Advances in Difference Equations
Volume 2010 (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

Received 12 August 2009; Revised 10 December 2009; Accepted 10 January 2010

Academic Editor: Binggen Zhang

Copyright © 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.

Abstract

Yu. V. Nesterenko has proved that 𝜁 ( 3 ) = 𝑏 0 + 𝑎 1 | / | 𝑏 1 + ⋯ + 𝑎 𝜈 | / | 𝑏 𝜈 + ⋯ , 𝑏 0 = 𝑏 1 = 𝑎 2 = 2 , 𝑎 1 = 1 , 𝑏 2 = 4 , 𝑏 4 𝑘 + 1 = 2 𝑘 + 2 , 𝑎 4 𝑘 + 1 = 𝑘 ( 𝑘 + 1 ) , 𝑏 4 𝑘 + 2 = 2 𝑘 + 4 , and 𝑎 4 𝑘 + 2 = ( 𝑘 + 1 ) ( 𝑘 + 2 ) for 𝑘 ∈ ℕ ; 𝑏 4 𝑘 + 3 = 2 𝑘 + 3 , 𝑎 4 𝑘 + 3 = ( 𝑘 + 1 ) 2 , and 𝑏 4 𝑘 + 4 = 2 𝑘 + 2 , 𝑎 4 𝑘 + 4 = ( 𝑘 + 2 ) 2 for 𝑘 ∈ ℕ 0 . 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 𝜁 ( 3 ) . 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 𝜁 ( 3 ) in continued fraction. Let us consider a difference equation

𝑥 𝜈 + 1 − 𝑏 𝜈 + 1 𝑥 𝜈 − 𝑎 𝜈 + 1 𝑥 𝜈 − 1 = 0 , ( 1 . 1 ) with 𝜈 ∈ ℕ 0 . We denote by

𝑃 𝜈 𝑏 0 , 𝑎 1 , 𝑏 1 , … , 𝑎 𝜈 , 𝑏 𝜈 + ∞ 𝜈 = − 1 , 𝑄 𝜈 𝑏 0 , 𝑎 1 , 𝑏 1 , … , 𝑎 𝜈 , 𝑏 𝜈 + ∞ 𝜈 = − 1 ( 1 . 2 ) the solutions of this equation with initial values

𝑃 − 1 = 1 , 𝑄 − 1 = 0 , 𝑃 0 𝑏 0 = 𝑏 0 , 𝑄 0 𝑏 0 = 1 . ( 1 . 3 )

Then

𝑃 𝜈 𝑏 0 , 𝑎 1 , 𝑏 1 , … , 𝑎 𝜈 , 𝑏 𝜈 𝑄 𝜈 𝑏 0 , 𝑎 1 , 𝑏 1 , … , 𝑎 𝜈 , 𝑏 𝜈 + ∞ 𝜈 = 0 ( 1 . 4 ) is a sequence of convergents of the continued fraction

𝑏 0 + 𝑎 1 | | | | 𝑏 1 𝑎 + ⋯ + 𝜈 | | | | 𝑏 𝜈 + … . ( 1 . 5 ) Accoding to the famous result of R. Apéry [10],

𝜁 ( 3 ) = l i m 𝜈 → ∞ 𝑣 𝜈 𝑢 𝜈 , ( 1 . 6 ) where { 𝑢 𝜈 } + ∞ 𝜈 = 0 and { 𝑣 𝜈 } + ∞ 𝜈 = 0 are solutions of difference equation

( 𝜈 + 1 ) 3 𝑥 𝜈 + 1 − 3 4 𝜈 3 + 5 1 𝜈 2 𝑥 + 2 7 𝜈 + 5 𝜈 + 𝜈 3 𝑥 𝜈 − 1 = 0 ( 1 . 7 ) with initial values 𝑢 0 = 1 , 𝑢 1 = 5 , 𝑣 1 = 0 , 𝑣 1 = 6 . The equality (1.6) is equivalent to the equality

