Notes on Number Theory and Discrete Mathematics ISSN 1310–5132 Vol. 20, 2014, No. 2, 79–91
Nesterenko-like rational function, useful to prove the Ap´ery’s theorem Anier Soria Lorente Department of Basic Sciences Granma University, Cuba e-mail:
[email protected]
Abstract: In this paper, a brief introduction to the Ap´ery’s result and to the so called phenomenon of Ap´ery’s is given. Here, a modification of the Nesterenko’s rational function, from which new diophantine approximations to ζ(3) are deduced, is presented. Moreover, as a consequence we deduce the corresponding Ap´ery-Like recurrence relation as well as a new continued fraction expansion and a new series expansion for ζ(3). Keywords: Riemann zeta function, Ap´ery’s approximants, Recurrence relation, Continued fraction expansion, Irrationality. AMS Classification: Primary: 11B37, 30B70, 14G10, 11J72, 11M06; Secondary: 37B20, 11A55, 11J70, 11Y55, 11Y65.
1
Introduction
One of the more interesting open problems into the number theory has to do with the arithmetical nature of the values of the Riemann zeta function [16] Z X 1 (−1)k−1 1 logk−1 x = dx, ζ (k) ≡ nk (k − 1)! 0 1 − x n≥1
(1)
in the positive integers k ∈ N\ {1}. As it is well known, at the Journees Arithmetiques held at Marseille-Luminy in June 1978, Roger Ap´ery [5, 11, 28] gave an elementary proof of the following result, credited as Ap´ery’s theorem. Theorem 1.1. The number ζ (3) is irrational. 79
However due to the complexity of the Ap´ery’s proof, there was a general disagreement amongst the mathematicians present there as to the validity of the proof presented. Two months later a complete exposition of the proof was presented at the International Congress of Mathematicians in Helsinki in August 1978 by H. Cohen. This proof based on the lecture by Ap´ery but also contained some ideas of Cohen and Don Zaiger [11, 28]. Theorem 1.2. (Criterion for irrationality) If there is a δ > 0 and a sequence (vn /un )n≥0 of rational numbers such that vn /un 6= x and x − vn < 1 , n = 0, 1, . . . , un u1+δ n then x is irrational. From now a sketch of the Ap´ery’s proof will take place. The same one had as a fundamental ingredient the following recurrence relation [6, 12, 28] (n + 2)3 yn+2 − (2n + 3)(17n2 + 51n + 39)yn+1 + (n + 1)3 yn = 0,
n ≥ 0,
(2)
which is satisfied by the numerators an and denominators bn of the diophantine approximations to ζ (3) with the initial conditions a0 = 0,
a1 = 6,
b0 = 1,
b1 = 5,
or by the explicit representation of the sequences in question X n + k 2 n2 X n + k 2 n2 γn,k , and an ≡ bn ≡ k k k k 0≤k≤n 0≤k≤n where γn,k
X 1 X (−1)j−1 n + j −1 n−1 = + . 3 3 j 2j j j 1≤j≤n 1≤j≤k
Observe that, from (2) we deduce that an bn−1 − an−1 bn =
6 . n3
This leads to an example of a non-simple continued fraction expansion ζ (3) =
6| 1 | 64 | n6 | − − − ··· − − ··· , | 5 | 117 | 535 | (2n + 1) (17n2 + 17n + 5)
Then, seeing that ζ (3) − a0 /b0 = ζ (3), it is induced that X a 6 n ζ (3) − = = O b−2 . n 3 bn k b k−1 bk k≥n+1 It is easy to verify by the recurrence relation (2) and Poincar´e’s theorem [22, 23] that √ 4 bn = O ($n ) , where $ = 2+1 . 80
(3)
Moreover, by the prime number theorem, it can be shown that Y log n Y ln ≡ p[ log p ] ≤ n = O e(1+)n , ∀ > 0. p≤n
(4)
p≤n
Therefore, setting vn = 2an ln3 ∈ Z and un = 2bn ln3 ∈ Z, we obtain un = O ($n e3n ) and ζ (3) − vn = O u−(1+δ) , n un where
