Theorems

Question 1

Dirichlet’s Theorem

Answer 1

If k > 0 and gcd(h, k) = 1, then there are infinitely many primes in the arithmetic progression nk + h for n = 1, 2, …

Question 2

Prime Number Theorem

Answer 2

For large x, the ratio 𝜋(x) / (x/logx) approaches 1, i.e.
𝜋(x)
———- ⟶ 1 as x ⟶ ∞
(x/logx)

Question 3

Properties of divisibility

Answer 3

Let a, b, d, m, n be arbitrary integers unless otherwise indicated. Then

(1) d|m and d|n ⇒ d|(am + bn)
(2) ad|an and a ≠ 0 ⇒ d|n
(3) d|n and n ≠ 0 ⇒ |d| ≤ |n|
(4) d|n and n|d ⇒ |d| = |n|
(5) d|n and nd≠ 0 ⇒ (n/d) | n

Question 4

Common divisor

Answer 4

Given any two integers a and b, there is a common divisor of the form d = ax + by where x, y ∈ Z. Moreover, every common divisor of a and b divides this d.

Question 5

Euclid’s Lemma

Answer 5

If a|bc and gcd(a, b) = 1, then a|c.

Question 6

Prime numbers

Answer 6

(1) Every integer n > 1 is a prime number of a product of prime numbers.
(2) There are infinitely many prime numbers.
(3) If p is a prime and p ∤ a, then (p, a) = 1
(4) If p is a prime and p|ab, then p|a or p|b. More generally if p|a…a, then p divides at least one of the factors.

Question 7

Fundamental Theorem of Arithmetic

Answer 7

Every integer n > 1 can be represented as a product of prime factors in only one way, apart from the order of the factors.

Question 8

∞
∑ 1 / pₙ
n=1

Answer 8

Let pₙ denote the nth prime. The infinite series
              ∞
              ∑  1 / pₙ
            n=1
diverges.

Question 9

Quotient and remainder terms

Answer 9

Given positive integers a and b with b > 0, there exists a unique pair of integers q and r such that a = bq + r with 0 ≤ r < b. Moreover, r = 0 if and only if b|a.

Question 10

Euclidean Algorithm

Answer 10

We are given positive integers and b where b∤a. Let r₀ = a and
r₁ = b, and apply the division repeatedly to obtain a set of remainders r₁, …, rₙ₊₁ defined successively by the relations
r₀ = r₁q₁ + r₂ 0 < r₂ < r₁
r₁ = r₂q₂ + r₃ 0 < r₃ < r₂
… …
rₙ₋₂ = rₙ₋₁qₙ₋₁ + rₙ 0 < rₙ < rₙ₋₁
rₙ₋₁ = rₙqₙ + rₙ₊₁ rₙ₊₁ = 0
Then rₙ = gcd(a, b)

Question 11

∑ 𝜇(d)

d|n

Answer 11

If n ≥ 1, we have
∑ 𝜇(d) = [1/n] = 1 if n = 1
d|n = 0 if n > 1

Question 12

∑ 𝜙(d)

d|n

Answer 12

If n ≥ 1, we have
∑ 𝜙(d) = n
d|n

Question 13

A relation connecting 𝜙 and 𝜇

Answer 13

If n ≥ 1, we have
𝜙(n) = ∑ 𝜇(d) n/d
d|n

Question 14

Product formula for 𝜙(n)

Answer 14

For n ≥ 1, we have
𝜙(n) = n ∏ (1 - 1/p)
p|n

Question 15

Properties of 𝜙(n)

Answer 15

(a) 𝜙(pᵅ) = pᵅ - pᵅ⁻¹ for prime p and a ≥ 1
(b) 𝜙(mn) = 𝜙(m)𝜙(n)d / 𝜙(d), where d = (m, n)
(c) 𝜙(mn) = 𝜙(m)𝜙(n) if (m, n) = 1
(d) a|b ⇒ 𝜙(a)|𝜙(b)
(e) 𝜙(n) is even for n ≥ 3. Moreover if n has r distinct odd prime factors, then 2 ͬ |𝜙(n).

Question 16

Properties of Dirichlet Convolution

Answer 16

Let f, g, and k be arithmetical functions. Then

a) f ∗ g = g ∗ f
(b) (f ∗ g) ∗ k = f ∗ (g ∗ k
i. e. Dirichlet convolution is commutative and associative.

Question 17

Identity function on an arithmetical function

Answer 17

Let f be an arithmetical function. Then for all f, we have

f ∗ I = I ∗ f = f

Question 18

Dirichlet Inverse

Answer 18

If f is an arithmetical function with f(1) ≠ 0, then there exists a unique arithmetical function f⁻¹ such that
f ∗ f⁻¹ = f⁻¹ ∗ f = I
The arithmetical function f⁻¹ is called the Dirichlet inverse of f. Moreover, f⁻¹ is given by the recursion formula
f⁻¹(1) = 1 / f(1)
f⁻¹(n) = - (1 / f(1)) ∑ f(n/d) f⁻¹(d) for n > 1
d|n
d

Question 19

Set of arithmetical functions

Answer 19

The set of all arithmetical functions f with f(1) ≠ 0 forms an abelian group under the operation ∗.

Question 20

Mobius Inversion Formula

Answer 20

The equation
              f(n) =  ∑  g(d)
                       d|n
implies
             g(n) =  ∑  f(d) 𝜇(n/d)
                       d|n
The converse holds.

Question 21

Von-Mangoldt formula for log

Answer 21

If n ≥ 1, we have
log n = ∑ Λ(d)
d|n

Question 22

Von-Mangoldt formula in terms of log and 𝜇

Answer 22

If n ≥ 1, we have
Λ(n) = ∑ 𝜇(d) log(n/d) = - ∑ 𝜇(d) log d
d|n d|n

Question 23

When is f(1) = 1?

Answer 23

If f is multiplicative, then f(1) = 1.