𝜁 ( 3 ) = 𝑏 ∨ 0 + 𝑎 ∨ 1 | | | | 𝑏 ∨ 1 + 𝑎 ∨ 2 | | | | 𝑏 ∨ 2 𝑎 + ⋯ + ∨ 𝜈 | | | | 𝑏 ∨ 𝜈 + … ( 1 . 8 ) with

𝑏 ∨ 0 = 0 , 𝑏 ∨ 1 = 5 , 𝑎 ∨ 1 = 6 , 𝑏 ∨ 𝜈 + 1 = 3 4 𝜈 3 + 5 1 𝜈 2 + 2 7 𝜈 + 5 , 𝑎 ∨ 𝜈 + 1 = − 𝜈 6 , ( 1 . 9 ) where 𝜈 ∈ ℕ . Nesterenko in [9] has offered the following expansion of the number 2 𝜁 ( 3 ) in continued fraction:

1 | | 2 𝜁 ( 3 ) = 2 + | | 2 + 2 | | | | 4 + 1 | | | | 3 + 4 | | | | 2 . . . , ( 1 . 1 0 ) with

𝑏 0 = 𝑏 1 = 𝑎 2 = 2 , 𝑎 1 = 1 , 𝑏 2 𝑏 = 4 , ( 1 . 1 1 ) 4 𝑘 + 1 = 2 𝑘 + 2 , 𝑎 4 𝑘 + 1 = 𝑘 ( 𝑘 + 1 ) , 𝑏 4 𝑘 + 2 = 2 𝑘 + 4 , 𝑎 4 𝑘 + 2 = ( 𝑘 + 1 ) ( 𝑘 + 2 ) ( 1 . 1 2 ) for 𝑘 ∈ ℕ ;

𝑏 4 𝑘 + 3 = 2 𝑘 + 3 , 𝑎 4 𝑘 + 3 = ( 𝑘 + 1 ) 2 , 𝑏 4 𝑘 + 4 = 2 𝑘 + 2 , 𝑎 4 𝑘 + 4 = ( 𝑘 + 2 ) 2 ( 1 . 1 3 ) for 𝑘 ∈ ℕ 0 .

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

1 | | 𝜁 ( 3 ) = 1 + | | 4 + 4 | | | | 4 + 1 | | | | 3 + 4 | | | | 2 … , ( 2 . 1 ) with

𝑏 0 = 1 , 𝑎 1 = 1 , 𝑏 1 = 𝑎 2 = 𝑏 2 = 4 . ( 2 . 2 ) Furthermore, to avoid confusion in notations, we denote below 𝑎 𝜈 , 𝑏 𝜈 for the fraction (2.1) by 𝑎 ⋀ 𝜈 , 𝑏 ⋀ 𝜈 . Let 𝑃 ∨ − 1 = 1 , 𝑄 ∨ − 1 = 0 ,

𝑃 ∨ 𝜈 = 𝑃 𝜈 𝑏 ∨ 0 , 𝑎 ∨ 1 , 𝑏 ∨ 1 , … , 𝑎 ∨ 𝜈 , 𝑏 ∨ 𝜈 , 𝑄 ∨ 𝜈 = 𝑄 𝜈 𝑏 ∨ 0 , 𝑎 ∨ 1 , 𝑏 ∨ 1 , … , 𝑎 ∨ 𝜈 , 𝑏 ∨ 𝜈 , ( 2 . 3 ) where values 𝑎 ∨ 𝜈 , 𝑏 ∨ 𝜈 are specified in (1.9), and 𝜈 ∈ ℕ 0 . Then

