Abdeljalil Saghe
email: saghe007@gmail.com

Nonstandard analysis : The black magic matrix with the field of Omicran-reals

Keywords: Nonstadard analysis, Omicran-reals, Abraham Robinson, Bernoulli numbers, Euler-Maclaurin, Riemann zeta function, hyppereals, infinitesimal

teaching and research institutions in France or abroad, or from public or private research centers.

Introduction

This paper is organized as follows, we restart in the first section of preliminary results by some notions for define the field of Omicran-reals [START_REF] Saghe | The field of Omicran-reals A new approach to nonstandard analysis[END_REF], in the next section we define the exact limit, and we determine the relationship between the limit and the exact limit of a real sequence, we present some examples and we give necessary conditions for the existence of the exact limit in the section 5. In the object to calculate the exact limit of a series, we define the beautiful matrix ψ (n) which is called the black magic matrix, by using the coefficients of this matrix we determine the Bernoulli numbers in the section 10. Finally, we obtain the standard Euler-Maclaurin [START_REF] Borwein | Computational Strategies for the Riemann Zeta Function[END_REF] formula which give us an expansion of the Riemann zeta function ζ(s).

Preliminary results

• The binomial coefficient is defined as :

C k n = n! k!(n-k)! . • The Bernoulli numbers are given below :              B 0 = 1 B 0 + 2B 1 = 0 B 0 + 3B 1 + 3B 2 = 0 B 0 + 4B 1 + 6B 2 + 4B 3 = 0 . . . . . . . . . B 0 + C 1 n B 1 + .... + C n-1 n B n-1 = 0
We can verify that B 2k+1 = 0, for every natural k ≥ 1.

• Stirling's formula :

n! ∼ √ 2πn( n e ) n .
• An important result of the Stirling's formula is given by :

| B 2n |∼ 4 √ πn( n πe ) 2n .
Definition 1 Let u be a map from ]0, 1] to R, such that :

• There exists a map u is defined on D ′ (0, 1), and holomorphic in a neighborhood of 0.

• There exists ε > 0, such that ∀x ∈]0, ε[ we have a : u(x) = u(x).

The map u is called a metalic map, and u is a metalic extension of u.

Notations :

We note :

• ∆ 1 = { u, u is a metalic map }.

• A 2 = { 1 u , u ∈ ∆ 1 ∀x ∈]0, 1] u(x) = 0 and lim u(

1 n ) = 0 }.
• ∆ 2 = {v / v :]0, 1] -→ R , there exists 1/u ∈ A 2 and ε > 0 such that v /]0,ε] = ( 1 u

) /]0,ε] }.

• ∆ = ∆ 1 ∪ ∆ 2 , (∆, +, .) is a unitary ring.

• Let I 0 the set of maps defined in ]0, 1] and zero on ]0, ε] ( for 0 < ε ≤ 1 ).

The set I 0 is a maximal ideal of ∆, then ∆/I 0 is a field.

• We define ∆( 1 n ) = { h( 1 n ) n≥1 , h ∈ ∆ }. • Let ∼ the equivalence relation on ∆( 1 n ) defined by : g( 1 n ) n≥1 ∼ h( 1 n ) n≥1 ⇔ ∃n 0 / ∀n ≥ n 0 , h( 1 n ) = g( 1 n
).

• The equivalence class of (f

( 1 n )) n≥1 modulo ∼ is noted (f ( 1 n )) n≥1 .
• Let ∆( 1 n ) the set of the equivalence classes (f ( 1 n )) n≥1 . • Let ∆ 0 = ∆/I 0 , and g the equivalence class of g.

If it has not ambiguity, we replace g by g.

• The field of Omicran-reals is defined as follows [START_REF] Saghe | The field of Omicran-reals A new approach to nonstandard analysis[END_REF] :

O = ∆ 0 (δ) = { g(δ), g ∈ ∆ 0 },
where δ is an indeterminate.

Remark 1

• We define on O the following total order :

g(δ) ≤ h(δ) ⇔ g( 1 n ) ≤ h( 1 n ) f rom a certain rank n 0 .
• The field of Omicran-reals is an extension of R, in addition, the total order relation defined on O extends the usual order relation ≤.