Question 24

Properties of functions with f(1) = 1

Answer 24

Given an arithmetical function f with f(1) = 1,
(a) f is multiplicative if and only if
f(p₁ ͣ ¹ … pₖ ͣ ᵏ) = f(p₁ ͣ ¹) … f(pₖ ͣ ᵏ)
for all primes pᵢ and all integers aᵢ ≥ 1
(b) If f is multiplicative, then f is completely multiplicative if and
only if
f(p ͣ ) = f(p) ͣ
for all primes p and all integers a ≥ 1

Question 25

Dirichlet product of multiplicative functions

Answer 25

If f and g are multiplicative functions, then the Dirichlet product f ∗ g is also multiplicative. Note that the Dirichlet product of two completely multiplicative functions is not always completely multiplicative. Also, if both g and f ∗ g are multiplicative, then f is multiplicative.

Question 26

Dirichlet inverse of a multiplicative function

Answer 26

If g is multiplicative, then g⁻¹, the Dirichlet inverse of g, is multiplicative.

Question 27

Inverse of a completely multiplicative function

Answer 27

Let f be a multiplicative function. Then f is a completely multiplicative function if and only if f⁻¹(n) = 𝜇(n) f(n) for all n ≥ 1.

Question 28

∑ 𝜇(d) f(d) | d|n

Answer 28

If f is multiplicative, then ∑ 𝜇(d) f(d) = ∏ (1 - f(p)) d|n p|n

Question 29

Product formula for 𝜙⁻¹(n)

Answer 29

𝜙⁻¹(n) = ∏ (1 - p) | p|n

Question 30

∑ λ(d) | d|n

Answer 30

For every n ≥ 1, we have ∑ λ(d) = 1 if n is a square d|n 0 otherwise

Question 31

𝜎ₐ⁻¹(n)

Answer 31

For n ≥ 1, we have that 𝜎ₐ⁻¹(n) = ∑ d ͣ 𝜇(d) 𝜇(n/d) d|n

Answer 32

Let 𝛼 and 𝛽 be arithmetical functions. Then we have 𝛼 ⚬ (𝛽 ⚬ F) = (𝛼 ∗ 𝛽) ⚬ F where F : (0, ∞) ⟶ C such that F(x) = 0 for 0 < x < 1

Question 33

(𝛼 ○ F)(m) for integers m

Answer 33

Let F : (0, ∞) ⟶ C such that F(x) = 0 for 0 < x < 1 and let 𝛼 be an arithmetical function. Also let G : (0, ∞) ⟶ C such that G(x) = 0 for 0 < x < 1 and G(x) = (𝛼 ○ F)(x) = ∑ 𝛼(n) F(x/n) n≤x If F(x) = 0 for all nonintegral x (i.e. x is not an integer), we find that (𝛼 ○ F)(m) = (𝛼 ∗ F)(m) for all integers m.

Question 34

Dirichlet convolution identity function on ○

Answer 34

The identity function for Dirichlet convolution is also a left identity on the operation ○.

Question 35

Generalised Inversion Formula

Answer 35

``` Let 𝛼 be an arithmetical function with Dirichlet inverse 𝛼⁻¹. Then the equation G(x) = ∑ 𝛼(n) F(x/n) n≤x implies F(x) = ∑ 𝛼⁻¹(n) G(x/n) n≤x The converse also holds. ```

Question 36

Generalised Mobius Inversion Formula

Answer 36

``` If 𝛼 is a completely multiplicative function, we have G(x) = ∑ 𝛼(n) F(x/n) n≤x if and only if F(x) = ∑ 𝜇(n) 𝛼(n) G(x/n) n≤x ```

Question 37

Arithmetic on derivatives of arithmetical functions

Answer 37

If f and g are arithmetical functions, we have (a) (f + g)' = f' + g' (b) (f ∗ g)' = f' ∗ g + f ∗ g' (c) (f⁻¹)' = -f' ∗ (f ∗ f)⁻¹ as long as f(1) ≠ 0