𝑄 ∨ 0 = 1 , 𝑃 ∨ 0 = 𝑏 ∨ 0 = 0 , 𝑄 ∨ 1 = 𝑏 ∨ 1 = 5 , 𝑃 ∨ 1 = 𝑎 ∨ 1 = 6 , 𝑏 ∨ 2 = 1 1 7 , 𝑎 ∨ 2 𝑃 = − 1 , ∨ 2 = 𝑏 ∨ 2 𝑃 ∨ 1 + 𝑎 ∨ 2 𝑃 ∨ 0 = 7 0 2 , 𝑄 ∨ 2 = 𝑏 ∨ 2 𝑄 ∨ 1 + 𝑎 ∨ 2 𝑄 ∨ 0 = 5 8 4 . ( 2 . 4 ) Let 𝑃 ⋀ − 1 = 1 , 𝑄 ⋀ − 1 = 0 ,

𝑃 ∧ 𝜈 = 𝑃 𝜈 𝑏 ∧ 0 , 𝑎 ∧ 1 , 𝑏 ∧ 1 , … , 𝑎 ∧ 𝜈 , 𝑏 ∧ 𝜈 , 𝑄 ∧ 𝜈 = 𝑄 𝜈 𝑏 ∧ 0 , 𝑎 ∧ 1 , 𝑏 ∧ 1 , … , 𝑎 ∧ 𝜈 , 𝑏 ∧ 𝜈 , ( 2 . 5 ) where 𝜈 ∈ ℕ 0 , 𝑎 ⋀ 𝜈 ∶ = 𝑎 𝜈 , 𝑏 ⋀ 𝜈 ∶ = 𝑏 𝜈 , and values 𝑎 𝜈 , 𝑏 𝜈 are specified in (2.2), (1.12), and (1.13). We calculate first 𝑃 ⋀ 𝑘 and 𝑄 ⋀ 𝑘 for 𝑘 = 0 , . . . , 6 .

Since 𝑃 ⋀ − 1 = 1 , 𝑄 ⋀ − 1 = 0 , it follows from (2.2) that

𝑃 ∧ 0 = 𝑏 0 = 1 , 𝑄 ∧ 0 𝑃 = 1 , ∧ 1 = 𝑏 ∧ 1 𝑃 ∧ 0 + 𝑎 ∧ 1 𝑃 ∧ − 1 = 5 , 𝑄 ∧ 1 = 𝑏 ∧ 1 𝑄 ∧ 0 + 𝑎 ∧ 1 𝑄 ∧ − 1 𝑃 = 4 , ∧ 2 = 𝑏 ∧ 2 𝑃 ∧ 1 + 𝑎 ∧ 2 𝑃 ∧ 0 = 2 4 = 4 𝑃 ∨ 1 , 𝑄 ( 2 . 6 ) ∧ 2 = 𝑏 ∧ 2 𝑄 1 + 𝑎 ∧ 2 𝑄 0 = 2 0 = 4 𝑄 ∨ 1 𝑃 , ( 2 . 7 ) ∧ 3 = 𝑏 ∧ 3 𝑃 ∧ 2 + 𝑎 ∧ 3 𝑃 ∧ 1 = 7 7 , 𝑄 ∧ 3 = 𝑏 ∧ 3 𝑄 ∧ 2 + 𝑎 ∧ 3 𝑄 ∧ 1 𝑃 = 6 4 , ∧ 4 = 𝑏 ∧ 4 𝑃 ∧ 3 + 𝑎 ∧ 4 𝑃 ∧ 2 = 2 5 0 , 𝑄 ∧ 4 = 𝑏 ∧ 4 𝑄 ∧ 3 + 𝑎 ∧ 4 𝑄 ∧ 2 𝑃 = 2 0 8 , ∧ 5 = 𝑏 ∧ 5 𝑃 ∧ 4 + 𝑎 ∧ 5 𝑃 ∧ 3 = 1 1 5 4 , 𝑄 ∧ 5 = 𝑏 ∧ 5 𝑄 ∧ 4 + 𝑎 ∧ 5 𝑄 ∧ 3 𝑃 = 9 6 0 , ( 2 . 8 ) ∧ 6 = 𝑏 ∧ 6 𝑃 ∧ 5 + 𝑎 ∧ 6 𝑃 ∧ 4 = 1 2 × 7 0 2 = 1 2 𝑃 ∨ 2 𝑄 , ( 2 . 9 ) ∧ 6 = 𝑏 ∧ 6 𝑄 ∧ 5 + 𝑎 ∧ 6 𝑄 ∧ 4 = 1 2 × 5 8 4 = 1 2 𝑄 ∨ 2 . ( 2 . 1 0 ) Let 𝑘 ∈ ℕ , 𝑘 ≥ 2 ,

