Hindawi Publishing Corporation
Advances in Difference Equations
Volume 2010, Article ID 143521, 11 pages
doi:10.1155/2010/143521
Research Article
Elementary Proof of Yu. V. Nesterenko Expansion
of the Number Zeta(3) in Continued Fraction
Leonid Gutnik
Moscow State Institute of Electronics and Mathematics, Russia
Correspondence should be addressed to Leonid Gutnik, gutnik@gutnik.mccme.ru
Received 12 August 2009; Revised 10 December 2009; Accepted 10 January 2010
Academic Editor: Binggen Zhang
Copyright q 2010 Leonid Gutnik. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
Yu. V. Nesterenko has proved that ζ3 b0 a1 |/|b1 · · · aν |/|bν · · · , b0 b1 a2 2,
a1 1, b2 4, b4k1 2k 2, a4k1 kk 1, b4k2 2k 4, and a4k2 k 1k 2 for k ∈ N;
b4k3 2k 3, a4k3 k 12 , and b4k4 2k 2, a4k4 k 22 for k ∈ N0 . His proof is based
on some properties of hypergeometric functions. We give here an elementary direct proof of this
result.
1. Foreword
Applications of difference equations to the Number Theory have a long history. For example,
one can find in this journal several articles connected with the mentioned applications see
1–8. The interest in this area increases after Apéry’s discovery of irrationality of the number
ζ3. This paper is inspired by Yu. V. Nesterenko’s work 9. My goal is to give an elementary
direct proof of his expansion of the number ζ3 in continued fraction. Let us consider a
difference equation
xν1 − bν1 xν − aν1 xν−1 0,
1.1
with ν ∈ N0 . We denote by
{Pν b0 , a1 , b1 , . . . , aν , bν }∞
ν−1 ,
{Qν b0 , a1 , b1 , . . . , aν , bν }∞
ν−1
1.2
the solutions of this equation with initial values
P−1 1,
Q−1 0,
P0 b0 b0 ,
Q0 b0 1.
1.3
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Then
Pν b0 , a1 , b1 , . . . , aν , bν Qν b0 , a1 , b1 , . . . , aν , bν ∞
1.4
ν0
is a sequence of convergents of the continued fraction
b0 a1 |
aν |
··· ... .
|b1
|bν
1.5
Accoding to the famous result of R. Apéry 10,
vν
,
ν → ∞ uν
ζ3 lim
1.6
∞
where {uν }∞
ν0 and {vν }ν0 are solutions of difference equation
ν 13 xν1 − 34ν3 51ν2 27ν 5 xν ν3 xν−1 0
1.7
with initial values u0 1, u1 5, v1 0, v1 6. The equality 1.6 is equivalent to the
equality
ζ3 b0∨
a∨1 a∨2 a∨ |
∨ ∨ · · · ν∨ . . .
b
b
|bν
2
1
1.8
with
b0∨ 0,
b1∨ 5,
a∨1 6,
∨
bν1
34ν3 51ν2 27ν 5,
a∨ν1 −ν6 ,
1.9
where ν ∈ N. Nesterenko in 9 has offered the following expansion of the number 2ζ3 in
continued fraction:
2ζ3 2 1| 2| 1| 4|
...,
|2 |4 |3 |2
1.10
with
b0 b1 a2 2,
b4k1 2k 2, a4k1 kk 1,
a1 1,
b4k2 2k 4,
b2 4,
1.11
a4k2 k 1k 2
1.12
for k ∈ N;
b4k3 2k 3,
for k ∈ N0 .
a4k3 k 12 ,
b4k4 2k 2,
a4k4 k 22
1.13
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The halved convergents of continued fraction 1.10 compose a sequence containing
convergents of continued fraction 1.8. I give an elementary proof of Yu. V. Nesterenko
expansion in Section 2 .
2. Elementary Proof of Yu. V. Nesterenko Expansion
Instead of expansion 1.10 with 1.11, it is more convenient for us to prove the equivalent
expansion
ζ3 1 1| 4| 1| 4|
...,
|4 |4 |3 |2
2.1
with
b0 1,
a1 1,
b1 a2 b2 4.