Question 38

Selberg Identity

Answer 38

For n ≥ 1, we have Λ(n) log n + ∑ Λ(n) Λ(n/d) = ∑ µ(d) log²(n/d) d|n d|n

Question 39

f(t) = O(g(t)) for t ≥ a

Answer 39

``` f(t) = O(g(t)) for t ≥ a implies that x x ∫ f(t) dt = O( ∫ g(t) dt ) a a for x ≥ a ```

Question 40

Euler Summation Formula

Answer 40

If f has a continuous derivative f' on the interval [y, x], where 0 < y < x, then x x ∑ f(n) = ∫ f(t) dt + ∫ (t - [t])f'(t) dt + f(x)([x] - x) - f(y)([y] - y) y

Question 41

Euler Summation formula if f has a continuous derivative on | x > 0

Answer 41

Let f have a continuous derivative on x > 0. Then x x ∑ f(n) = ∫ f(t) dt + f(x)([x] - x) + f(1) + ∫ (t - [t])f'(t) dt n≤x 1 1

Question 42

Consequences of Euler's Summation Formula

Answer 42

If x ≥ 1 then we have: (a) ∑ 1/n = log x + 𝛾 + O(1/x) n≤x where 𝛾 is Euler's constant defined by the equation n 𝛾 = lim ( ∑ 1/k - log n ) n→∞ k=1 (b) ∑ 1/nˢ = x¹⁻ˢ / (1-s) + ζ(s) + O(x⁻ˢ) if s > 0, s ≠ 1 n≤x (c) ∑ 1/nˢ = O(x¹⁻ˢ) if s > 1 n>x (d) ∑ nᵅ = xᵅ⁺¹ / (a + 1) + O(xᵅ) if a ≥ 0 n≤x

Question 43

Dirichlet's Asymptotic Formula (weak)

Answer 43

For all x ≥ 1, we have ∑ d(n) = x log x + O(x) n≤x

Question 44

Dirichlet's Asymptotic Formula

Answer 44

For all x ≥ 1, we have ∑ d(n) = x log x + (2𝛾 - 1)x + O(√x) n≤x

Question 45

Average order of 𝜎(n)

Answer 45

For all x ≥ 1, we have ∑ 𝜎(n) = 1/2 ζ(2) x² + O(x log x) n≤x

Question 46

Average order of 𝜎ₐ(n) for a > 0

Answer 46

If x ≥ 1 and a > 0, a ≠ 1, we have ∑ 𝜎ₐ(n) = ( ζ(a + 1) / (a + 1) ) x ͣ ⁺¹ + O(x ᷩ ) n≤x where 𝛽 = max{1, a}

Question 47

Average order of 𝜎ₐ(n) for a < 0

Answer 47

Write a = -𝛽, so 𝛽 > 0. Let 𝛿 = max{0, 1 - 𝛽}. Then if x > 1, we have ∑ 𝜎ₐ(n) = ζ(𝛽 + 1) x + O(xᵟ) if 𝛽 ≠ 1 n≤x ζ(2) x + O(log x) if 𝛽 = 1

Question 48

Average order of 𝜙(n)

Answer 48

For x > 1, we have ∑ 𝜙(n) = (3/𝜋²) x² + O(x log x) n≤x

Question 49

lim 1/x ∑ 𝜇(n) | x→∞ n≤x

Answer 49

lim 1/x ∑ 𝜇(n) = 0 x→∞ n≤x This is equivalent to the PNT.

Question 50

lim 1/x ∑ Λ(n) | x→∞ n≤x

Answer 50

lim 1/x ∑ Λ(n) = 1 x→∞ n≤x This is equivalent to the PNT.

Question 51

Partial sums of a Dirichlet product

Answer 51

If h = f ∗ g, let H(x) = ∑ h(n), F(x) = ∑ f(n), G(x) = ∑ g(n) n≤x n≤x n≤x Then H(x) = ∑ f(n) G(x/n) = ∑ g(n) F(x/n) n≤x n≤x

Question 52

∑ F(x/n) | n≤x

Answer 52

If F(x) = ∑ f(n) n≤x, we have ∑ ∑ f(d) = ∑ f(n) [x/n] = ∑ F(x/n) n≤x d|n n≤x n≤x

Question 53

∑ 𝜇(n) [x/n] | n≤x

Answer 53

We have ∑ 𝜇(n) [x/n] = 1 n≤x

Question 54

∑ Λ(n) [x/n] | n≤x

Answer 54

We have ∑ Λ(n) [x/n] = (log [x])! n≤x

Question 55

∞ | ∑ 𝜇(n) / n | n=1

Answer 55

``` For all x ≥ 1, we have ∞ | ∑ 𝜇(n) / n | ≤ 1 n=1 with equality holding only when x < 2. ```

Question 56

Legendre's Identity

Answer 56

``` For every x ≥ 1, we have [x]! = ∏ pᵅ⁽ᵖ⁾ where ∞ 𝛼(p) = ∑ [x/pᵐ] m=1 ```

Question 57

(log[x])! when x ≥ 2

Answer 57

If x ≥ 2, we have (log[x])! = x log x - x + O(log x) and hence ∑ Λ(n) [x/n] = x log x - x + O(log x) n≤x

Question 58

When x ≥ 2, ∑ [x/p] log p p≤x

Answer 58

When x ≥ 2, ∑ [x/p] log p = x log x + O(x) p≤x

Question 59

Dirichlet's Hyperbola Method

Answer 59

Let F(x) = ∑ f(n), G(x) = ∑ g(n) and H(x) = ∑ (f ∗ g)(n), n≤x n≤x n≤x so that H(x) = ∑ ∑ f(d) g(n/d) = ∑ f(d) g(q) n≤x d|n q, d qd≤x If a, b ∈ R such that ab = x, than ∑ f(d) g(q) = ∑ f(n) G(x/n) + ∑ g(n) F(x/n) - F(a) G(b) q, d n≤a n≤b qd≤x

Question 60

𝜓(x) / x - θ(x) / x

Answer 60

For x > 0, we have 0 ≤ 𝜓(x) / x - θ(x) / x ≤ (log x)² / (2 √x log2) This implies that lim ( 𝜓(x) / x - θ(x) / x ) = 0 x→∞ Hence if either 𝜓(x) / x or θ(x) / x tends to a limit, the so does the other and the limits are equal.

Question 61

Abel's Identity

Answer 61

For any arithmetical function a(n), let A(x) = ∑ a(n) n≤x where A(x) = 0 if x < 1. Assume f has a continuous derivative on the interval [y, x], where 0 < y < x. Then we have x ∑ a(n) f(n) = A(x) f(x) - A(y) f(y) - ∫ A(t) f'(t) dx y < n ≤ x y

Question 62

Abel's Identity where f has a continuous derivative on x > 0

Answer 62

For any arithmetical function a(n), let A(x) = ∑ a(n) n≤x where A(x) = 0 if x < 1. Assume f has a continuous derivative on x > 0. Then we have x ∑ a(n) f(n) = A(x) f(x) - ∫ A(t) f'(t) dt n≤x 1

Question 63

θ(x) and 𝜋(x) in terms of integrals

Answer 63

``` For x ≥ 2, we have x θ(x) = 𝜋(x) log x - ∫ 𝜋(t) / t dt 2 and x 𝜋(x) = θ(x) / logx + ∫ θ(t) / (t log²(t)) dt 2 ```

Question 64

lim θ(x) / x | x→∞

Answer 64

lim θ(x) / x = 1 x→∞ This is equivalent to the PNT.

Question 65

lim 𝜓(x) / x | x→∞

Answer 65

lim 𝜓(x) / x = 1 x→∞ This is equivalent to the PNT.

