A Proof of the Riemann's Hypothesis

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A Proof of the Riemann’s Hypothesis

Charaf Ech-Chatbi

To cite this version:

Charaf Ech-Chatbi. A Proof of the Riemann’s Hypothesis. 2019. �hal-02139903v2�

A Proof of the Riemann’s Hypothesis

Charaf ECH-CHATBI ^∗ Tuesday 14 May 2019

Abstract

We present a proof of the Riemann’s Zeta Hypothesis, based on asymp- totic expansions and operations on series. We use the symmetry property presented by Riemann’s functional equation to extend the proof to the whole set of complex numbersC. The advantage of our method is that it only uses undergraduate maths which makes it accessible to a wider audience.

Keywords: Riemann Hypothesis; Riemann zeta function; Zeta Zeros;

Asymptotic distribution of the Prime Numbers; Millennium Problems.

1 The Riemann Hypothesis

The prime number theorem determines the average distribution of the primes. The Riemann hypothesis tells us about the deviation from the average. Formulated in Riemann’s 1859 paper, it asserts that all the ’non- obvious’ zeros of the zeta function are complex numbers with real part 1/2.

2 Riemann Zeta Function

For a complex number s, the Zeta function is defined as the following infinite sum:

ζ(s) =











Z¹(s) =P+∞

n=1 1

n^s, if<(s)>1

1 1− ¹

2s−1

Z²(s),whereZ²(s) =P+∞

n=1 (−1)ⁿ⁺¹

n^s if

<(s)>0, s6= 1 + _ln(2)^2kπ,kinteger

Z²0(s)

ln(2) , ifs= 1 +_ln(2)^2kπ,knonzero integer

(1) Riemann extended the domain of definition of the zeta function to the whole complex domain using the analytical continuation. The Dirichlet

∗Maybank. One Raffles Quay, North Tower Level 35. 048583 Singapore. Email:

charaf@melix.net or charaf.chatbi@gmail.com. Lnikedin/Twitter: @sharafmouzak. The opin- ions of this article are those of the author and do not reflect in any way the views or business of his employer.

eta function η(s) = Z²(s) is convergent when <(s) > 0 and it is used as the analytical continuation of the zeta function on the domain where

<(s)>0. When<(s)≤0, the Riemann zeta functional equation is used as the analytical continuation for the zeta function.

In his famous paper, Riemann[3] has shown that the Zeta has a func- tional equationζ(1−s) = 2^1−sπ^−scos ^πs₂

Γ(s)ζ(s). Using this last one, Riemann has shown that the non trivial zeros of ζ are located sym- metrically with respect to the line <(s) = 1/2, inside the critical strip 0<<(s)<1. Riemann has stated that all these non trivial zeros are very likely located on the critical line<(s) = 1/2 itself. In 1896, Hadamard and De la Vall´ee Poussin independently proved thatζ(s) has no zeros of the forms= 1 +it fort∈R.

Therefore, to prove this “Riemann Hypothesis” (RH), it is sufficient to prove thatζ has no zero on the right hand side 1/2<<(s)<1 of the critical strip.

3 Proof of the Riemann Hypothesis

Let’s take a complex number s suchs =a0+ib0. Unless we explicitly mention otherwise, let’s suppose thata0>0 andb0>0.

3.1 Case One: ¹₂ < a₀ ≤1

In this casesis a zero ofZ²(s) =P+∞

n=1 (−1)ⁿ⁺¹

n^s . We are going to develop the sequenceZN¹(s) =PN

n=1 1

n^s as follows:

ForN≥1

ZN¹ =

N

X

n=1

1

n^s (2)

=

N

X

n=1

1

n^a0+ib0 (3)

=

N

X

n=1

n^−ib0

n^a0 (4)

=

N

X

n=1

exp −ib0ln (n)

n^a0 (5)

=

N

X

n=1

cos(b0ln (n))−isin(b0ln (n))

n^a0 (6)

=

N

X

n=1

cos(b0ln (n)) n^a0 −i

+∞

X

n=1

sin(b0ln (n))

n^a0 (7)

(8)

Let’s define the sequencesUn,Vnas follows: Forn≥1 Un = cos(b0ln (n))

n^a0 (9)

Vn = sin(b0ln (n))

n^a0 (10)

Let’s define the seriesAn,BnandZnas follows:

An =

n

X

k=1

cos(b0ln (k))

k^a0 (11)

Bn =

n

X

k=1

sin(b0ln (k))

k^a0 (12)