• The number δ is infinitesimal [START_REF] Saghe | The field of Omicran-reals A new approach to nonstandard analysis[END_REF]. ( we verify that 0 < δ < ε for every real ε > 0 )

3 The exact limit Proposition 1 The map ϕ defined as :

ϕ : (∆( 1 n ), +, .) -→ (O, +, .) (g( 1 n )) n≥1 -→ g(δ) is a ring isomorphism.
If we want to define a new concept more precise than the limit that allows to give the value taken by the sequence (f ( 1 n )) n≥1 on ad infinitum, then this concept (called exact limit) is dependent on the values taken by (f ( 1 n )) n≥1 from a certain rank n 0 . Intuitively, the equivalence class (f ( 1 n )) n≥1 is a only concept can give these values independently n 0 . On other hand, if f is an element of ∆, then we can identify the equivalence class (f ( 1 n )) n≥1 by f (δ) from the above proposition, so, we deduce that we can define the new concept as follow :

Definition 2 Let f ∈ ∆. The Omicran-real ϕ(f ( 1 n ) n≥1 ) = f (δ) is called the exact limit of the sequence (f ( 1 n )) n≥1 . We note : lim exact f ( 1 n ) = f (δ).
Remark 2 We remark that lim exact = ϕ • s, where s is a canonical surjection defined as :

s : (∆( 1 n ), +, .) -→ (∆( 1 n ), +, .) (f ( 1 n )) n≥1 -→ (f ( 1 n )) n≥1
.

Definition 3 (The projection) Let f a metalic function, and x ∈ O such that x = f (δ).

If we find an element

x * ∈ R such that | x -x * | ≤ | x -y |, ∀y ∈ R, then the real x * is called the projection of x onto R. Remark 3 The distance from x to R is d R (x) = inf y∈R | x -y |=| x -x * |. Example 1 • δ * = 0. • ( 1 δ 2 + 1 ) * = 1.
Theorem 1 Let f a metalic function, and

x ∈ O / x = f (δ).
The projection x * of x onto R exist and unique, in addition :

x * = lim n→+∞ f ( 1 n ).
For more details see [START_REF] Saghe | Nonstandard analysis : The field of Omicran-reals and its applications[END_REF].

Theorem 2 Let f the metalic map, and x = f (δ).

Then : ∀ε > 0 real, we have :

| x -x * |≤ ε
x * is a unique element of R which verify this property.

For more details see [START_REF] Saghe | Nonstandard analysis : The field of Omicran-reals and its applications[END_REF].

Theorem 3 (of the exact limit)

If f ∈ ∆ 1 , then the real lim f ( 1 n ) is the projection of lim exact f ( 1 n
) onto R, so we get :

( lim exact f ( 1 n )) * = lim f ( 1 n ).
For more details see [START_REF] Saghe | Nonstandard analysis : The field of Omicran-reals and its applications[END_REF].

4 The calcul of the exact limit

To calcul the exact limit of the sequence

(x n ), it's sufficient to find a function f ∈ ∆, such that f ( 1 n ) =
x n , from a certain rank , and we get lim

exact x n = f (δ). If lim x n = l,
where l is a real number, then f is a metalic function, in the other case, we have f ∈ ∆ 2 . We assume that the f is a metalic function. (if not, we consider the function defined on the interval ]0, ε[ by g

(x) = 1 f (x) ) Example 2 • Let x n = 1 n(2n-1)
, consider f the function on ]0, 1]defined as :

f (x) = 1, while x ∈ [ 1 2 , 1]; x 2 2-x , while x ∈]0, 1 2 [.

,

we can verify that f ( 1 n ) = x n , from a certain rank. Let f be a map defined on D ′ (0, 1) as

f (z) = z 2 2-z , while z ∈ D(0, ε) 1,
else.

, where 0 < ε << 1, we can verify that f is a holomorphic function on a neighborhood of 0, and f (x) = f (x), for every x ∈]0, ε[.

Then f is an element of ∆ 1 , and we can verify the exact limit of (x n ), we get :

lim exact x n = f (δ) = δ 2 2 -δ . • lim exact sin( 1 n ) = sin(δ), for f (x) = sin(x) and f (z) = sin(z).
• lim exact sin(2πn) = 0, for f (x) = 0 and f (z) = 0.

We can't choose f (x) = sin( 2π x ), because the function x -→ sin( 2π x ) is not element of ∆. • We can verify that there does not exist an element f ∈ ∆, such that f ( 1 n ) = (-1) n from a certain rank, then we can't define the exact limit lim exact (-1) n .

• Generally, from the proprieties of the elements of ∆, we can show that if (x n ) does not keep a constant sign from a certain rank, then this sequence not admits a exact limit, for example, if x n = (-1) n n , then lim x n = 0, but we can't define the exact limit of (x n ).

5 Necessary conditions for the existence of the exact limit

• Let (x n ) n≥1 be a sequence of real numbers, we assume that the exact limit of (x n ) n≥1 exists, then there exists a function f ∈ ∆ such that lim

exact x n = f (δ). If f ∈ ∆ 1 , then f is a metalic function, let f be a metalic extension of f , we have f ( 1 n ) = f ( 1 n ) =
x n from a certain rank, since f is holomorphic at 0, then the limit of (x n ) exists and we have lim x n = f (0), finally, we deduce that the existence of the exact limit implies the existence of the limit, and lim x n = f (0). Generally, we get :

lim exact x n = f (δ) =⇒ lim x n = f (0), while f ∈ ∆ 1 ; ±∞, while f ∈ ∆ 2 .
• Generally, the reciprocal of the above implication is not true, we can find a convergent sequence which does not have a exact limit (for example :

x n = (-1) n n ).
• If a sequence (x n ) has a exact limit, then (x n ) keeps a constant sign from a certain rank.