𝑃 ∗ 𝑘 = 𝑃 ∧ 4 𝑘 − 2 2 ( 𝑘 + 1 ) ! , 𝑄 ∗ 𝑘 = 𝑄 ∧ 4 𝑘 − 2 . 2 ( 𝑘 + 1 ) ! ( 2 . 1 1 ) We want to to prove that if 𝑘 ∈ ℕ , then

𝑃 ∗ 𝑘 = 𝑃 ∨ 𝑘 , 𝑄 ∗ 𝑘 = 𝑄 ∨ 𝑘 . ( 2 . 1 2 ) 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:

𝑥 𝜈 + 1 − 𝑏 ∨ 𝜈 + 1 𝑥 𝜈 − 𝑎 ∨ 𝜈 + 1 𝑥 𝜈 − 1 𝑥 = 0 , ( 2 . 1 3 ) 𝜈 + 1 − 𝑏 ∧ 𝜈 + 1 𝑥 𝜈 − 𝑎 ∧ 𝜈 + 1 𝑥 𝜈 − 1 = 0 , ( 2 . 1 4 ) with 𝜈 ∈ ℕ 0 . Then 𝑥 𝜈 = 𝑃 ∨ 𝜈 , 𝑥 𝜈 = 𝑄 ∨ 𝜈 , with 𝜈 ∈ ( − 1 , + ∞ ) ∩ ℤ representing a fundamental system of solutions of (2.13), and 𝑥 𝜈 = 𝑃 ⋀ 𝜈 , 𝑥 𝜈 = 𝑄 ⋀ 𝜈 with 𝜈 ∈ ( − 1 , + ∞ ) ∩ ℤ representing a fundamental 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

𝑋 𝜈 + 1 = 𝐴 ∨ 𝜈 𝑋 𝜈 , 𝑋 ( 2 . 1 5 ) 𝜈 + 1 = 𝐴 ∧ 𝜈 𝑋 𝜈 , ( 2 . 1 6 ) where

𝑋 𝜈 = 𝑥 𝜈 − 1 𝑥 𝜈 𝐴 , ( 2 . 1 7 ) ∨ 𝜈 = 𝑎 0 1 ∨ 1 + 𝜈 𝑏 ∨ 1 + 𝜈 , 𝐴 ∧ 𝜈 = 𝑎 0 1 ∧ 1 + 𝜈 𝑏 ∧ 1 + 𝜈 , ( 2 . 1 8 ) and 𝜈 ∈ ℕ 0 . Let

𝑈 ∨ 𝜈 = 𝑃 ∨ 𝜈 − 1 𝑄 ∨ 𝜈 − 1 𝑃 ∨ 𝜈 𝑄 ∨ 𝜈 𝑈 , ( 2 . 1 9 ) ∧ 𝜈 = 𝑃 ∧ 𝜈 − 1 𝑄 ∧ 𝜈 − 1 𝑃 ∧ 𝜈 𝑄 ∧ 𝜈 , ( 2 . 2 0 ) with 𝜈 ∈ ℕ 0 be fundamental matrices of solutions of systems (2.15) and (2.16), respectively. Therefore,

𝑈 ∧ 𝜈 = 𝐴 ∧ 𝜈 − 1 𝑈 ∧ 𝜈 − 1 , 𝑈 ∨ 𝜈 = 𝐴 ∨ 𝜈 − 1 𝑈 ∨ 𝜈 − 1 ( 2 . 2 1 ) for 𝜈 ∈ ℕ . In view of (2.18) and (2.21), d e t ( 𝑈 𝜈 ) = − 𝑎 𝜈 d e t ( 𝑈 𝜈 − 1 ) , and therefore,