Zn¹ = An−iBn (13)

When we are dealing with complex numbers, it is always insightful to work with the norm. So let’s develop further the squared norm of the serieZN¹ as follows:

Z_N¹

2 = A²_N+B_N² (14)

= N

X

n=1

cos(b0ln (n)) n^a0

² +

N

X

n=1

sin(b0ln (n)) n^a0

² (15) So

A²N = N

X

n=1

cos(b0ln (n)) n^a0

²

(16)

=

N

X

n=1

cos²(b0ln (n)) n^2a0 + 2

N

X

n=1 n−1

X

k=1

cos(b0ln (k))cos(b0ln (n))

k^a0n^a0 (17)

=

N

X

n=1

cos²(b0ln (n)) n^2a0 + 2

N

X

n=1

cos(b0ln (n)) n^a0

n−1

X

k=1

cos(b0ln (k))

k^a0 (18)

=

N

X

n=1

cos²(b0ln (n)) n^2a0 + 2

N

X

n=1

cos(b0ln (n)) n^a0

An−cos(b0ln (n)) n^a0

(19)

= −

N

X

n=1

cos²(b0ln (n)) n^2a0 + 2

N

X

n=1

cos(b0ln (n))An

n^a0 (20)

And the same calculation forBN

BN² = ^N

X

n=1

sin(b0ln (n)) n^a0

²

(21)

=

N

X

n=1

sin²(b0ln (n)) n^2a0 + 2

N

X

n=1 n−1

X

k=1

sin(b0ln (k))sin(b0ln (n))

k^a0n^a0 (22)

=

N

X

n=1

sin²(b0ln (n)) n^2a0 + 2

N

X

n=1

sin(b0ln (n)) n^a0

n−1

X

k=1

sin(b0ln (k))

k^a0 (23)

=

N

X

n=1

sin²(b0ln (n)) n^2a0 + 2

N

X

n=1

sin(b0ln (n)) n^a0

Bn−sin(b0ln (n)) n^a0

(24)

= −

N

X

n=1

sin²(b0ln (n)) n^2a0 + 2

N

X

n=1

sin(b0ln (n))Bn

n^a0 (25)

Hence we have the new expression of square norm ofZN¹:

Z_N¹

2= 2

N

X

n=1

cos(b0ln (n))An

n^a0 + 2

N

X

n=1

sin(b0ln (n))Bn

n^a0 −

N

X

n=1

sin²(b0ln (n)) + cos²(b0ln (n))

n^2a0 (26)

which simplies even further to:

ZN¹

2= 2

N

X

n=1

cos(b0ln (n))An

n^a0 + 2

N

X

n=1

sin(b0ln (n))Bn

n^a0 −

N

X

n=1

1 n^2a0 (27) Let’s now defineFn andGn as follows:

Fn = cos(b0ln (n))An

n^a0 (28)

Gn = sin(b0ln (n))Bn

n^a0 (29)

Therefore

A²N = 2

N

X

n=1

Fn−

N

X

n=1

cos²(b0ln (n))

n^2a0 (30)

B_N² = 2

N

X

n=1

Gn−

N

X

n=1

sin²(b0ln (n))

n^2a0 (31)

We will use the formula (13) and writeZ_N² as the following:

ZN²(s) = ZN¹(s)− 1 2^s−1Z¹N

2(s) (32)

= AN−iBN− 1 2^s−1

AN

2 −iBN 2

(33)

= AN− 1 2^s−1AN

2 −i

BN− 1 2^s−1BN

2

(34)

= AN,N−iBN,N (35)

Where

AN,N =

N 2

X

k=1

cos(b0ln (k))−2^1−a0cos(b0 ln (2k))

k^a0 +

N

X

k=^N₂+1

cos(b0ln (k)) k^a0 (36) And

BN,N =

N 2

X

k=1

sin(b0ln (k))−2^1−a0sin(b0 ln (2k))

k^a0 +

N

X

k=^N₂+1

sin(b0ln (k)) k^a0 (37) We note here that we are using the same partial sum of the Dirichlet se- rieη(s) =P

n≥1 (−1)ⁿ⁺¹

n^s that is conditionally convergent when<(s)>0.