In addition, if x n > 0 from a certain rank, then lim exact

x n > 0.

• If a sequence (x n ) has a exact limit, from the proprities of the elements of ∆, we can show that the sequence (x n+1 -x n ) keeps a constant sign from a certain rank.

Theorem 4 Let (a n ) be a real sequence, and f is a holomorphic function on D(0, ε) -{0} such that :

• f (]0, ε[) ⊆ R. • f ( 1 n ) = a n from a certain rank. • f is bounded on D(0, ε) -{0}.
Then the sequence (a n ) has a exact limit, and we have :

lim exact a n = f (δ).
Proof. 0 is an artificial sigularity of f .

The case of a series

Let s n = n k=1 a k is a convergent series, where (a k ) is a sequence of real numbers, to calculate the exact limit of (s n ) it's sufficient to find a holomorphic function f in a neighborhood of 0 such that: On other hand :

               f ( 1 n0 ) = s n0 = a 1 + a 2 + ... + a n0 f ( 1 n1 ) = s n1 = a 1 + a 2 + ... + a n0 + a n1 f ( 1 n2 ) = s n2 = a 1 + a 2 + ... + a n0 + a n1 + a n2 . . . . . . . . . f ( 1 n ) = s n = a 1 + a 2 +
( * )                f ( 1 n0 ) = s n0 = α + a n0 f ( 1 n1 ) = s n1 = α + a n0 + a n1 f ( 1 n2 ) = s n2 = α + a n0 + a n1 + a n2 . . . . . . . . . f ( 1 n ) = s n = α + a n0 +
lim exact s n = f (δ), we note : f (δ) = lim exact n k=1 a k = 1 δ k=1 a k = ℵ0 k=1 a k . The system (*) is equivalent to :         f ( 1 n0 ) f ( 1 n0+1 ) . . . f ( 1 n )         =       α α . . . α       +       1 0 0 . . 0 1 1 0 . . . . . . 0 . . . . . . . 0 1 . . . 1 0 1 1 . . 1 1             a 1 a 2 . . . a n      
Let T n be a matrix defined as :

T n =       1 0 0 . . 0 1 1 0 . . . . . . 0 . . . . . . . 0 1 . . . 1 0 1 1 . . 1 1      
Then T n is invertible, and we have :

T -1 n =       1 0 0 . . 0 -1 1 0 . . . 0 -1 . 0 . . . . . . . 0 0 . . -1 1 0 0 0 . 0 -1 1       Then :       a n0 a n0+1 . . . a n       =         f ( 1 n0 ) -f ( 1 n0 ) + f ( 1 n0+1 ) . . . -f ( 1 n-1 ) + f ( 1 n )         -       α 0 . . . 0      
We deduce that :

( * * ) a n = f ( 1 n ) -f ( 1 n -1
), f rom a certain rank n 0

We assume that lim exact a n exists, in addition the sequence (s n ) is convergent, then (a n ) converges to 0, we can find a holomorphic function on a neighborhood of 0, such that : a n = g( 1 n ) from a certain rank, in this case we have lim exact a n = g(δ). We can show that : g(0) = 0, from (**), ∃p such that ∀n ≥ p, we have :

g( 1 n ) = f ( 1 n ) -f ( 1 n -1
).

Since f and g are a holomorphic functions on a neighborhood of 0, we deduce that ∃ε > 0 such that :

g(z) = f (z) -f ( z 1 -z ), ∀z ∈ D(0, ε).
On other hand, we have f ( 1 n0 ) = α = n0-1 k=1 a k , this result is true for every p ≥ n 0 , then f (0) = +∞ k=1 a k . Finally, we deduce the following theorem :

Theorem 5 Let g be a metalic function, and (s n ) n≥1 the convergent series defined as

s n = n k=1 g( 1 k
).

If the exact limit of (s n ) n≥1 exists, then there exists a function f holomorphic at 0 such that f (δ) = lim

exact n k=1 g( 1 k
).

This function is given by :

f (0) = +∞ k=1 g( 1 k ), g(z) = f (z) -f ( z 1-z ), on a neighborhood of 0.
Remark 4 (Calculating of the finite sum)

If f (δ) = lim exact n k=1 a k , then f ( 1 n ) = n k=1 a k , from a certain rank n 0 . Example 3 • lim exact n k=1 1 k(k + 1) = 1 1 + δ .
And we have :

n k=1 1 k(k + 1) = 1 1 + 1 n , ∀n ≥ 1.
7 The calcul of the exact limit of a k

Let (s n ) be a series defined as

s n = n k=1 g( 1 k
), we assume that the series is convergent, and g is metalic, then holomorphic on a neighborhood of 0. The existence of the exact limit of (s n ) implies that there exists a holomorphic function f on a neighborhood of 0 which verify :

g(z) = f (z) -f ( z 1 -z
), on the disk D(0, ε).