𝑈 d e t ∧ 𝜈 = ( − 1 ) 𝜈 𝑈 d e t ∧ 0 𝜈 𝑘 = 1 𝑎 ∧ 𝑘 = ( − 1 ) 𝜈 𝜈 𝑘 = 1 𝑎 ∧ 𝑘 . ( 2 . 2 2 ) Hence

𝑃 ∧ 𝜈 − 1 𝑄 ∧ 𝜈 − 1 − 𝑃 ∧ 𝜈 𝑄 ∧ 𝜈 = ( − 1 ) 𝜈 𝜈 ∏ 𝑘 = 1 𝑎 ∧ 𝑘 𝑄 ∧ 𝜈 𝑄 ∧ 𝜈 − 1 ( 2 . 2 3 ) (see [11]).

Further, we have

𝑈 ∨ 0 = 1 0 0 1 , 𝑈 ∨ 1 = 0 1 6 5 , 𝑈 ∨ 2 = , 𝑈 6 5 7 0 2 5 8 4 ∧ 0 = 1 0 0 1 , 𝑈 ∧ 1 = 1 1 5 4 , 𝑈 ∧ 2 = , 𝑈 5 4 2 4 2 0 ∧ 3 = 2 4 2 0 7 7 6 4 , 𝑈 ∧ 4 = , 𝑈 7 7 6 4 2 5 0 2 0 8 ∧ 5 = 2 5 0 2 0 8 1 1 5 4 9 6 0 , 𝑈 ∧ 6 = , 𝑈 1 1 5 4 9 6 0 8 4 2 4 7 0 0 8 ( 2 . 2 4 ) ∨ 1 𝑈 ∧ 2 − 1 = 1 4 𝑈 − 2 4 5 0 1 , ( 2 . 2 5 ) ∨ 2 𝑈 ∧ 6 − 1 = 1 9 6 − 3 6 5 0 8 . ( 2 . 2 6 ) Let 𝑘 ∈ ℕ , 𝑘 ≥ 2 . Then, in view of (2.20),

𝐴 ∧ 4 𝑘 − 6 = 𝑎 0 1 ∧ 4 ( 𝑘 − 2 ) + 3 𝑏 ∧ 4 ( 𝑘 − 2 ) + 3 = 0 1 ( 𝑘 − 1 ) 2 , 𝐴 2 𝑘 − 1 ∧ 4 𝑘 − 5 = 𝑎 0 1 ∧ 4 ( 𝑘 − 2 ) + 4 𝑏 ∧ 4 ( 𝑘 − 2 ) + 4 = 𝑘 0 1 2 , 𝐴 2 𝑘 − 2 ∧ 4 𝑘 − 4 = 𝑎 0 1 ∧ 4 ( 𝑘 − 1 ) + 1 𝑏 ∧ 4 ( 𝑘 − 1 ) + 1 = 𝑘 0 1 2 , 𝐴 − 𝑘 2 𝑘 ∧ 4 𝑘 − 3 = 𝑎 0 1 ∧ 4 ( 𝑘 − 1 ) + 2 𝑏 ∧ 4 ( 𝑘 − 1 ) + 2 = 𝑘 0 1 2 . + 𝑘 2 𝑘 + 2 ( 2 . 2 7 ) Let 𝑌 𝑘 = 𝑋 4 𝑘 − 6 for 𝑘 ∈ [ 2 , + ∞ ) ∩ ℤ . In view of (2.16) and (2.18),

𝑌 𝑘 + 1 = 𝐵 ∧ 𝑘 𝑌 𝑘 , 𝑈 ( 2 . 2 8 ) ∧ 4 𝑘 − 2 = 𝐵 ∧ 𝑘 𝑈 ∧ 4 𝑘 − 6 , ( 2 . 2 9 ) where, as before, 𝑘 ∈ [ 2 , + ∞ ) ∩ ℤ ,