We are not changing the partial sum by rearraging the terms of the Dirich- let serie. The representationZ_N²(s) = AN,N −iBN,N is conserving the same terms of the original partial sum of the Dirichlet seriePN

n=1 (−1)ⁿ⁺¹

n^s . Therefore,AN,N and−BN,N are the real part and the imaginary part of the partial sum of the Dirichlet serie. Hence

N→∞lim AN,N−iBN,N = lim

N→∞

N

X

n=1

(−1)ⁿ⁺¹

n^s =Z²(s) =η(s) (38) Therefore

Conclusion. sis azetazero,ζ(s) = 0, if and only if

N→∞lim AN,N = 0and lim

N→∞BN,N = 0 (39)

Equally,sis azetazero, ζ(s) = 0, if and only if

N→∞lim A²N,N = 0and lim

N→∞BN,N² = 0 (40)

Remark. The rearragement used in the Riemann rearrangement theorem to make a conditionally convergent serie converge to any number including {+∞,−∞} is not conserving the same terms of the partial sum of the original series.

For simplification, let’s define the following sequences: ForN≥1:

Un,N=

(cos(b0ln (n))−2^1−a0cos(b0(ln (2n)))

n^a0 , ifn≤ ^N₂

cos(b₀ln (n))

n^a0 ifn > ^N₂ (41)

And

Vn,N =

(sin(b0ln (n))−2^1−a0sin(b0(ln (2n)))

n^a0 , ifn≤ ^N₂

sin(b₀ln (n))

n^a0 ifn > ^N₂ (42)

And

An,N =









 Pn

k=1

cos(b₀ln (k))−2^1−an0cos(b₀ln (2k))

k^a0 , ifn≤ ^N₂

P^N₂

k=1

cos(b₀ln (k))−2^1−a0cos(b₀ln (2k))

k^a0 +Pn

k=^N₂+1

cos(b₀ln (k))

k^a0 ifn > ^N₂

AN,N ifn≥N

(43)

And

Bn,N =









 Pn

k=1

sin(b₀ln (k))−2^1−a0sin(b₀ln (2k))

k^a0 , ifn≤^N₂

P^N₂

k=1

sin(b₀ln (k))−2^1−a0sin(b₀ln (2k))

k^a0 +Pn

k=^N₂+1

sin(b₀ln (k))

k^a0 ifn >^N₂

BN,N ifn≥N

(44) Let’s develop further the squared norm of the serieZ_N² as follows:

ZN²

2=A²N,N+BN,N² (45)

= ^N

X

n=1

Un,N

² +

^N X

n=1

Vn,N

²

(46) So

A²_N,N = N

X

n=1

Un,N

²

(47)

=

N

X

n=1

Un,N+ 2

N

X

n=1 n−1

X

k=1

UnUk,N (48)

=

N

X

n=1

U_n,N² + 2

N

X

n=1

Un,N n−1

X

k=1

Uk,N (49)

=

N

X

n=1

U_n,N² + 2

N

X

n=1

Un,N

An,N−Un,N

(50)

= −

N

X

n=1

Un,N² + 2

N

X

n=1

Un,NAn,N (51)

(52) And the same calculation forBN,N

B_N,N² = ^N

X

n=1

Vn

²

(53)

=

N

X

n=1

Vn,N+ 2

N

X

n=1 n−1

X

k=1

VnVk,N (54)

=

N

X

n=1

V_n,N² + 2

N

X

n=1

Vn,N n−1

X

k=1

Vk,N (55)

=

N

X

n=1

V_n,N² + 2

N

X

n=1

Vn,N

Bn,N−Vn,N

(56)

= −

N

X

n=1

V_n,N² + 2

N

X

n=1

Vn,NBn,N (57) (58)

Let’s now define the sequences (Fn,N) and (Gn,N) as follows:

Fn,N = Un,NAn,N (59)

Gn,N = Vn,NBn,N (60)

Hence we have the new expression of square norm ofZ_N²:

ZN²

2= 2

N

X

n=1

Fn,N+ 2

N

X

n=1

Gn,N−

N

X

n=1

Un,N² +Vn,N²

(61) which simplies even further to:

N

X

n=1

Un,N² +Vn,N²

=

N 2

X

n=1

Un,N² +Vn,N²

+

N

X

n=^N₂+1

Un,N² +Vn,N²

(62)

=

N

X

n=^N₂+1

1 n^2a0 (63)

+

N 2

X

n=1

cos(b0ln (n))−2^1−a0cos(b0(ln (2n)))2

+

sin(b0ln (n))−2^1−a0sin(b0(ln (2n)))2

n^2a0 (64)

=

N

X

n=^N₂+1

1 n^2a0 (65)