Let : g(z) = +∞ n=0 β n z n , and f (z) = +∞ n=0 α n z n
, where (α n ) and (β) n are a real sequences. To determine f means to find the sequence (α n ), we get :

f (z) = α 0 + α 1 z + α 2 z 2 + ..... + α n z n + o(z n ).
And :

f ( z 1-z ) = f (z + z 2 + .... + z n + o(z n )) = α 0 + α 1 (z + ... + z n + o(z n )) + .... + α n (z + ... + z n + o(z n )) n + o(z n ) = α 0 + α 1 z + (α 1 + α 2 )z 2 + (α 1 + 2α 2 + α 3 )z 3 + (α 1 + 3α 2 + 3α 3 + α 4 )z 4 +(α 1 + 4α 2 + 6α 3 + 4α 4 + α 5 )z 5 + (α 1 + 5α 2 + 10α 3 + 10α 4 + 5α 5 + α 6 )z 6 +... + (α 1 + C 1 n-1 α 2 + C 2 n-1 α 3 + ... + C n-2 n-1 α n-1 + α n )z n + o(z n ). Since: g(z) = f (z) -f ( z 1-z )
, we deduce that :

β 0 = β 1 = 0 β k = -α 1 -C 1 k-1 α 2 -C 2 k-1 α 3 -.... -C k-2 k-1 α k-1 , ∀2 ≤ k ≤ n.
Remark 5 Let g be a holomorphic function on a neighborhood of 0, if the exact limit of the series n k=1 g( 1 k ) exists, then β 0 = β 1 = 0, which implies g(0) = g ′ (0) = 0, we deduce that the necessary condition of the existence of the exact limit of

n k=1 g( 1 k ) is : g(z) = z 2 g 1 (z)
, where g 1 is a holomorphic function on a neighborhood of 0. Now, from the above results, we deduce that :

               β 0 = β 1 = 0 β 2 = -α 1 β 3 = -α 1 -2α 2 β 4 = -α 1 -3α 2 -
β n = -α 1 -(n -1)α 2 -... -C k n-1 α k+1 -... -C n-2 n-1 α n-1 .
Then :

         β 2 β 3 β 4 . . . . β n          =          -1 0 0 . 0 -1 -2 0 . 0 -1 -3 -3 . . . . . 0 . . . . . . . . . 0 . . . . . 0 -1 -(n -1) . . . . -(n -1)                   α 1 α 2 α 3 . . . . α n-1         
Consider the matrix defined as :

M n =          -1 0 0 . 0 -1 -2 0 . 0 -1 -3 -3 . . . . . 0 . . . . . . . . . 0 . . . . . 0 -1 -n . . . . -n         
We have : det(M n ) = (-1) n n!, then M n is invertible, and we have :

         β 2 β 3 β 4 . . . . β n          = M n-1          α 1 α 2 α 3 . . . . α n-1          .
Then, the above system admits a unique solution (α 1 , α 2 , ..., α n-1 ).

If lim sup n | α n | = 1 R > 0, then the function f (z) = +∞ n=0
α n z n is holomorphic on the disk D(0, R), and the exact limit

lim exact n k=1 g( 1
k
) exists, in addition, we have :

f (δ) = lim exact n k=1 g( 1 k ) = lim exact n k=1 a k ,
we get :

( lim exact n k=1 a k ) * = +∞ k=1 a k .
We note : lim exact a n = α n δ n . Then :

( lim exact a n ) * = +∞ n=1 a n , ( α n δ n ) * = +∞ n=1 a n .
8 The black magic matrix

Let g be a metalic function and f a holomorphic function in a neighborhood of 0. We assume that the series

n k=1 g( 1 k ) admits the exact limit f (δ). Let (α n ) and (β n ) be a real sequences such that f (z) = α 0 + +∞ k=1 α k z k and g(z) = +∞ k=0 β k z k . We have : lim exact n k=1 g( 1 k ) = f (δ).
And :

         β 2 β 3 β 4 . . . . β n          = M n-1          α 1 α 2 α 3 . . . . α n-1          . Definition 4
The black magic matrix of order n is defined as

ψ (n) = M -1 n . We obtain :          α 1 α 2 α 3 . . . . α n-1          = ψ (n-1)          β 2 β 3 β 4 . . . . β n          .
The real α 0 is given by α

0 = f (0) = +∞ k=1 g( 1 k
). Remark 6 We can verify that :

m k=1 g( 1 k ) = α 0 + lim n→+∞ 1 m 1 m 2 1 m 3 ... 1 m n-1 ψ (n-1)      β 2 β 3 β 4 . . . β n      ,
from a certain rank m 0 .