log α − 3 = 0.080529 . . . > 0, log α + 3 which, proves the Ap´ery’s theorem, by virtue of criterion for irrationality. Ap´ery’s irrationality proof of ζ (3) operates with the faster convergent series δ=
ζ (3) =
5 X (−1)n−1 , 2 n≥1 n3 2n n
first obtained by A. A. Markov in 1890 [18]. The aforementioned result, in a beginning somewhat polemic, inspired to several mathematicians to construct different methods to explain the irrationality of aforesaid constant [6, 7, 8, 10, 19, 20, 24, 26, 27, 29]. Surprisingly, these methods conduce to the same sequences of diophantine approximations (3) to ζ (3) (named Ap´ery’s approximants or Ap´ery’s diophantine approximations). This fact names Ap´ery’s phenomenon. In several papers Ap´ery’s phenomenon it has been interpreted in different ways. For example: shortly after Ap´ery announced his proof, Beukers produced an elegant and entirely different proof of the irrationality of ζ (2) and ζ (3). In the case of ζ (2), Beukers in [7, 8] considers a double integrals defined by Z 1Z 1 (1 − y)n Ln (x) dxdy = θn ζ (2) − ϑn , 1 − xy 0 0 where n ∈ N and Ln (x) is the Legendre-type polynomial, orthogonal with respect to the Lebesgue measure on (0, 1), given by X n k 1 dn n k n+k n Ln (z) = z (1 − z) = (−1) z , n! dz n k k 0≤k≤n and θn , ϑn ∈ Q , for all n. An estimation of the latter linear form shows that it tends to 0 as n approaches infinity fast enough to yield the irrationality of ζ (2). Beukers also gives an analogous argument for the case of ζ (3) [7, 8, 20] by using the following triple integral instead Z 1Z 1 log xy bn ζ (3) − an = − Ln (x) Ln (y) dxdy (5) 0 0 1 − xy Z 1Z 1Z 1 (xyz (1 − x) (1 − y) (1 − z))n = dxdydz. (1 − (1 − xy) z)n+1 0 0 0 Thus, Beukers shows that previous expression coincides with O ($−n ), which proves the Ap´ery’s theorem. 81
In [9], Beukers considered a rational approximation problem in an intent to formulate Ap´ery’s proof more natural and introduced the rational function Rn (z) ≡
(n − z + 1)2n , (−z)2n+1
(6)
from which by computing the partial fraction expansion, deduces the Ap´ery’s rational approximants. Here, (·)n denotes the Pochhammer symbol [13], also called the shifted factorial, defined by Y (z)n ≡ (z + j) , n ≥ 1, (z)0 = 1, 0≤j≤n−1
which in terms of the gamma function is given by (z)n =
Γ (z + n) , Γ (z)
n = 0, 1, 2, . . .
Sorokin, in virtue of a approximation problem, obtained Ap´ery’s sequences as well as Beukers’s error-term sequence (5) [26, 29]. After this, Nesterenko in [20], inspired by Gutnik’s work [14] considered the following modification of the Beukers’s rational function (6) Nn (z) ≡
(−z)2n , (z + 1)2n+1
(7)
and proved the following expression for the error-term sequence Z π 2 X d 1 Nn (k) = Nn (ν) dν, bn ζ (3) − an = − dk 2πi L sin πν k≥0 where d Nn (z) = 2Nn (z) dz
X 1 1 − t − k 1≤k≤n+1 t + k 0≤k≤n−1 X
! ,
and L is the vertical line Re z = C, 0 < C < n + 1, oriented from top to bottom. Indeed, the use of Laplace’s method allowed him to estimate the above contour integral (7) yielding the behavior O ($−n ). Moreover, he proposed the following continued fraction expansion 2ζ (3) = 2 +
1| 2| 1| 4| 2| 6| 4| + + + + + + + ··· , |2 |4 |3 |2 |4 |6 |5
where the numerators an , n ≥ 2, and denominators bn , n ≥ 1, are definded by b4k+1 = 2k + 2,
a4k+1 = k (k + 1) ,
a4k+2 = (k + 1) (k + 2) , 2
a4k+3 = (k + 1) ,
b4k+2 = 2k + 4,
b4k+3 = 2k + 3,
b4k = 2k,
a4k = (k + 1)2 .