+

N 2

X

n=1

1 + 2^2−2a0−2^2−a0cos(b0ln (n))cos(b0(ln (2n)))−2^2−a0sin(b0ln (n))sin(b0(ln (2n)))

n^2a0 (66)

=

N

X

n=1

1 n^2a0 (67) +

N 2

X

n=1

2^2−2a0−2^2−a0

cos(b0ln (n))cos(b0(ln (2n))) + sin(b0ln (n))sin(b0(ln (2n)))

n^2a0 (68)

=

N

X

n=1

1 n^2a0 +

N 2

X

n=1

2^2−2a0−2^2−a0

cos(b0ln (2))

n^2a0 (69)

(70) We have 2a0>1, hence the partial sumsPN

n≥1U_n,N² andPN n≥1V_n,N² converge absolutely.

We have

A²N,N = 2

N

X

n=1

Fn,N−

N

X

n=1

Un,N² (71)

B²_N,N = 2

N

X

n=1

Gn,N−

N

X

n=1

V_n,N² (72)

Proof Strategy. The idea is to prove that in the case of a complexsthat is in the right hand side of the critical strip ¹₂ < a0 ≤1 and that is aζ zero, that the limitlimn→∞A²n,n= +/− ∞ORthe limitlimn→∞Bn,n² = +/− ∞. This will create a contradiction. Because ifsis aζzero then the limn→∞A²n,n should be0and thelimn→∞Bn,n² should be0. And therefore the sequences(PN

n=1Fn,N)N≥1and(PN

n=1Gn,N)N≥1should converge and their limits should be: limn→∞PN

n=1Fn,N = ¹₂limN→∞PN

n=1U_n,N² <

+∞andlimn→∞PN

n=1Gn,N= ¹₂limN→∞PN

n=1V_n,N² <+∞.

We need the following lemmas during different stages of the proof.

Lemma 3.1. There is exist two constantsK1≥0andK2≥0such that forn≥1andN≥1we have: |An,N| ≤K1+K2n^1−a0.

Proof. We will make use of the lemma 3.5 in [1]. Let’s N ≥ 1. Let’s n≥1.

Case: n≤ ^N₂

An,N=

n

X

k=1

cos(b0ln (n))−2^1−a0cos(b0(ln (2n)))

k^a0 (73)

=

n

X

k=1

1−2^1−a0cos(b0ln(2))

cos(b0ln (k)) + 2^1−a0sin(b0ln(2))

sin(b0ln (k))

k^a0 (74)

= 1−2^1−a0cos(b0ln(2))

n

X

k=1

cos(b0ln (k))

k^a0 + 2^1−a0sin(b0ln(2))

n

X

k=1

sin(b0ln (k)) k^a0 (75)

(76) Therefore using the lemma 3.5 in [1]:

|An,N| ≤

1−2^1−a0cos(b0ln(2))

n

X

k=1

cos(b0ln (k)) k^a0

(77)

+

2^1−a0sin(b0ln(2))

n

X

k=1

sin(b0ln (k)) k^a0

(78)

≤

1−2^1−a0cos(b0ln(2))

K1+K2n^1−a0

+

2^1−a0sin(b0ln(2))

K1+K2n^1−a0

(79)

≤

1−2^1−a0cos(b0ln(2)) +

2^1−a0sin(b0ln(2))

K1+K2n^1−a0 (80) And this prove the case.

Case: n > ^N₂ We have:

n

X

k=^N₂+1

cos(b0ln (n)) k^a0 =

n

X

k=1

cos(b0ln (n)) k^a0 −

N 2

X

k=1

cos(b0ln (n))

k^a0 (81) (82)

And An,N =

N 2

X

k=1

cos(b0ln (n))−2^1−a0cos(b0(ln (2n)))

k^a0 +

n

X

N 2+1

cos(b0ln (n)) k^a0 (83)

(84) Therefore using the first case ofn≤ ^N₂ and the lemma 3.5 in [1]:

N 2

X

k=1

cos(b0ln (n))−2^1−a0cos(b0(ln (2n))) k^a0

≤

K1+K2(N 2)^1−a0

(85)

≤

K1+K2n^1−a0 , as (N

2)^1−a0 ≤n^1−a0 (86) And

n

X

k=1

cos(b0ln (n)) k^a0 −

N 2

X

k=1

cos(b0ln (n)) k^a0

≤

n

X

k=1

cos(b0ln (n)) k^a0

+

N 2

X

k=1

cos(b0ln (n)) k^a0

(87)

≤

K1+ K2n^1−a0 +

K1+K2(N 2)^1−a0

(88)

≤K1+ K1+ (K2+ K2)n^1−a0, as (N

2)^1−a0≤n^1−a0 (89) Therefore

|An,N| ≤K1+ K1+ (K2+ K2)n^1−a0 (90) Which proves our case.