2.2
Furthermore, to avoid confusion in notations, we denote below aν , bν for the fraction 2.1 by
∨
∨
1, Q−1
0,
a∧ν , bν∧ . Let P−1
Pν∨ Pν b0∨ , a∨1 , b1∨ , . . . , a∨ν , bν∨ ,
Qν∨ Qν b0∨ , a∨1 , b1∨ , . . . , a∨ν , bν∨ ,
2.3
where values a∨ν , bν∨ are specified in 1.9, and ν ∈ N0 . Then
Q0∨ 1,
P0∨ b0∨ 0,
Q1∨ b1∨ 5,
P2∨ b2∨ P1∨ a∨2 P0∨ 702,
P1∨ a∨1 6,
b2∨ 117,
a∨2 −1,
Q2∨ b2∨ Q1∨ a∨2 Q0∨ 584.
2.4
∧
∧
Let P−1
1, Q−1
0,
Pν∧ Pν b0∧ , a∧1 , b1∧ , . . . , a∧ν , bν∧ ,
Qν∧ Qν b0∧ , a∧1 , b1∧ , . . . , a∧ν , bν∧ ,
2.5
where ν ∈ N0 , a∧ν : aν , bν∧ : bν , and values aν , bν are specified in 2.2, 1.12, and 1.13.
We calculate first Pk∧ and Qk∧ for k 0, ..., 6.
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∧
∧
1, Q−1
0, it follows from 2.2 that
Since P−1
P0∧ b0 1,
∧
5,
P1∧ b1∧ P0∧ a∧1 P−1
Q0∧ 1,
∧
Q1∧ b1∧ Q0∧ a∧1 Q−1
4,
2.6
P2∧ b2∧ P1∧ a∧2 P0∧ 24 4P1∨ ,
Q2∧ b2∧ Q1 a∧2 Q0 20 4Q1∨ ,
P3∧ b3∧ P2∧ a∧3 P1∧ 77,
2.7
Q3∧ b3∧ Q2∧ a∧3 Q1∧ 64,
P4∧ b4∧ P3∧ a∧4 P2∧ 250,
Q4∧ b4∧ Q3∧ a∧4 Q2∧ 208,
P5∧ b5∧ P4∧ a∧5 P3∧ 1154,
Q5∧ b5∧ Q4∧ a∧5 Q3∧ 960,
2.8
P6∧ b6∧ P5∧ a∧6 P4∧ 12 × 702 12P2∨ ,
2.9
Q6∧ b6∧ Q5∧ a∧6 Q4∧ 12 × 584 12Q2∨ .
2.10
Let k ∈ N, k ≥ 2,
Pk∗ ∧
P4k−2
2k 1!
,
Qk∗ ∧
Q4k−2
2k 1!
.
2.11
We want to to prove that if k ∈ N, then
Pk∗ Pk∨ ,
Qk∗ Qk∨ .
2.12
Note that if k 1, 2, then 2.12 follows from 2.6–2.10. Therefore, we can consider only
k ∈ 3, ∞ ∩ Z. Let us consider the following difference equations:
∨
xν − a∨ν1 xν−1 0,
xν1 − bν1
2.13
∧
xν1 − bν1
xν − a∧ν1 xν−1 0,
2.14
with ν ∈ N0 . Then xν Pν∨ , xν Qν∨ , with ν ∈ −1, ∞ ∩ Z representing a fundamental system
of solutions of 2.13, and xν Pν∧ , xν Qν∧ with ν ∈ −1, ∞ ∩ Z representing a fundamental
Advances in Difference Equations
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system of solutions of 2.14. Making use of standard interpretation of a difference equation
as a difference system, we rewrite the equalities 2.13 and 2.14, respectively in the form
Xν1 A∨ν Xν ,
2.15
Xν1 A∧ν Xν ,
2.16
where
xν−1
Xν A∨ν
0
xν
1
A∧ν
,
∨
a∨1ν b1ν
2.17
,
0
1
∧
a∧1ν b1ν
,
2.18
and ν ∈ N0 . Let
Uν∨
∨
∨
Qν−1
Pν−1
Pν∨
Uν∧
Qν∨
∧
∧
Qν−1
Pν−1
Pν∧
2.19
,
Qν∧
2.20
,
with ν ∈ N0 be fundamental matrices of solutions of systems 2.15 and 2.16, respectively.
Therefore,
∧
Uν∧ A∧ν−1 Uν−1
,
∨
Uν∨ A∨ν−1 Uν−1
2.21
for ν ∈ N. In view of 2.18 and 2.21, detUν −aν detUν−1 , and therefore,
ν
ν
det Uν∧ −1ν det U0∧
a∧k −1ν
a∧k .
k1
2.22
k1
Hence
∧
Pν−1
∧
Qν−1
see 11.