9 The magical properties of ψ (n)

• Property 1. The matrix ψ (n) is given by ψ

(n) = M -1 n , where M n [i, j] = -C j-1 i , if 1 ≤ j -1 ≤ i ≤ n, 0,
else. We deduce that the matrix ψ (n) is invertible and lower triangular.

• Property 2. We have ψ (n) i,i = -1 i , then tr(ψ (n) ) = -H(n) and det(ψ (n) ) = (-1) n n!
, where (H(n)) n≥1 is the harmonic series which defined as

H(n) = n i=1 1 i .
Proof. The matrix M n is lower triangular and Sp(M n ) = {-i, f or 1 ≤ i ≤ n}.

Then : ψ (n) is lower triangular and we get Sp(ψ (n) ) = { -1 i , f or 1 ≤ i ≤ n}. • Property 3. For every 1 ≤ i ≤ n -1, we have :

ψ (n) i+1,i = 1 2 .
Proof. We have :

δ i+1,i = n k=1 M n [i + 1, k]ψ n k,i .
Then :

i+1 k=i M n [i + 1, k]ψ (n) k,i = 0.
We deduce that :

ψ (n) i+1,i = - M n [i + 1, i]ψ (n) i,i M n [i + 1, i + 1] , ψ (n) i+1,i = - C i-1 i+1 ψ (n) i,i C i i+1 , Then : ψ (n) i+1,i = 1 2 .
• Property 4. For every (m, p) ∈ N 2 , such that :2 ≤ m, and 2m + p ≤ n, we have :

ψ (n) 2m+p,1+p = 0.
In particular, for every 2 ≤ m ≤ n 2 , we get :

ψ (n)
n,n-2m+1 = 0. Proof. We can see the demonstration in the following pages.

• Property 5. For every 1 ≤ m ≤ n -1, we have :

ψ (n) m,m ψ (n) m+1,m-1 = 1 12 .
Then :

ψ (n) m+1,m-1 = -m 12 .
Proof. We have :

ψ (n) M n = I n .
Then :

n k=1 ψ (n) i,k M n [k, j] = δ ij .
In particular :

n k=1 ψ (n) m+1,k M n [k, m -1] = δ m+1,m-1 .
Then :

m+1 k=m-1 ψ (n) m+1,k M n [k, m -1] = 0.
Which implies :

ψ (n) m+1,m-1 M n [m -1, m -1] + ψ (n) m+1,m M n [m, m -1] + ψ (n) m+1,m+1 M n [m + 1, m -1] = 0.
Then :

-(m -1)ψ (n) m+1,m-1 - m(m -1) 4 + m(m -1) 6 = 0.
Finally, we get :

ψ (n) m+1,m-1 = -m 12 .
• Property 6. For every (i, j) ∈ N 2 , such that 1 ≤ i, j ≤ n, we have :

ψ (n+1) i,j = ψ (n) i,j .
Prof. From the definition of M n , we have :

M n+1 =     0 M n . . . 0 X n -(n + 1)     ,
where

X n = -(C 0 n+1 , C 1 n+1 , ..., C n-1 n+1 ). To prove that ψ (n+1) i,j = ψ (n)
i,j , it is sufficient to show that there exists a row vector Y n such that :

ψ (n+1) =     0 ψ (n)
. . .

0 Y n -1/(n + 1)     .
On other hand, we have :

M n+1 ψ (n+1) = I n+1 , then :     0 M n . . . 0 X n -(n + 1)         0 ψ (n) . . . 0 Y n -1/(n + 1)     = I n+1 , which implies :     0 M n ψ (n) . . . 0 X n ψ (n) -(n + 1)Y n 1     = I n+1 .
Finally, we deduce that :

X n ψ (n) -(n + 1)Y n = 0 n .
Then, we can choose Y n as a form Y n = X n ψ (n) n + 1 , and we get :

ψ (n+1) =       0 ψ (n) . . . 0 1 n + 1 X n ψ n -1 n+1      
.

Finally, we deduce that : ψ

( n+1) i,j = ψ (n) i,j , for every 1 ≤ i, j ≤ n.
Remark 7 From the above result, we deduce that ψ

(i) i,j = ψ (n) i,j , for every 1 ≤ i, j ≤ n. We note ψ (n) i,j = ψ i,j .
• Property 7. For every 1 < i ≤ n, we have :

Then :

n i=1 C i = C 1 + C 2 + ... + C n =     -1 0 . . . 0     .
Where C 1 , C 2 , ..., C n are the column vectors of the matrix ψ (n) .

Prof. We know that :

ψ n M n = I n , then n k=1 ψ n [i, k]M n [k, 1] = δ i1 .
We deduce that :

n k=1 ψ 1,k M n [k, 1] = 1, n k=1 ψ i,k M n [k, 1] = 0, if i = 1 . Then : n k=1 ψ 1,k = -1, n k=1 ψ i,k = 0, if i = 1.
.