Lemma 3.2. For a large n, N and N0 such N0 ≤n < n+ 1≤ ^N₂ We can write An,N = E1(n) +γn such that limn→+∞γn = 0 and E1(n) = n^1−a0

X1sin b0ln (n)

+X2cos b0ln (n)

where X1 and X2 two real constants defined as follows:

X1 =

a3b0+b3(1−a0)

(b0)²+ (1−a0)² (91) X2 =

a3(1−a0)−b3b0

(b0)²+ (1−a0)² (92) a3 = 1−2^1−a0cos(b0ln (2)) (93) b3 = 2^1−a0sin(b0ln (2)) (94) Proof. We have limN→+∞AN,N = 0. Let’s >0. Let’sN0 be such that for each N ≥N0: |AN,N| ≤. Let’s take n, N and N0 suchN0 ≤n <

n+ 1≤^N₂. We have|AN,N| ≤. Therefore:

We have the notationddn= cos(b0ln (n))−2^1−a0cos(b0(ln (2n))) . We are going to use the lemma 3.2 in [1]. Let’s develop further:

AN,N−An,N =

N 2

X

k=n+1

ddk

k^a0 +

N

X

k=^N₂+1

cos(b0ln (k))

k^a0 (95)

=

N 2

X

k=n+1

ddk

k^a0 −

N 2

X

k=n+1

Z k+1 k

e1(t)dt+

N

X

k=^N₂+1

cos(b0ln (k)) k^a0 −

N

X

k=^N₂+1

Zk+1 k

e(t)dt (96)

+

N 2

X

k=n+1

Z k+1 k

e1(t)dt+

N

X

k=^N₂+1

Zk+1 k

e(t)dt (97)

=

N 2

X

k=n+1

ddk

k^a0 −

N 2

X

k=n+1

Z k+1 k

e1(t)dt+

N

X

k=^N₂+1

cos(b0ln (k)) k^a0 −

N

X

k=^N₂+1

Zk+1 k

e(t)dt (98)

+ Z N

n+1

e1(t)dt+ Z N

N 2+1

e(t)dt (99)

=−E1(n+ 1) +

N 2

X

k=n+1

ddk

k^a0 −

N 2

X

k=n+1

Z k+1 k

e1(t)dt+

N

X

k=^N₂+1

cos(b0ln (k)) k^a0 −

N

X

k=^N₂+1

Zk+1 k

e(t)dt (100)

+E1(N) +E(N)−E(N

2 + 1) (101) Therefore from the lemma 3.2 in [1]:

|An,N−E1(n+ 1)| ≤ |AN,N|+

N 2

X

k=n+1

ddk

k^a0 −

N 2

X

k=n+1

Z k+1 k

h(t)dt

(102)

+

N

X

k=^N₂+1

cos(b0ln (k)) k^a0 −

N

X

k=^N₂+1

Z k+1 k

g(t)dt

+

E1(N) +E(N)−E(N 2 + 1)

(103)

≤+K5

a0

1 n^a0 − 1

(^N₂)^a0

+K1

a0

1

(^N₂)^a0 − 1 (N^a0)

+

E1(N) +E(N)−E(N 2 + 1)

(104) (105) We haveN0 ≤n < n+ 1≤^N₂. So

|An,N−E1(n+ 1)| ≤+K5

a0

1 n^a0 + 1

n^a0

+K1

a0

1 n^a0 + 1

n^a0

+

E1(N) +E(N)−E(N 2 + 1)

(106)

≤+ 2K1+K5

a0

1 n^a0 +

E1(N) +E(N)−E(N 2 + 1)

(107) Quick calculation gives us: E1(N) +E(N)−E(^N₂) = 0. We apply the

lemma 3.8 in [1] to the functionE, we have limN→+∞E(^N₂+ 1)−E(^N₂) = 0. Therefore: limN→+∞E1(N) +E(N)−E(^N₂) = 0.

|An,N−E1(n+ 1)| ≤+ 2K1+K5

a0

1 n^a0 +

E1(N) +E(N)−E(N 2 + 1)

(108)