ν
−
Pν∧
Qν∧
a∧k
k1
−1ν ∧ ∧
Qν Qν−1
2.23
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Further, we have
U0∨
1 0
U1∨
0 1
U2∨
6
5
,
,
6 5
702 584
1 1
5 4
∧
∧
∧
, U1 , U2 ,
U0 0 1
5 4
24 20
24 20
77 64
∧
∧
, U4 ,
U3 77 64
250 208
250 208
1154 960
∧
∧
, U6 ,
U5 1154 960
8424 7008
∨ ∧ −1 1 −24 5
,
U1 U2
4
0 1
,
0 1
1 0
U2∨
−1
U6∧
1
96
−36 5
0
2.24
2.25
8
2.26
.
Let k ∈ N, k ≥ 2. Then, in view of 2.20,
A∧4k−6
0
0
A∧4k−4 0
1
1
∧
a∧4k−12 b4k−12
0
,
,
0
1
0
1
2.27
,
k2 − k 2k
1
k2 2k − 2
1
k − 12 2k − 1
∧
a∧4k−11 b4k−11
0
0
1
∧
a∧4k−24 b4k−24
A∧4k−3
∧
a∧4k−23 b4k−23
A∧4k−5
1
k2 k 2k 2
.
Let Yk X4k−6 for k ∈ 2, ∞ ∩ Z. In view of 2.16 and 2.18,
Yk1 Bk∧ Yk ,
2.28
∧
∧
U4k−2
Bk∧ U4k−6
,
2.29
where, as before, k ∈ 2, ∞ ∩ Z,
Bk∧
A∧4k−3 A∧4k−4 A∧4k−5 A4k−6
5kk − 13
12kk 1k − 13
k 12k2 − 15k 5
.
kk 1 29k2 − 36k 12
2.30
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∧
is a fundamental
In view of 2.22, 2.2, 1.12, 1.13, 2.29, and 2.28, the matrix U4k−6
0 for k ∈
matrix of solutions of system 2.28. The substitution Zk Ck Yk , with detCk /
2, ∞ ∩ Z, transforms the system 2.28 into the system
Zk1 Dk Zk ,
2.31
with Dk Ck1 Bk∧ Ck −1 for k ∈ 2, ∞ ∩ Z. We prove now that if we take k ∈ 3, ∞ ∩ Z,
and Ck Hk−1 , where
1
H1 4
−24 5
0
2.32
,
1
Hk 12k 2k 1ck 1 −5k 2ck 1
−k − 13 ck
0
2.33
,
−1
with k ∈ 2, ∞ ∩ Z and ck −2k − 13 k 1! , then we obtain the equality Dk A∨k−1 .
So, let k ∈ 3, ∞ ∩ Z. Then, in view of 2.33,
12k 1kck
−5k 1ck
0
−k − 23 ck − 1
.
2.34
bk∨ 34k − 13 51k − 12 27k − 1 5 34k 3 − 51k2 27k − 5,
a∨k −k − 16 , 2.35
Hk−1 In view of1.9
where k ∈ 3, ∞ ∩ Z. Hence, in view of 2.19,
A∨k−1
0
1
−k − 16 34k3 − 51k2 27k − 5
2.36
.
In view of 2.34–2.36,
A∨k−1 Hk−1
1
−k − 16 34k3 − 51k2 27k − 5
0
0
×
12k 1kck
−5k 1ck
0
−k − 23 ck − 1
−k − 23 ck − 1
−k − 16 12k 1kck k − 16 5k 1ck − bk∨ k − 26 ck − 1.
.
2.37
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In view of 2.30 and 2.33,
Hk Bk∧
12k 2k 1ck 1 −5k 2ck 1
0
×
5kk − 13
12kk 1k − 13
−k − 13 ck
k 12k2 − 15k 5
kk 1 29k2 − 36k 12
0
−ck12kk 1k − 16
2.38
k 2ck 1kk 1 −k2
.