The purpose of this paper is to present a modification of the Nesterenko’s rational function (7) defined by (−z)2n−1 (n − z − 1) Fn (z) ≡ . (8) (z + 1)2n+1 82
From which we deduce new rational approximants that prove Ap´ery’s theorem. This modification, consists in deleting one of the zeroes of the rational function (7), in particular z = n − 1. Moreover, as a consequence we deduce the corresponding Ap´ery-Like recurrence relation for these rational approximants as well as a new continued fraction expansion for ζ(3). Also, we show a new series expansion for ζ(3), inspired by Arves´u’s work in [6].
2
Ap´ery-like recurrence relation
Lemma 2.1. The following relation is valid rn = −2−1
X d Fn (z) = qn ζ (3) − pn , dz z≥0
n = 0, 1, . . . ,
(9)
being X
qn =
(n)
bk
and
pn =
0≤k≤n (n)
X
(n)
(3)
bk Hk + 2−1
1≤k≤n
X
(n)
(2)
ak Hk ,
(10)
1≤k≤n
(n)
where the coefficients bk and ak are given in (11)-(12), respectively. Moreover, nqn ∈ Z and (r) 2nln3 pn ∈ Z. Here, we denote for Hk to the harmonic number k of order r. In fact, let us expand the function Fn (z) on the sum of partial fractions Fn (z) =
(n) (n) X ak bk + 2, z + k + 1 (z + k + 1) 0≤k≤n 0≤k≤n
X
where (n) bk
, k = 0, . . . , n, = (z + k + 1) Fn (z) z=−k−1 2 2 n+k n = (n + k)−1 k k 2 2 2 1 n+k−1 n 1 n+k−1 n−1 n = + , n k k n k k−1 k 2
(11)
and (n) ak
d 2 = (z + k + 1) Fn (z) , k = 0, . . . , n, dt z=−k−1 (n) = 2bk 2Hk − Hn+k−1 − Hn−k − 2−1 (n + k)−1 .
Since Fn (z) = O (z −2 ) as z → ∞, then we infer X (n) X ak = Res Fn (z) = − Res Fn (z) = 0. 0≤k≤n
0≤k≤n
z=∞
z=−k−1
83
(12)
Thus, we have rn =
(n) (n) X X ak bk −1 + 2 (z + k + 1)3 (z + k + 1)2 z≥0 0≤k≤n z≥0 0≤k≤n
X X
X X b(n) X X a(n) −1 k k + 2 3 2 l l 0≤k≤n l≥k+1 0≤k≤n l≥k+1 ! ! X (n) X X 1 X (n) X X 1 −1 = +2 ak bk − − l3 l2 0≤k≤n 0≤k≤n l≥1 l≥1 1≤l≤k 1≤l≤k X (n) X 1 X (n) X 1 X (n) X 1 −1 = − − 2 ak , bk b l3 1≤k≤n k 1≤l≤k l3 l2 1≤k≤n 0≤k≤n 1≤l≤k l≥1
=
which coincides with (9) by considering the expressions given in (10). Clearly, from the expres(n) (n) sions (11)–(12) it follows that nbk ∈ Z and nln ak ∈ Z, k = 0, . . . , n. Therefore, using lnj
k X 1 ∈ Z, ij i=1
k = 0, 1, . . . , n,
j ∈ Z+ ,
we deduce that nqn ∈ Z and 2nln3 pn ∈ Z as required. The lemma is completely proved. Before continuing with the following result let us define ρ1,n ≡ 2(n + 2)(384n8 + 2592n7 + 7516n6 + 12228n5 + 12255n4 + 7838n3 + 3166n2 + 730n + 73), ρ2,n ≡ 2(288n7 + 1776n6 + 4682n5 + 6677n4