𝐵 ∧ 𝑘 = 𝐴 ∧ 4 𝑘 − 3 𝐴 ∧ 4 𝑘 − 4 𝐴 ∧ 4 𝑘 − 5 𝐴 4 𝑘 − 6 = 5 𝑘 ( 𝑘 − 1 ) 3 𝑘 1 2 𝑘 2 − 1 5 𝑘 + 5 1 2 𝑘 ( 𝑘 + 1 ) ( 𝑘 − 1 ) 3 𝑘 ( 𝑘 + 1 ) 2 9 𝑘 2 − 3 6 𝑘 + 1 2 . ( 2 . 3 0 ) In view of (2.22), (2.2), (1.12), (1.13), (2.29), and (2.28), the matrix 𝑈 ⋀ 4 𝑘 − 6 is a fundamental matrix of solutions of system (2.28). The substitution 𝑍 𝑘 = 𝐶 𝑘 𝑌 𝑘 , with d e t ( 𝐶 𝑘 ) ≠ 0 for 𝑘 ∈ [ 2 , + ∞ ) ∩ ℤ , transforms the system (2.28) into the system

𝑍 𝑘 + 1 = 𝐷 𝑘 𝑍 𝑘 , ( 2 . 3 1 ) with 𝐷 𝑘 = 𝐶 𝑘 + 1 𝐵 ⋀ 𝑘 ( 𝐶 𝑘 ) − 1 for 𝑘 ∈ [ 2 , + ∞ ) ∩ ℤ . We prove now that if we take 𝑘 ∈ [ 3 , + ∞ ) ∩ ℤ , and 𝐶 𝑘 = 𝐻 𝑘 − 1 , where

𝐻 1 = 1 4 𝐻 − 2 4 5 0 1 , ( 2 . 3 2 ) 𝑘 = 1 2 ( 𝑘 + 2 ) ( 𝑘 + 1 ) 𝑐 ( 𝑘 + 1 ) − 5 ( 𝑘 + 2 ) 𝑐 ( 𝑘 + 1 ) 0 − ( 𝑘 − 1 ) 3 𝑐 ( 𝑘 ) , ( 2 . 3 3 ) with 𝑘 ∈ [ 2 , + ∞ ) ∩ ℤ and 𝑐 ( 𝑘 ) = ( − 2 ( 𝑘 − 1 ) 3 ( 𝑘 + 1 ) ! ) − 1 , then we obtain the equality 𝐷 𝑘 = 𝐴 ∨ 𝑘 − 1 . So, let 𝑘 ∈ [ 3 , + ∞ ) ∩ ℤ . Then, in view of (2.33),

𝐻 𝑘 − 1 = 1 2 ( 𝑘 + 1 ) 𝑘 𝑐 ( 𝑘 ) − 5 ( 𝑘 + 1 ) 𝑐 ( 𝑘 ) 0 − ( 𝑘 − 2 ) 3 𝑐 ( 𝑘 − 1 ) . ( 2 . 3 4 ) In view of(1.9)

𝑏 ∨ 𝑘 = 3 4 ( 𝑘 − 1 ) 3 + 5 1 ( 𝑘 − 1 ) 2 + 2 7 ( 𝑘 − 1 ) + 5 = 3 4 𝑘 3 − 5 1 𝑘 2 + 2 7 𝑘 − 5 , 𝑎 ∨ 𝑘 = − ( 𝑘 − 1 ) 6 , ( 2 . 3 5 ) where 𝑘 ∈ [ 3 , + ∞ ) ∩ ℤ . Hence, in view of (2.19),

𝐴 ∨ 𝑘 − 1 = 0 1 − ( 𝑘 − 1 ) 6 3 4 𝑘 3 − 5 1 𝑘 2 + 2 7 𝑘 − 5 . ( 2 . 3 6 ) In view of (2.34)–(2.36),