• Property 8. For every 1 ≤ i ≤ n, we have :

n k=1 (-1) k ψ i,k = (-1) i+1 .
Which implies :

n i=1 (-1) i-1 C i = C 1 -C 2 + ... + (-1) n-1 C n =     -1 1 . . . (-1) n     .
Where C 1 , C 2 , ..., C n are the column vectors of the matrix ψ (n) .

Prof. From the example 3, we have lim

exact n k=1 -1 k(k + 1) = -1 1 + δ . Then : lim exact n k=1 g( 1 k ) = f (δ), for g(z) = -z 2 1 + z = +∞ k=2 (-1) k z k and f (z) = -1 1 + z = +∞ k=0 (-1) k+1 z k .
By using the Property 9, we deduce that :

         1 -1 1 . . . . (-1) n+1          = ψ (n)          -1 1 -1 . . . . (-1) n         
, finally, we deduce that :

n k=1 (-1) k ψ i,k = (-1) i+1 .
• Property 9. Let g be a metalic function, such that g(z) = +∞ k=0 β k z k on a neighborhood of 0, we assume that the series

n k=1 g( 1 k
) is convergent and admits a exact limit, then there exists a holomorphic function on a neighborhood of 0 f , such that lim

exact n k=1 g( 1 k ) = f (δ), there exits a real sequence (α n ) such that f (z) = +∞ k=0 α k z k
on a neighborhood of 0, the real α is given by :

α 0 = +∞ k=1 g( 1 k ) and 
        α 1 α 2 α 3 . . . . α n-1          = ψ (n-1)          β 2 β 3 β 4 . . . . β n         
, and we have :

β 0 = β 1 = 0.
Example 4

(n = 2) ψ (2) = -1 0 1/2 -1/2 (n = 3) ψ (3) = -1 0 0 1/2 -1/2 0 -1/6 1/2 -1/3 (n = 5) ψ (5) =     -1 0 0 0 0 1/2 -1/2 0 0 0 -1/6 1/2 -1/3 0 0 0 -1/4 1/2 -1/4 0 1/30 0 -1/3 1/2 -1/5     (n = 8) ψ (8) =           -1 0 0 0 0 0 0 0 1/2 -1/2 0 0 0 0 0 0 -1/6 1/2 -1/3 0 0 0 0 0 0 -1/4 1/2 -1/4 0 0 0 0 1/30 0 -1/3 1/2 -1/5 0 0 0 0 1/12 0 -5/12 1/2 -1/6 0 0 -1/42 0 1/6 0 -1/2 1/2 -1/7 0 0 -1/12 0 7/24 0 -7/12 1/2 -1/8           (n = 12) ψ (12) =                  -1 0 0 0 0 0 0 0 0 0 0 0 1/2 -1/2 0 0 0 0 0 0 0 0 0 0 -1/6 1/2 -1/3 0 0 0 0 0 0 0 0 0 0 -1/4 1/2 -1/4 0 0 0 0 0 0 0 0 1/30 0 -1/3 1/2 -1/5 0 0 0 0 0 0 0 0 1/12 0 -5/12 1/2 -1/6 0 0 0 0 0 0 -1/42 0 1/6 0 -1/2 1/2 -1/7 0 0 0 0 0 0 -1/12 0 7/24 0 -7/12 1/2 -1/8 0 0 0 0 1/30 0 -2/9 0 7/15 0 -2/3 1/2 -1/9 0 0 0 0 3/20 0 -1/2 0 7/10 0 -3/4 1/2 -1/10 0 0 -5/66 0 1/2 0 -1 0 1 0 -5/6 1/2 -1/11 0 0 -5/12 0 11/8 0 -11/6 0 11/8 0 -11/12 1/2 -1/12                 
Theorem 6 (To calculate the coefficients of (ψ i,j ) by induction) For every 1 ≤ j ≤ n, we have :

ψ n+1,j = X n n + 1     ψ 1,j ψ 2,j . . . ψ n,j     ,
where

X n = -(C 0 n+1 , C 1 n+1 , ..., C n-1 n+1
), and we have ψ n+1,n+1 = -1 n+1 .

Proof. There exists a row vector Y n = (y 1 , y 2 , ..., y n ), such that :

ψ (n+1) =     0 ψ (n) . . . 0 Y n -1 n+1     .
On other hand, Y n = Xnψ (n) n+1 , then :

y j = ψ n+1,j = Y n e j = X n n + 1 ψ (n) e j = X n n + 1     ψ 1,j ψ 2,j . . . ψ n,j     ,
where (e 1 , e 2 , ..., e n ) is a canonical base of R n .