ρ3,n and
+ 5390n3 + 2448n2 + 628n + 69), ≡ 2 96n5 + 360n4 + 478n3 + 285n2 + 88n + 11 ,
ρ2,n z + (2n2 + 5n + 6)ρ3,n z 2 + ρ3,n z 3 − ρ1,n Sn (z) ≡ , (k − n) (k − n + 1) (z + n + 2)2 (n + 1)2
which is the output of the so-called algorithm of creative telescoping due to W. Gosper and D. Zeilberger [1, 2, 3, 4, 25]. The algorithm is realised in different computer algebra systems, in particular, in Maple and Mathematica. Proposition 2.2. The sequences (pn )n≥1 , (qn )n≥1 and (rn )n≥1 verify the following Ap´ery-like recurrence relation (n + 2)4 24n3 + 30n2 + 16n + 3 yn+2 − 4(n + 1)(204n6 + 1173n5 + 2668n4 + 3065n3 + 1905n2 + 634n + 86yn+1 )yn+1 + n4 24n3 + 102n2 + 148n + 73 yn = 0,
84
n ≥ 1. (13)
Proof. In fact, firstly let us prove that (rn )n≥0 verify the previous Ap´ery-like recurrence relation (13). For such purpose, it is enough to check (n + 2)4 24n3 + 30n2 + 16n + 3 Fn+2 − 4(n + 1)(204n6 + 1173n5 + 2668n4 + 3065n3 + 1905n2 + 634n + 86yn+1 )Fn+1 + n4 24n3 + 102n2 + 148n + 73 Fn = Sn (z + 1) Fn (z + 1) − Sn (z) Fn (z) ,
n ≥ 1. (14)
Thus, we have (n + 2)4 24n3 + 30n2 + 16n + 3 rn+2 − 4(n + 1)(204n6 + 1173n5 + 2668n4 + 3065n3 + 1905n2 + 634n + 86yn+1 )rn+1 + n4 24n3 + 102n2 + 148n + 73 rn X d = −2−1 [Sn (z + 1) Fn (z + 1) − Sn (z) Fn (z)] dz z≥0 = −2−1 Sn0 (0) Fn (0) − 2−1 Sn (0) Fn0 (0) = 0. Since Fn (0) = Fn0 (0) = 0 for all n ≥ 1. (n) For the sequence (qn )n≥1 , observe that bk = 0 for k < 0 and k > n. In consequence, using (14) we have (n + 2)4 24n3 + 30n2 + 16n + 3 qn+2 − 4(n + 1)(204n6 + 1173n5 + 2668n4 + 3065n3 + 1905n2 + 634n + 86yn+1 )qn+1 + n4 24n3 + 102n2 + 148n + 73 qn i Xh = S˜n (z + 1) − S˜n (z) (z + k + 1)2 k∈Z
where S˜n (z) = Sn (z) Fn (z) and i Xh S˜n (z + 1) − S˜n (z) (z + k + 1)2 k∈Z
, z=−k−1
= z=−k−1
Xh k∈Z
i ˜ ˜ Sn (j − k) − Sn (j − k − 1) j 2
= 0. j=0
Finally, the sequence (pn = qn ζ (3) − rn )n≥1 satisfies the Ap´ery’s recurrence relation (2) as a linear combination of the sequences (qn )n≥1 and (rn )n≥1 . The proposition is completely proved. Notice that, from the Proposition 2.2 we have that the characteristic equation for (13) is t − 34t + 1 and their zeros are t1 = $, and t2 = $−1 respectively. Hence, from Poincar´e’s 2
85