−k − 13 ckkk 1 29k2 − 36k 12
Since
− k 2k 1ck 1k3 −ck − 1k − 23 ,
− k − 13 ckkk 1 29k2 − 36k 12 − k − 16 5k 1ck
− 34k3 − 51k2 27k − 5 k − 13 k 1ck
2.39
− 34k3 − 51k2 27k − 5 k − 23 ck − 1,
it follows from 2.35, 2.37, and 2.38 that
A∨k−1 Hk−1 Hk Bk∧
2.40
for k ∈ 3, ∞ ∩ Z. We prove by induction now the following equality:
∧
Uk∨ Hk U4k−2
,
2.41
for any k ∈ N. In view of 2.25 and 2.32, the equality 2.41 holds for k 1. In view of 2.26
and 2.33, the equality 2.41 hold for k 2. Let k ∈ 3, ∞ ∩ Z and 2.41 holds for k − 1.
Then, in view of 2.29, 2.40, and 2.21,
∧
∧
∧
∨
Hk Bk U4k−6
A∨k−1 Hk−1 U4k−6
A∨k−1 Uk−1
Uk∨ .
Hk U4k−2
2.42
So, the equality 2.41 holds for any k ∈ N. In view of 2.41,
∧
,
Pk∨ 2k 1!−1 P4k−2
∧
Qk∨ 2k 1!−1 Q4k−2
2.43
for k ∈ 2, ∞ ∩ Z. Since
Pν∨ ν!3 vν ,
Qν∨ ν!3 uν
2.44
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for vν and uν in 1.6 and ν ∈ N0 , it follows from 2.43 and 2.44, that
∧
2k 1k!4 vk ,
P4k−2
∧
Q4k−2
2k 1k!4 uk .
2.45
As it is well known, for any ε > 0 there exist C1 ε > 0 and C2 ε > 0 such that
√ 4k1−ε
√ 4k1ε
C1 ε 1 2
< |uk | < C2 ε 1 2
,
2.46
√ 4k1−ε
√ 4k1ε
< |vk | < C2 ε 1 2
,
C1 ε 1 2
2.47
vk C2 ε
C1 ε
ζ3
−
<
<
√ 8k1ε √ 8k1−ε .
uk
1 2
1 2
2.48
We apply 2.23 now. Let k ∈ 2, ∞∩Z. In view of 2.2, 1.12–1.13, and 2.45, if η 1, 2, 3,
then
0≤
4k−2η
κ1
aκ ≤
4k1
aκ ≤ a4k−1 a4k a4k1 × k3 k 13
κ1
4k3 k 13
k
4k−2
aκ
κ1
a4κ−5 a4κ−4 a4κ−3 a4κ−2
2.49
κ2
4k3 k 13
k
κ − 12 κ2 κ − 1κκκ 1 2k!8 k 14 ,
κ2
4k 12 k!8 u2k Q4k−2 2 < Q4k−3η Q4k−2η .
2.50
In view of 2.23, 2.50, and 2.49, if θ 1, 2, 3
θ P
P4k−2
P4k−2η P4k−2θ 4k−3η
≤
−
−
Q
Q4k−1θ η1 Q4k−3η Q4k−2η 4k−2
3 P
√ 8k−1o1
P4k−2η k 12 4k−3η
≤
−
≤
1
2
,
≤3
Q4k−3η Q4k−2η 2u2k
η1
2.51
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when k → ∞. In view of 2.45, 2.48, and 2.51, there exist C3 ε > 0 and C4 ε > 0 such
that
∧
P4k−2θ
C4 ε
C3 ε
ζ3
−
<
<
∧
√ 8k1ε √ 8k1−ε ,
Q4k−2 1 2
1 2
2.52
where θ 0, 1, 2, 3. So, the equality 2.1 is proved. In view of 2.23,
ζ3 −
P0∧
Q0∧
∞
−1ν−1 dν ,
2.53
ν1
where
ν
0 < dν a∧k
k1
Q∧ν Qν−1 ∧
.
2.54
Further, we have
dν1
aν1 ∧ Qν−1 ∧
∧
< 1.
∧
dν
bν1 Qν∧ a∧ν1 Qν−1
2.55
∧
∧
Hence, the series 2.53 is the series of Leibnitz type. Therefore, P2k−1
/Q2k−1
decreases, when
∧
∧
k increases in N, and P2k /Q2k increases, when k increases in N.
Acknowledgment
The author would like to express his thanks to the reviewer of this article for his efforts, his
criticism, his advices, and indications of misprints. Ravi P. Agarwal had expressed a useful
suggestion, which the author realized in foreword and references. He is grateful to him in
this connection.
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