Question 66

Relating the PNT to the asymptotic value of the nth prime

Answer 66

Let pₙ denote the nth prime. Then the following asymptotic relations are logically equivalent. lim (𝜋(x) log x) / x = 1 x→∞ lim (𝜋(x) log 𝜋(x)) / x = 1 x→∞ lim pₙ / (n log n) = 1 n→∞

Question 67

Order of magnitude of 𝜋(n)

Answer 67

For every integer n ≥ 2, we have | 1/6) (n / log n) < 𝜋(n) < 6 (n / log n

Question 68

Order of magnitude of pₙ

Answer 68

For n ≥ 1, the nth prime satisfies the inequalities | 1/6 n log n < pₙ < 12 (n log n + n log(12/e) )

Question 69

Shapiro's Tauberian Theorem

Answer 69

``` Let {a(n)} be a non-negative sequence such that ∑ a(n) [x/n] = x log x + O(x) n≤x for all x ≥ 1. Then (a) For x ≥ 1, we have ∑ a(n) / n = log x + O(1) n≤x (b) There exists a constant B > 0 such that ∑ a(n) ≤ Bx n≤x for all x ≥ 1. (c) There exists a constant A > 0 and an x₀ > 0 such that ∑ a(n) ≥ A(x) n≤x for all x ≥ x₀ ```

Question 70

Shapiro's Theorem on Λ(n)

Answer 70

``` For all x ≥ 1, we have ∑ Λ(n) / n = log x + O(1) n≤x Also, there exist positive constants c₁ and c₂ such that ∑ Λ(n) = 𝜓(x) ≤ c₁x n≤x for all x ≥ 1 and ∑ Λ(n) = 𝜓(x) ≥ c₂x n≤x for all sufficiently large x. ```

Question 71

Shapiro's Theorem on θ(n)

Answer 71

``` For all x ≥ 1, we have ∑ log p / p = log x + O(1) p≤x Also, there exist positive constants c₁ and c₂ such that ∑ Λ₁(n) = θ(x) ≤ c₁x n≤x for all x ≥ 1 and ∑ Λ₁(n) = θ(x) ≥ c₂x n≤x for all sufficiently large x. Here, Λ₁(n) = log p if n is a prime p 0 otherwise ```

Question 72

∑ 𝜓(x/n) and ∑ θ(x/n) | n≤x n≤x

Answer 72

``` We have ∑ 𝜓(x/n) = x log x - x + O(log x) n≤x and ∑ θ(x/n) = x log x + O(x) n≤x ```

Question 73

An asymptotic formula for the partial sums ∑ 1/p p≤x

Answer 73

There is a constant A such that ∑ 1/p = log log x + A + O(1 / log x) p≤x for all x ≥ 2.

Question 74

Relating M(x) to the PNT

Answer 74

``` We have lim ( M(x) / x - H(x) / (x log x) ) = 0 x→∞ Hence the PNT implies lim ( M(x) / x ) = 0 x→∞ Also, M(x) = o(x) as x ⟶ ∞ implies 𝜓(x) ∼ x as x ⟶ ∞ and so lim ( M(x) / x ) = 0 x→∞ also implies the PNT. ```

Question 75

Relating ∑ 𝜇(n) / n n≤x to the PNT

Answer 75

``` If A(x) = ∑ 𝜇(n) / n n≤x the relation A(x) = o(1) as x ⟶ ∞ implies the PNT. In other words, the PNT is a consequence of the statement that ∞ ∑ 𝜇(n) / n n≤x converges and has sum 0. ```

Question 76

Selberg's Asymptotic Formula

Answer 76

For x > 0, we have 𝜓(x) log x + ∑ Λ(n) 𝜓(x/n) = 2x log x + O(x) n≤x

Question 77

Deducing Selberg's Asymptotic Formula

Answer 77

Let F be a real or complex valued function defined on (0, +∞), and let G(x) = log x ∑ F(x/n) n≤x Then F(x) log x + ∑ F(x/n) Λ(n) = ∑ 𝜇(d) G(x/d) n≤x d≤x

Question 78

If N ≥ 1 and 𝜎 ≥ c > 𝜎ₐ, what is ∞ | ∑ f(n) / n⁻ˢ | n=N

Answer 78

If N ≥ 1 and 𝜎 ≥ c > 𝜎ₐ, we have ∞ ∞ | ∑ f(n) / n⁻ˢ | ≤ N^-(𝜎 - c) ∑ |f(n)| / nᶜ n=N n=N

Question 79

For a Dirichlet series F(n), what is lim F(𝜎 + it) 𝜎→+∞

Answer 79

``` Let ∞ F(s) = ∑ f(n) / nˢ n=1 Then lim F(𝜎 + it) = f(1) 𝜎→+∞ uniformly for -∞ < t < ∞ ```

Question 80

Uniqueness Theorem

Answer 80

Given 2 Dirichlet series ∞ ∞ F(s) = ∑ f(n) / nˢ and G(s) = ∑ g(n) / nˢ n=1 n=1 both absolutely convergent for 𝜎 > 𝜎ₐ, if F(s) = G(s) for each s is an infinite sequence {sₖ} such that 𝜎ₖ ⟶ ∞ as k ⟶ ∞, then f(n) = g(n) for all n.

Question 81

Half-planes with F(s) never zero

Answer 81

``` Let F(s) = ∑ f(n) / nˢ and assume that F(s) ≠ 0 for some s with 𝜎 > 𝜎ₐ. Then there is a half-plane 𝜎 ≥ c > 𝜎ₐ in which F(s) is never zero. ```

Question 82

Dirichlet convolution in half-plane of convergence

Answer 82

Given two functions F(s) and G(s) represented by Dirichlet series ∞ ∞ F(s) = ∑ f(n) / nˢ for 𝜎>a and G(s) = ∑ g(n) / nˢ for 𝜎>b n=1 n=1 In the half-plane where both series converge absolutely, we have ∞ F(s) G(s) = ∑ h(n) / nˢ n=1 where h = f ∗ g, the Dirichlet convolution of f and g: h(n) = ∑ f(d) g(n/d) d|n Conversely, if F(s)G(s) = ∑ 𝛼(n)/nˢ for all s in a sequence {sₖ} with 𝜎ₖ ⟶ ∞ as k ⟶ ∞, then 𝛼 = f ∗ g.