theorem [22, 23] it’s has the behavior qn = O ($n ) and rn = O ($−n ), as n goes to infinity, for the two linearly independent solutions, respectively. Then, assuming that ζ (3) = p/q, where p, q ∈ Z+ , we have that 2qnln3 rn = 2pnln3 qn −2qnln3 pn , is an integer distinct from zero. Therefore, by the prime numbers theorem we have (4) and as a consequence we deduce that 1 ≤ 2qnln3 |rn | = O (ln3 $−n ), wich is a contradiction, because e3 $−1 = 0, 591263 . . . < 1. Thus, the Ap´ery’s theorem 1.1 is proven. Theorem 2.3. Let (pn )n≥−1 and (qn )n≥−1 be two sequences of numbers such that q−1 = 0, p−1 = q0 = 1 and pn qn−1 − pn−1 qn 6= 0 for n = 0, 1, 2, . . .. Then there exists a unique irregular continued fraction b1 | b2 | b3 | bn | a0 + + + + ··· + + ··· , (15) | a1 | a2 | a3 | an whose n-th numerator is pn and n-th denominator is qn , for each n ≥ 0. More precisely (see [15, p. 31]) a0 = p0 , a1 = q1 , b1 = p1 − p0 q1 , pn qn−2 − pn−2 qn pn−1 qn − pn qn−1 an = , bn = , n = 0, 1, 2, . . . pn−1 qn−2 − pn−2 qn−1 pn−1 qn−2 − pn−2 qn−1 Theorem 2.4. Two irregular continued fractions a0 +
b1 | b2 | b3 | bn | + + + ··· + + ··· , | a1 | a2 | a3 | an
a00 +
b01 | b02 | b03 | b0n | + + + · · · + + ··· , | a01 | a02 | a03 | a0n
are equivalent if and only if there exists a sequence of non-zero (cn )n≥0 with c0 = 1 such that (see [15, p. 31] ) a0n = cn an ,
n = 0, 1, 2, . . . ,
b0n = cn cn−1 bn ,
n = 1, 2, . . .
(16)
Then, using the previous theorem we deduce the following result. Theorem 2.5. The following irregular continued fraction expansion for ζ (3) is verify ζ (3) =
7| −146 | −38864 | P4 | Pn | + + + + ··· + + ··· , | 6 | 827 | Q3 | Q4 | Qn
where Pn = −(n − 2)4 (n − 1)4 24n3 − 186n2 + 484n − 423 × 24n3 − 42n2 + 28n − 7 , and Qn = 4(n − 1) × 204n6 − 1275n5 + 3178n4 − 3999n3 + 2667n2 − 910n + 126 . 86
3
New series expansion for ζ (3)
We define ϕn (z) ≡ An (z) − Bn (z) log z
and ψn (z) ≡ ϕn (z) log z,
where An (z) and Bn (z) polynomials of degree exactly n defined by X (n) X (n) An (z) ≡ ak z n and Bn (z) ≡ bk z n . 0≤k≤n
Thus, applying the identity Z
0≤k≤n
1
xi logj xdx = (−1)j j!
0
we have
1 , (i + 1)j+1
(17)
1
Z
xz ϕn (x) dx,
Fn (z) = 0
(18) Gn (z) =
d Fn (z) = dz
Then, from (8) and (18) we induce Z 1 xk ϕn (x) dx = 0,
Z
1
xz ψn (x) dx.
0
k = 0, . . . , n − 1,
0
(19) Z
1
xk ψn (x) dx = 0,
k = 0, . . . , n − 2.