Remark 8 For every n ≥ 1, we have :

X n = (0, X n-1 ) + (X n-1 , -n).
10 The relationship between ψ i,j and the Bernoulli numbers

The Bernoulli numbers are defined as :

             B 0 = 1 B 0 + 2B 1 = 0 B 0 + 3B 1 + 3B 2 = 0 B 0 + 4B 1 + 6B 2 + 4B 3 = 0 . . . . . . . . . B 0 + C 1 n B 1 + .... + C n-1 n B n-1 = 0
Then :

M n     B 0 B 1 . . . B n-1     =     -1 0 . . . 0     .
We deduce that :

ψ (n)     -1 0 . . . 0     =     B 0 B 1 . . . B n-1     .
Finally, we get the following theorem :

• Property 10. For every natural number k ≥ 1, we have :

ψ k,1 = -B k-1 .
Then, the first column of ψ (n) is given by : 

   ψ 1,1 ψ 2,1 . . . ψ n,1     = -     B 0 B 1 . . . B n-1     .
Remark 9 From the above theorem, we deduce that ψ 2k+2,1 = 0, for every natural number k ≥ 1, because B 2k+1 = 0.

• Property 11. For every k ∈ N, and s ∈ N * , we have :

ψ k+s,s = - B k k! k-1 i=1 (s + i). Proof. Let        0 . . . x s . . . x n       
be a s th column of the matrix ψ (n) . To find this column it's sufficient to determine the values of the real numbers (x i ) which verify :

-(C 0 j , C 1 j , C j-1 j , 0, ..., 0)        0 . . . x s . . . x n        = δ s,j .
On other hand, we know that B 2k+1 = 0 for every natural number k ≥ 1, then to prove that ψ s+2k+1,s = 0 it's sufficient to show that the above column has the following form :

         0 . . . x s x s+1 . . . x n          =          0 . . . α s B 0 α s+1 B 1 . . . α n B n-s          ,
where α s , α s+1 , ..., α n are a real numbers. Then :

(C 0 j , C 1 j , C j-1 j , 0, ..., 0)        0 . . . α s B 0 . . . α n B n-s        = -δ s,j .
If s = j, and s > j, this product is zero. If s < j, we find :

α s B 0 C s-1 j + α s+1 B 1 C s j + .... + α j B j-s C j-1 j = 0.
From the property 10, we know that :

C 0 j-s+1 B 0 + C 1 j-s+1 B 1 + .... + C j-s j-s+1 B j-s = 0.
To find the real numbers (α i ), it's sufficient to find λ ∈ R, such that :

α s+k B k C s+k-1 j = λC k j-s+1 B k , f or s + k ≤ j.
Then :

α s+k = λ C k j-s+1 C s+k-1 j , α s+k = λ k! (s + k -1)!(j -s + 1)! j! .
In the case of k = 0, we get :

α s = λ (s -1)!(j -s + 1)! j! .
Then : λ = j!α s (s -1)!(j -s + 1)! .

On other hand, we know that B 0 = 1, then α s = -1 s , and we get :

λ = - j! s!(j -s + 1)! .
We replace λ by his value, and we define the real (α i ) as :

α s+k = - (s + k -1)! s!k! = - 1 k! (s + 1)(s + 2)...(s + k -1).
Finally, we deduce that :

ψ k+s,s = - B k k! (s + 1)(s + 2)...(s + k -1).
Corollary 1 For every (l, m) ∈ N 2 such that m ≥ 2 et 2m ≤ l, we have :

ψ l,l-2m+1 = 0.
Remark 10 From the above results, we deduce that the Property 4 is true, and we can show the following theorem.

Theorem 7 The matrix of the black magic ψ (n) = (ψ i,j ) 1≤i,j≤n is given by :

ψ i,j = - C j i Bi-j i , if i ≥ j 0, else.
11 The black magic matrix with the Riemann zeta function

Important lemma

Lemma 1 The radius of convergence of the series

+∞ k=1 ψ k,s-1 z k , is zero.
Proof. The radius of convergence of the series

+∞ k=1 ψ k,s-1 z k , is given by : 1 R = lim k→+∞ k | ψ k,s-1 |.
We have :

ψ k,s-1 = - B k-s+1 (k -s + 1)! s(s + 1)(s + 2)...(k -1) = - B k-s+1 (k -s + 1)! (k -1)! (s -1)! .
For k = 2m + s -1, we get :

ψ 2m+s-1,s-1 = - B 2m (s -1)! (2m + s -2)! (2m)! . Since 2m! ∼ √ 4πm( 2m e ) 2m and | B 2m |∼ 4 √ πm( m πe ) 2m , then : | B 2m | 2m! ∼ 2( 1 2π ) 2m .
We deduce that :

2m+s-1 | B 2m | 2m! ∼ 1 2π .
On other hand, we have :

(2m + s -1)! ∼ 2π(2m + s -1)( 2m + s -1 e ) 2m+s-1 .
Then :

(2m + s -1)! (s -1)! ∼ 2π(2m + s -1) (s -1)! ( 2m + s -1 e ) 2m+s-1 .
We deduce that :