Question 83

Euler Product

Answer 83

Let f be a multiplicative arithmetical function such that the series ∑f(n) is absolutely convergent. Then the sum of the series can be expressed as an absolutely convergent infinite product ∞ ∑ f(n) = ∏ (1 + f(p) + f(p²) + ...) n=1 p extended over all primes. If f is completely multiplicative, then ∞ ∑ f(n) = ∏ 1 / (1 - f(p)) n=1 p

Question 84

Product formulae for Dirichlet series in half plane of convergence

Answer 84

Assume ∑ f(n)/nˢ converges absolutely for 𝜎 > 𝜎ₐ. If f is multiplicative, we have ∞ ∑ f(n) / nˢ = ∏ (1 + f(p)/pˢ + f(p²)/p²ˢ + ...) if 𝜎 > 𝜎ₐ n=1 p and if f is completely multiplicative, we have ∞ ∑ f(n) / nˢ = ∏ (1 - f(p)/p)⁻¹ if 𝜎 > 𝜎ₐ n=1

Question 85

Modulus of Dirichlet series with bounded partial sums

Answer 85

Let s₀ = 𝜎₀ + it₀ and assume that the Dirichlet series ∑ f(n)/nˢ⁰ has bounded partial sums, say | ∑ f(n) / nˢ⁰ | ≤ M n≤x for all x ≥ 1. Then for each 𝜎 > 𝜎₀, we have | ∑ f(n) / nˢ | ≤ 2Ma^(𝜎₀ - 𝜎) (1 + (|s - s₀|) / (𝜎 - 𝜎₀) ) a

Question 86

Convergent/ divergent Dirichlet series for s = 𝜎₀ + it₀

Answer 86

If the series ∑ f(n)/nˢ converges for s = 𝜎₀ + it₀, then it also converges for all s with 𝜎 > 𝜎₀. If it diverges for s = 𝜎₀ + it₀, then it diverges for all s with 𝜎 < 𝜎₀.

Question 87

Bounds on 𝜎ₐ and 𝜎_c

Answer 87

For any Dirichlet series with 𝜎_c finite, we have | 0 ≤ 𝜎ₐ - 𝜎_c ≤ 1

Question 88

𝜎ₐ and 𝜎_c on ∞ ∑ (-1)ⁿ / nˢ n=1

Answer 88

``` The series ∞ ∑ (-1)ⁿ / nˢ n=1 converges if 𝜎 > 0 but only converges absolutely for 𝜎ₐ = 1. Therefore 𝜎_c = 0 and 𝜎ₐ = 1. ```

Question 89

Analytic functions

Answer 89

Let {fₙ} be a sequence of functions analytic on an open subset S of the complex plane, and assume that {fₙ} converges uniformly on every compact subset of S to a limit function f. Then f is analytic on S and the sequence of derivatives {fₙ'} converges uniformly on every compact subset of S to the derivative of f'.

Question 90

Dirichlet series on compact rectangles

Answer 90

A Dirichlet series converges uniformly on every compact rectangle lying interior to the half-plane of convergence 𝜎 > 𝜎_c.

Question 91

Derivative of Dirichlet series

Answer 91

``` The sum function ∞ F(s) = ∑ f(n) / nˢ n=1 of a Dirichlet series is analytic in its half-plane of convergence 𝜎 > 𝜎_c and its derivative F'(s) is represented in this half-plane by the Dirichlet series ∞ F'(s) = - ∑ (f(n) log n) / nˢ n=1 obtained by differentiating term by term ```

Question 92

Landau's Theorem

Answer 92

Let F(s) be represented in the half-plane 𝜎 > c by the Dirichlet series ∞ F(s) = ∑ f(n) / nˢ n=1 where c is finite and assume f(n) ≥ 0 for all n ≥ n₀. If F(s) is analytic in some disc about the point s = c, then the Dirichlet series converges in the half-plane 𝜎 > c - 𝜀 for some 𝜀 > 0. Consequently, if the Dirichlet series has a finite abscissa of convergence 𝜎_c, then F(s) has a singularity on the real axis at the point s = 𝜎_c.

Question 93

Half-plane of a Dirichlet series as an exponential

Answer 93

Let F(s) = ∑ f(n)/nˢ be absolutely convergent for 𝜎 > 𝜎ₐ and assume that f(1) ≠ 0. If F(s) ≠ 0 for 𝜎 > 𝜎₀ > 𝜎ₐ, then for 𝜎 > 𝜎₀ we have F(s) = exp( G(s) ) with ∞ G(s) = log(f(1)) + ∑ (f' ∗ f⁻¹)(n) / ( (log n)nˢ ) n=1 where f⁻¹ is the Dirichlet inverse of f and f'(n) = f(n) log n.

Question 94

∞ ∑ f(n) g(n) / n^(a+b) n=1

Answer 94

Given two Dirichlet series F(s) = ∑ f(n)/nˢ and G(s) = ∑ g(n)/nˢ with abscissae of absolute convergence 𝜎₁ and 𝜎₂ respectively. Then for a > 𝜎₁ and b > 𝜎₂ we have T ∞ lim (1 / 2T) ∫ F(a + it) G(b - it) dt = ∑ f(n) g(n) / n^(a+b) T→∞ -T n=1

Answer 95

If F(s) = ∑ f(n)/nˢ converges absolutely for 𝜎 > 𝜎ₐ, then for 𝜎 > 𝜎ₐ we have T ∞ lim (1 / 2T) ∫ |F(𝜎 + it)|² dt = ∑ |f(n)|² / n^(2𝜎) T→∞ -T n=1 In particular, if 𝜎 > 1, we have (a) T ∞ lim (1 / 2T) ∫ |ζ(𝜎 + it)|² dt = ∑ 1 / n^(2𝜎) = ζ(2𝜎) T→∞ -T n=1 (b) T ∞ lim (1 / 2T) ∫ |ζ⁽ᵏ⁾(𝜎 + it)|² dt = ∑ log²ᵏ(n) / n^(2𝜎) = ζ⁽²ᵏ⁾(2𝜎) T→∞ -T n=1 (c) T ∞ lim (1 / 2T) ∫ |ζ(𝜎 + it)|⁻² dt = ∑ 𝜇²(n) / n^(2𝜎) = ζ(2𝜎) / ζ(4𝜎) T→∞ -T n=1 (d) T ∞ lim (1 / 2T) ∫ |ζ(𝜎 + it)|⁴ dt = ∑ d²(n) / n^(2𝜎) = ζ⁴(2𝜎) / ζ(4𝜎) T→∞ -T n=1

