is sufficently large. This implies the asserted uniform convergence on K of the Dirichlet series Theorem. Let f(s = n= a n n s be a Dirichlet series which converges for some s C. Then the series converges uniformly in every angular region Ang(s, α, < α < π/. In particular lim f(s = f(s, s s when s approaches s within an angular region Ang(s, α. Proof. As in the proof of theorem 6.4 we may suppose s =. Set C α := / cos α. Let ε > be given. Since n= a n converges, there exists an n N, such that N a n < ε := n=m ε + C α for all N M n. With A Mn := n k=m a k, A M,M =, we have by the Abel summation formula N n=m a n n s = A MN N N + s n=m A Mn ( n s (n + s. From this, we get for all s Ang(, α, σ := Re(s, and N M n the estimate N n=m a n ε n s N s + ε N n s (n + s n=m N ( ε + ε C α n σ (n + σ n=m = ε + ε C α ( M σ N σ ε + ε C α = ε. This shows the uniform convergence of the Dirichlet series in Ang(, α. Therefore f is continuous in Ang(, α, which implies the last assertion of the theorem Definition. Let f(s = of f is defined by n= σ c := σ c (f := inf {Re(s : a n be a Dirichlet series. The abscissa of convergence ns a n converges }. ns n= 4

43 By theorem 6.4 this is the same as σ c = inf {Re(s : N a n is bounded for N } ns n= and it follows that the series converges to a holomorphic function in the halfplane H(σ c. Examples. Consider the three Dirichlet series ζ(s = n= n s, g(s := n= ( n n s, ζ (s = n= µ(n n s. We have σ a (ζ = σ a (g = σ a (/ζ =. Clearly σ c (ζ = and σ c (g =, since the partial sums N n= ( n are bounded. The abscissa of convergence σ c (/ζ is not known; of course σ c (/ζ. One conjectures that σ c (/ζ =, which is equivalent to the Riemann Hypothesis, which we will discuss in a later chapter. Remark. Multiplying the zeta series by s yields s ζ(s = n=. Hence (n s g(s = ( s ζ(s Theorem. If the Dirichlet series f(s = a n n= n s has a finite abscissa of convergence σ c, then for the abscissa of absolute convergence σ a the following estimate holds: σ c σ a σ c +. Proof. Without loss of generality we may suppose σ c =. Then every ε >. We have to show that n= a n n σ < for all σ >. To see this, write σ = + ε, ε >. Then a n n = a n σ n ε n +ε n= a n n ε converges for Since a n /n ε is bounded for n and n= /n+ε <, the assertion follows. Remarks. a It can be easily seen that σ c = implies σ a =. b The above examples show that the cases σ a = σ c and σ a = σ c + do actually occur. 4

44 6. Dirichlet series c That σ a and σ c may be different is quite surprising if one looks at the situation for power series: If a power series n= a nz n converges for some z, it converges absolutely for every z with z < z Theorem (Landau. Let f(s = n= a n n s be a Dirichlet series with non-negative coefficients a n and finite abscissa of absolute convergence σ a R. Then the function f, which is holomorphic in the halfplane H(σ a, cannot be continued analytically as a holomorphic function to any neighborhood of σ a. Proof. Assume to the contrary that there exists a small open disk D around σ a such that f can be analytically continued to a holomorphic function in H(σ a D, which we denote again by f. Then the Taylor series of f at the point σ := σ a + has radius of convergence >. Since f (k (σ = n= the Taylor series has the form ( log n k a n, n σ f(s = k= f (k (σ (s σ k = k! k= n= ( log n k a n (s σ k! n σ k. By hypothesis there exists a real σ < σ a such that the Taylor series converges for s = σ. We have f(σ = (log n k a n (σ σ k k= n= k! n σ (log n k (σ σ k = k! n= k= an n σ, where the reordering is allowed since all terms are non-negative. Now (log n k (σ σ k k= k! = e (log n(σ σ = n σ σ, hence we have a convergent series f(σ = n σ σ n= an n σ = n= a n n σ. Thus the abscissa of absolute convergence is σ < σ a, a contradiction. Hence the assumption is false, which proves the theorem. 4