0
Lemma 3.1. The following relation −1
Z
1
rn = −2
0
ψn (x) dx. 1−x
holds. Proof. In fact, we know that Z 1 Z 1 An (x) − An (1) ψn (x) −1 −1 −2 dx = −2 log xdx 1−x 0 1−x 0 Z 1 Bn (x) − Bn (1) −1 +2 log2 xdx 1−x 0 Z 1 log2 x +2−1 Bn (1) dx. 0 1−x
87
(20)
Therefore, using (1) and (17) we get Z 1 ψn (x) −1 dx = qn ζ (3) −2 0 1−x X (n) X Z −1 +2 ak 0≤k≤n
X
−1
−2
0≤k≤n
= qn ζ (3) −
X Z 1≤j≤k
X
(n)
bk
0≤k≤n
xj−1 log xdx
0
1≤j≤k (n) bk
1
1
xj−1 log2 xdx
0
X (n) X 1 X 1 −1 − 2 ak . j3 j2 1≤j≤k 0≤k≤n 1≤j≤k
Thus, the lemma is completely proved. Observe that, as a consequence of the conditions of orthogonality (19) and the Lemma 3.1 we deduce the following relations Z 1 Z 1 pn (x) ϕn (x) ϕn (x) dx = pn (1) dx, 1−x 0 0 1−x (21) Z 1 p˜n−1 (x) ψn (x) p˜n−1 (1) rn = −2−1 dx, 1−x 0 where pn (x) and p˜n−1 (x) are arbitrary polynomials of degree at most n and n − 1 respectively. Proposition 3.2. The following relations are valid i.) pn qn+1 − pn+1 qn = −
24n3 + 30n2 + 16n + 3 , 2n4 (n + 1)4
(22)
ii.) ζ (3) =
7 X 24n3 + 30n2 + 16n + 3 + , 6 n≥1 2n3 (n + 1)3 Θn
(23)
being
−n − 1, −n − 1, n + 1, n + 2
Θn = 4 F3
1
1, 1, 1
−n, −n, n, n + 1
× 4 F3
1 , 1, 1, 1
where r Fs denotes the ordinary hypergeometric series [13, 17, 21].
88
Proof. Notice that, using (21) we have Z 1 Z 1 ϕn+1 (x) Bn (x) ψn+1 (x) −1 −1 qn rn+1 = 2 An (1) dx − 2 dx 1−x 1−x 0 0 Z 1 ϕn (x) ϕn+1 (x) −1 dx. = 2 1−x 0 Moreover Z 1 Z 1 ϕn (x) ϕn+1 (x) An+1 (x) ϕn (x) −1 −1 dx = 2 dx 2 1−x 1−x 0 0 −1
Z
1
−2
0
Bn+1 (x) ψn (x) dx, 1−x
being Z 0
1
1
[An+1 (x) − An+1 (1)] ϕn (x) dx 1−x 0 X X Z 1 (n+1) = − ak xj−1 ϕn (x) dx
An+1 (x) ϕn (x) dx = 1−x
Z
0≤k≤n+1
1≤j≤k
0
(n+1)
= −an+1 Fn (n) , and −1
Z
−2
0
1
Z 1 ψn (x) Bn+1 (x) ψn (x) −1 dx = −2 Bn+1 (1) dx 1−x 0 1−x Z 1 [Bn+1 (x) − Bn+1 (1)] ψn (x) −1 −2 dx 1−x 0 X (n+1) X Z 1 −1 xj−1 ψn (x) dx + qn+1 rn . =2 bk 0≤k≤n+1
1≤j≤k
0
From where we get pn qn+1 − pn+1 qn = qn rn+1 − qn+1 rn (n+1)
= 2−1 b(n+1) Gn (n − 1) + 2−1 bn+1 Gn (n − 1) n (n+1)
(n+1)
+ 2−1 bn+1 Gn (n) − 2−1 an+1 Fn (n) , which corresponds with (22). Presently, having in account X pk pk+1 pn p1 = − − , qn q1 1≤k≤n−1 qk qk+1 and using (22) conjointly with pn p1 X ζ (3) = lim = − n→∞ qn q1 n≥1
pn qn+1 − pn+1 qn qn qn+1
We deduce (23). Therefore, the proposition is completely proved. 89
.
Acknowledgments The author expresses his sincerest thanks to the referees for their valuable suggestions.
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