2m+s-1 (2m + s -1)! (s -1)! ∼ (2m + s -1) 1 4m+2s-2 ( 2m + s -1 e ), 2m+s-1 (2m + s -1)! (s -1)! ∼ e 1 4m+2s-2 ln(2m+s-1) ( 2m + s -1 e ).
Finally, we get :

2m+s-1 | ψ 2m+s-1,s-1 | ∼ 1 2π e 1 4m+2s-2 ln(2m+s-1) ( 2m + s -1 e ), 2m+s-1 | ψ 2m+s-1,s-1 | ∼ m πe .
Then :

lim m→+∞ 2m+s-1 | ψ 2m+s-1,s-1 | = +∞.
Which implies that : lim k→+∞ k | ψ k,s-1 | = +∞. We deduce that the radius of convergence of the series +∞ k=1 ψ k,s-1 z k , is zero.

The Riemann zeta function

This section will be concerned with the Euler zeta series, which is the function

ζ(s) = +∞ n=1 1 n s ,
where s is a real number greater than 1. If s ∈ N \ {0, 1}, consider the real function g : x -→ x s , and the series s

N = N k=1 g( 1 k ).
Theorem 8 The series s N = N k=1 1 k s does not admits an exact limit.

Prof. We assume that the series (s n ) admits a exact limit, then there exists a holomorphic function f on a neighborhood of 0 such that :

f ( 1 N ) = s N , from a certain rank. If f (z) = +∞ k=0 α k z k , then :        α 1 α 2 . . . . . α n        = ψ (n)          0 . . . 1 0 . . . 0          = ψ (n) e s-1 .
Then : Unfortunately, this result is not true, because the function f is not holomorphic on a neighborhood of 0, from the lemma 1, then we obtain a contradiction and we deduce that the series (s n ) does not admits an exact limit.

       α 1 α 2 . . . . . α n        =        ψ 1,s-1 ψ 2,s-1 . . . . . ψ n,s-1       
12 The twelfth property of the matrix ψ (n)

From the above results, the formula ζ(s) =

ψ s-1,s-1 N s-1 - ψ s,s-1 N s - ψ s+1,s-1 N s+1 - 2M+s-1 k=s+2 ψ k,s-1 N k + E(M, N, s).
We deduce that : Then :

ζ(s) = N n=1 1 n s + 1 s -1 1 N s-1 - 1 2 1 N s -
ζ(s) = N -1 n=1 1 n s + 1 s -1 1 N s-1 + 1 2 1 N s + 2M-1 r=1 C s-1
r+s B r+1 (r + s)N r+s + E(M, N, s).

On other hand, we have B 2k+1 = 0 for every natural k ≥ 1, then :

ζ(s) = N -1 n=1 1 n s + 1 s -1 1 N s-1 + 1 2 1 N s + M m=1 C s-1
2m+s-1 B 2m (2m + s -1)N 2m+s-1 + E(M, N, s).

We get :

ζ(s) = N -1 n=1 1 n s + 1 s -1 1 N s-1 + 1 2 1 N s + M m=1 2m-1 i=0 (s + i)
B 2m (2m)!N 2m+s-1 + E(M, N, s).

Finally, we find the standard Euler-Maclaurin [START_REF] Borwein | Computational Strategies for the Riemann Zeta Function[END_REF]. formula applied to the zeta function ζ(s), where s is a natural number and s ≥ 2, then we deduce that the matrix of the black magic ψ (n) has a beautiful twelfth property which given as :

• Property 12.

By using the black magic matrix, we can represent the Euler-maclaurin formula as :

ζ(s) = N n=1
1 n s -< C s-1 , T M,N,s > +E(M, N, s), where : -< ., . > is the sclar product < x, y >= x i y i .

-C s-1 = ψ (2M+s) e s-1 is the (s-1)-th column of the matrix ψ (2M+s) .

-T M,N,s is the column vector defined as :T M,N,s =

     1 N 1 N 2

a 3 +

 3 ............................ + a n , where n 0 ∈ N. Consider the real α = n0-1 k=1 a k , we get :

  .................................. + a n .

3α 3 .

 3 ......................................... ................................................

, and α 0 = +∞ k=1 1 k

 1 s = ζ(s). Then : f (z) = ζ(s) + +∞ k=1 ψ k,s-1 z k .

  despite this, we can correct this equality by using a new term E(M, N, s) which is defined as :E(M, N, s) = ζ(s)s-1 N k + E(M, N, s).

ψ

  s-1 N k + E(M, N, s).For r = k -s, we obtain :ζ(s) = r+s,s-1N r+s + E(M, N, s).

We have :

The coefficients of the first column of ψ (10) are -1, 1/2, -1/6, 0, 1/30, 0, -1/42, 0, 1/30, 0 then :

Similarly, we deduce the following formulas :