Question 96

Integral formula for coefficients of a Dirichlet series

Answer 96

Assume the series F(s) = ∑ f(n)/nˢ converges absolutely for 𝜎 > 𝜎ₐ Then for 𝜎 > 𝜎ₐ and x > 0 we have T lim (1 / 2T) ∫ F(𝜎 + it) x^(𝜎 + it) dt = f(n) if x = n T→∞ -T 0 otherwise

Question 97

What is c+i∞ 1 / 2𝜋i ∫ aᶻ / z dz c-i∞

Answer 97

Let c > 0 and a be any postitive real number. Then we have c+i∞ 1 if a > 1 1 / 2𝜋i ∫ aᶻ / z dz = 1/2 if a = 1 c-i∞ 0 if 0 < a < 1 Moreover, we have c+iT | 1 / 2𝜋i ∫ aᶻ / z dz | ≤ a ͨ / (𝜋 T log(1/a)) if 0 < a < 1 c-iT c+iT | 1 / 2𝜋i ∫ 1/z dz - 1/2 | ≤ c / 𝜋T if a = 1 c-iT c+iT | 1 / 2𝜋i ∫ aᶻ / z dz - 1 | ≤ a ͨ / (𝜋 T log a) if a > 1 c-iT

Question 98

Perron's Formula

Answer 98

Let F(s) = ∑ f(n)/nˢ be absolutely convergent for 𝜎 > 𝜎ₐ, and let c > 0 and x > 0 be arbitrary. Then if 𝜎 > 𝜎ₐ - c, we have c+i∞ ∗ 1 / 2𝜋i ∫ F(s + z) xᶻ / z dz = ∑ f(n) / nˢ c-i∞ n≤x where ∑* means that the last term in the sum must be multiplies by 1/2 when x is an integer.

Question 99

Perron's Formula for s = 0

Answer 99

If c > 𝜎ₐ Perron's Formula is valid for s = 0 and we obtain the following integral representation for the partial sum of the coefficients: c+i∞ ∗ 1 / 2𝜋i ∫ F(z) xᶻ / z dz = ∑ f(n) c-i∞ n≤x

Question 100

Riemann-zeta function for 𝜎 > 0, x > 0 and s ≠ 1.

Answer 100

Suppose that 𝜎 > 0, x > 0 and s ≠ 1. Then for N an integer, we have ∞ ζ(s) = ∑ 1 / nˢ + N¹⁻ˢ / (s - 1) - s ∫ {t} / tˢ⁺¹ dt n≤N N

Question 101

Poles of the Riemann-zeta function

Answer 101

The Riemann-zeta function has a simple pole at s = 1 with residue 1, but is otherwise analytic in the half-plane 𝜎 > 0.

Question 102

Properties of the gamma function

Answer 102

For 𝜎 > 0, we have that s𝚪(s) = 𝚪(s + 1). Moreover 𝚪(1) = 1 and 𝚪(n) = (n - 1)! for all n ∈ N.

Question 103

Poles of the gamma function

Answer 103

The function 𝚪(s) has an analytic continuation as a meromorphic function on C, the only singularities being at s = -k ∈ Z ≤ 0. These singularities are simple poles with residues (-1)ᵏ / k!.

Question 104

Representations of the gamma function

Answer 104

𝚪(s) = lim (nˢ n!) / s(s + 1) ... (s + n) for s ≠ 0, -1, -2, ... ∞ 1 / 𝚪(s) = s exp(𝛾s) ∏ (1 + s/n) exp(-s/n) for all s n=1

Question 105

Functional equation for the gamma function

Answer 105

𝚪(s) 𝚪(1 - s) = 𝜋 / sin(𝜋s) for all s

Question 106

Multiplication formula for the gamma equation

Answer 106

For all s and all integers m ≥ 1, we have | 𝚪(s) 𝚪(s + 1/m) ... 𝚪(s + (m-1)/m) = (2𝜋)^((m-1)/2) m^(1/2 - ms) 𝚪(ms)

Question 107

Functional equation for the Riemann-zeta function

Answer 107

For all s, we have | ζ(s) = 2 (2𝜋)ˢ⁻¹ 𝚪(1 - s) sin(𝜋s / 2) ζ(1 - s)

Question 108

Riemann Hypothesis

Answer 108

If 0 < Re(s) < 1 and ζ(s) = 0, then Re(s) = 1/2

Question 109

∑ (x - n) a(n) | n≤x

Answer 109

For any arithmetical function a(n), let A(x) = ∑ a(n) (n≤x) where A(x) = 0 if x < 1. Then x ∑ (x - n) a(n) = ∫ A(t) dt n≤x 1

Question 110

L'Hopital's Rule for increasing linear piecewise functions

Answer 110

Let A(x) = ∑ a(n) (n≤x) and let A₁(x) = ∫₁ˣ A(t) dt. Assume a(n) ≥ 0 for all n. If we have the asymptotic formula A₁(x) ∼ Lxᶜ as x ⟶ ∞ for some c > 0 and L > 0, then we have A(x) ∼ cLxᶜ⁻¹ as x ⟶ ∞

Question 111

L'Hopital's rule on 𝜓(x)

Answer 111

We have 𝜓₁(x) = ∑ (x - n) Λ(n) n≤x and the asymptotic formula 𝜓₁(x) ∼ 1/2 xᶜ as x ⟶ ∞ implies 𝜓(x) ∼ x as x ⟶ ∞

Question 112

For c > 0, u > 0, for every k ≥ 1, what is c+i∞ 1 / 2𝜋i ∫ u⁻ᶻ / (z (z + 1) ... (z +k)) dz c-i∞

Answer 112

If c > 0 and u > 0, then for every integer k ≥ 1, we have c+i∞ 1 / 2𝜋i ∫ u⁻ᶻ / (z (z + 1) ... (z +k)) dz = (1 / k!) (1 - u)ᵏ if 0 < u ≤ 1 c-i∞ 0 if u > 1 the integral being absolutely convergent.

Question 113

Contour integral representation for 𝜓₁(x) / x²

Answer 113

If c > 1 and x ≥ 1, we have c+i∞ 𝜓₁(x) / x² = 1 / 2𝜋i ∫ xˢ⁻¹ / s(s + 1) (- ζ'(s) / ζ(s) ) ds c-i∞

Question 114

Contour integral representation for 𝜓₁(x) / x² - 1/2 (1 - 1/x)²

Answer 114

If c > 1 and x ≥ 1, we have c+i∞ 𝜓₁(x) / x² - 1/2 (1 - 1/x)² = 1 / 2𝜋i ∫ xˢ⁻¹ h(s) ds c-i∞ where h(s) = 1 / s(s + 1) (- ζ'(s) / ζ(s) - 1 / (s - 1) )

Question 115

Upper bounds for |ζ(s)| and |ζ'(s)| near the line 𝜎 = 1

Answer 115

For every A > 0, there exists a constant M (depending on A) such that |ζ(s)| ≤ M log t and |ζ'(s)| ≤ M log² t for all s with 𝜎 ≥ 1/2 satisfying 𝜎 > 1 - A / log t and t ≥ e

Question 116

Riemann-zeta formula ≥ 1 for 𝜎 > 1

Answer 116

If 𝜎 > 1, we have | ζ³(𝜎) |ζ(𝜎 + it)|⁴ |ζ(𝜎 + 2it)| ≥ 1

Question 117

Non-vanishing of ζ(1 + it)

Answer 117

We have ζ(1 + it) ≠ 0 for every real t

Question 118

Inequalities for | 1 / ζ(s) | and | ζ'(s) / ζ(s) |

Answer 118

``` There is a constant M > 0 such that | 1 / ζ(s) | < M log⁷ t and | ζ'(s) / ζ(s) | < M log⁹ t whenever 𝜎 ≥ 1 and t ≥ e. ```

Question 119

First Order Poles of f'(s) / f(s)

Answer 119

If f(s) has a pole of order k at s = 𝛼, then the quotient f'(s) / f(s) has a first order pole at s = 𝛼 with residue -k.

Question 120

Where is F(s) = - ζ'(s) / ζ(s) - 1 / (s - 1) analytic?

Answer 120

The function F(s) = - ζ'(s) / ζ(s) - 1 / (s - 1) is analytic at s = 1.

Question 121

Formula for 𝜓(x) ∼ x

Answer 121

For x ≥ 1, we have ∞ 𝜓₁(x) / x² - 1/2 (1 - 1/x)² = 1 / 2𝜋 ∫ h(1 + it) exp(it log x) dt -∞ where the integral ∞ ∫ | h(1 + it) | dt -∞ converges. Therefore, by the Riemann-Lebesgue Lemma, we have 𝜓₁(x) ∼ 1/2 x² and hence 𝜓(x) ∼ x as x ⟶ ∞

Question 122

Zero-free regions for ζ(s)

Answer 122

Assume 𝜎 ≥ 1/2. Then there exist constants A > 0 and C > 0 such that | ζ(𝜎 + it) | > C / log⁷ t whenever 1 - A / log⁹ t < 𝜎 ≤ 1 and t ≥ e This implies that ζ(𝜎 + it) ≠ 0 for such 𝜎 and t.

Question 123

Number of Dirichlet characters

Answer 123

There are exactly 𝜙(k) distinct Dirichlet characters modulo k.

Question 124

Number of Dirichlet characters

Answer 124

There are exactly 𝜙(k) distinct Dirichlet characters modulo k.

Question 125

Properties of Dirichlet characters

Answer 125

Let k be a positive integer. (i) Then we have that 𝝌(n) = 0 or 𝝌(n) = exp(2𝜋in / 𝜙(k)) for some m ∈ N. (ii) Let (n, k) = 1. Then the inverse of a Dirichlet character 𝝌 modulo k is given by the complex conjugate 𝝌 bar which is defined by _ ___ 𝝌(n) = 𝝌(n) Furthermore, 𝝌 bar is also a Dirichlet character modulo k.

Question 126

Sums over 𝝌(n)

Answer 126

Let k N. Then we have (i) k ∑ 𝝌(n) = 𝜙(k) if 𝝌 = 𝝌₀ n=1, (n,k)=1 0 if 𝝌 ≠ 𝝌₀ (i) ∑ 𝝌(n) = 𝜙(k) if n ≡ 1 mod k 𝝌 mod k 0 otherwise

Question 127

What is _ | ∑ 𝝌ᵣ(m) 𝝌ᵣ(n)

Answer 127

Let 𝝌₀, ..., 𝝌_(𝜙(k) - 1) denote the Dirichlet characters modulo k, where k ∈ N. Let m, n ∈ Z such that (n, k) = 1. Then we have 𝜙(k)-1 _ ∑ 𝝌ᵣ(m) 𝝌ᵣ(n) = 𝜙(k) if m ≡ n mod k r=0 0 otherwise

Question 128

∑ 𝝌(n) f(n)

Answer 128

Let 𝝌 be any non-principal character modulo k and let f be a non-negative function which has a continuous negative derivative f'(x) for all x ≥ x₀. Then if y ≥ x ≥ x₀, we have ∑ 𝝌(n) f(n) = O(f(x)) where we sum over x < n ≤ y. If in addition f(x) ⟶ 0 as x ⟶ ∞, then the infinite series ∞ ∑ 𝝌(n) f(n) n=1 converges, and we have, for x ≥ x₀, ∞ ∑ 𝝌(n) f(n) = ∑ 𝝌(n) f(n) + O(f(x)) n≤x n=1

Question 129

∑ 𝝌(n) / n

Answer 129

If 𝝌 is any non-principal character modulo k and if x ≥ 1, we have ∞ ∑ 𝝌(n) / n = ∑ 𝝌(n) / n + O(1 / x) n≤x n=1

Question 130

∑ (𝝌(n) log n) / n

Answer 130

If 𝝌 is any non-principal character modulo k and if x ≥ 1, we have ∞ ∑ (𝝌(n) log n) / n = ∑ (𝝌(n) log n) / n + O((log x) / x) n≤x n=1

Question 131

∑ 𝝌(n) / √n

Answer 131

If 𝝌 is any non-principal character modulo k and if x ≥ 1, we have ∞ ∑ 𝝌(n) / √n = ∑ 𝝌(n) / √n + O(1 / √x) n≤x n=1

Question 132

A(n) = ∑ 𝝌(d) | d|n

Answer 132

Let 𝝌 be any real-valued character modulo k and let A(n) = ∑ 𝝌(d) d|n Then A(n) ≥ 0 for all n and A(n) ≥ 1 if n is a square.

Question 133

Properties of B(n) such that A(n) = ∑ 𝝌(d) and B(n) = ∑ A(n) / √n d|n n≤x

Answer 133

For any real-valued non-principal character 𝝌 modulo k, let A(n) = ∑ 𝝌(d) and B(n) = ∑ A(n) / √n d|n n≤x Then (a) B(x) ⟶ ∞ as x ⟶ ∞. (b) B(x) = 2√x L(1, 𝝌) + O(1) for all x ≥ 1

Question 134

When is L(1, 𝝌) nonzero?

Answer 134

For any real-valued non-principal character 𝝌 modulo k, we have L(1, 𝝌) ≠ 0.

Question 135

``` Formula for ∑ log p / p p≤x p≡h mod k if k > 0 and (h, k) = 1 ```

Answer 135

If k > 0 and (h, k) = 1, we have for all x > 1, ∑ log p / p = (1 / 𝜙(k)) log x + O(1) p≤x p≡h mod k where the sum is extended over those primes p ≤ x which are congruent to h mod k.

Question 136

``` Formula for ∑ log p / p p≤x p≡h mod k for all x > 1 in terms of characters ```

Answer 136

For x > 1, we have ∑ log p / p = (1 / 𝜙(k)) log x p≤x 𝜙(k)-1 _ p≡h mod k + (1 / 𝜙(k)) ∑ 𝝌ᵣ(h) ∑ 𝝌ᵣ(p) log p / p r=1 p≤x + O(1)

Question 137

Formula for ∑ log p / p p≤x for all x > 1

Answer 137

For each x > 1 and 𝝌 ≠ 𝝌₀, we have ∑ 𝝌(p) log p / p = - L'(1, 𝝌) ∑ 𝜇(n) 𝝌(n) / n + O(1) p≤x p≤x

Question 138

Size of L(1, 𝝌) ∑ 𝜇(n) 𝝌(n) / n n≤x

Answer 138

For x > 1 and 𝝌 ≠ 𝝌₀, we have L(1, 𝝌) ∑ 𝜇(n) 𝝌(n) / n ≪ 1 n≤x

Question 139

Formula for L'(1, 𝝌) ∑ 𝜇(n) 𝝌(n) / n n≤x

Answer 139

If 𝝌 ≠ 𝝌₀ and L(1, 𝝌) = 0, we have L'(1, 𝝌) ∑ 𝜇(n) 𝝌(n) / n = log x + O(1) n≤x

Question 140

``` Formula for ∑ log p / p p≤x p≡h mod k for all x > 1 in terms of number of characters ```

Answer 140

Let N(k) denote the number of non-principal characters 𝝌 modulo k such that L(1, 𝝌) = 0. For x > 1, we have ∑ log p / p = ( (1 - N(k)) / 𝜙(k) ) log x + O(1) p≤x p≡h mod k

Question 141

Prime Number Theorem for arithmetic progressions

Answer 141

For (a, k) = 1, we have 𝜋ₐ(x) ∼ 𝜋(x) / 𝜙(k) ∼ (1 / 𝜙(k)) (x / log x) as x ⟶ ∞.

Question 142

Alternative formulation for the PNT for arithmetic progressions

Answer 142

If the relation 𝜋ₐ(x) ∼ 𝜋(x) / 𝜙(k) as x ⟶ ∞ holds for every integer a relatively prime to k, then 𝜋ₐ(x) ∼ 𝜋_b(x) as x ⟶ ∞ whenever (a, k) = (b, k) = 1. The converse is also true.

Question 143

Size of integral of a continuous function on a contour

Answer 143

Consider a contour 𝛾 and a continuous function f defined on 𝛾. Suppose that |f(z)| ≤ M for all z on 𝛾. Then | ∫ f(z) dz | ≤ ML 𝛾 where L is the length of 𝛾.

Question 144

Cauchy's Integral Formula

Answer 144

Let U be a domain. Let 𝛾 be a positively oriented simple closed contour with its image and interior lying entirely within U. Suppose that a is a point in the interior of 𝛾. If f is holomorphic on U, then f(a) = 1 / 2𝜋i ∫ f(z) / (z - a) dz 𝛾

Question 145

Laurent's Theorem

Answer 145

If f is holomorphic on an annulus A = {z ∈ C : R < |z - a| < S} for 0 ≤ R < s. Then there exists a sequence (bₙ), n ∈ Z, of complex numbers such that ∞ f(z) = ∑ bₙ(z - a)ⁿ n=-∞ for every z ∈ A. Such a series is referred to as a Laurent series. Moreover for any r with R < r < S and for any n ∈ Z, if 𝛾 is a circular closed contour with centre a and radius r, then bₙ = 1 / 2𝜋i ∫ f(w) / (w - a)ⁿ⁺¹ dw 𝛾

Question 146

Cauchy's Residue Theorem

Answer 146

If 𝛾 is a closed contour traversed anticlockwise, if f is a function which is holomorphic on a domain containing the image and interior of 𝛾, except for a finite number of isolated singularities on the interior (call them a₁, ..., aₖ), then k ∫ f(z) dz = 2𝜋i ∑ Res(f, a) 𝛾 j=1