𝐴 ∨ 𝑘 − 1 𝐻 𝑘 − 1 = 0 1 − ( 𝑘 − 1 ) 6 3 4 𝑘 3 − 5 1 𝑘 2 × + 2 7 𝑘 − 5 1 2 ( 𝑘 + 1 ) 𝑘 𝑐 ( 𝑘 ) − 5 ( 𝑘 + 1 ) 𝑐 ( 𝑘 ) 0 − ( 𝑘 − 2 ) 3 = 𝑐 ( 𝑘 − 1 ) 0 − ( 𝑘 − 2 ) 3 𝑐 ( 𝑘 − 1 ) − ( 𝑘 − 1 ) 6 1 2 ( 𝑘 + 1 ) 𝑘 𝑐 ( 𝑘 ) ( 𝑘 − 1 ) 6 5 ( 𝑘 + 1 ) 𝑐 ( 𝑘 ) − 𝑏 ∨ 𝑘 ( 𝑘 − 2 ) 6 . 𝑐 ( 𝑘 − 1 ) . ( 2 . 3 7 ) In view of (2.30) and (2.33),

𝐻 𝑘 𝐵 ∧ 𝑘 = 1 2 ( 𝑘 + 2 ) ( 𝑘 + 1 ) 𝑐 ( 𝑘 + 1 ) − 5 ( 𝑘 + 2 ) 𝑐 ( 𝑘 + 1 ) 0 − ( 𝑘 − 1 ) 3 × 𝑐 ( 𝑘 ) 5 𝑘 ( 𝑘 − 1 ) 3 𝑘 1 2 𝑘 2 − 1 5 𝑘 + 5 1 2 𝑘 ( 𝑘 + 1 ) ( 𝑘 − 1 ) 3 𝑘 ( 𝑘 + 1 ) 2 9 𝑘 2 = 0 − 3 6 𝑘 + 1 2 ( 𝑘 + 2 ) 𝑐 ( 𝑘 + 1 ) 𝑘 ( 𝑘 + 1 ) − 𝑘 2 − 𝑐 ( 𝑘 ) 1 2 𝑘 ( 𝑘 + 1 ) ( 𝑘 − 1 ) 6 − ( 𝑘 − 1 ) 3 𝑐 ( 𝑘 ) 𝑘 ( 𝑘 + 1 ) 2 9 𝑘 2 . − 3 6 𝑘 + 1 2 ( 2 . 3 8 ) Since

− ( 𝑘 + 2 ) ( 𝑘 + 1 ) 𝑐 ( 𝑘 + 1 ) 𝑘 3 = − 𝑐 ( 𝑘 − 1 ) ( 𝑘 − 2 ) 3 , − ( 𝑘 − 1 ) 3 𝑐 ( 𝑘 ) 𝑘 ( 𝑘 + 1 ) 2 9 𝑘 2 − 3 6 𝑘 + 1 2 − ( 𝑘 − 1 ) 6 5 ( 𝑘 + 1 ) 𝑐 ( 𝑘 ) = − 3 4 𝑘 3 − 5 1 𝑘 2 + 2 7 𝑘 − 5 ( 𝑘 − 1 ) 3 ( 𝑘 + 1 ) 𝑐 ( 𝑘 ) = − 3 4 𝑘 3 − 5 1 𝑘 2 + 2 7 𝑘 − 5 ( 𝑘 − 2 ) 3 𝑐 ( 𝑘 − 1 ) , ( 2 . 3 9 ) it follows from (2.35), (2.37), and (2.38) that

𝐴 ∨ 𝑘 − 1 𝐻 𝑘 − 1 = 𝐻 𝑘 𝐵 ∧ 𝑘 ( 2 . 4 0 ) for 𝑘 ∈ [ 3 , + ∞ ) ∩ ℤ . We prove by induction now the following equality:

𝑈 ∨ 𝑘 = 𝐻 𝑘 𝑈 ∧ 4 𝑘 − 2 , ( 2 . 4 1 ) 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),

𝐻 𝑘 𝑈 ∧ 4 𝑘 − 2 = 𝐻 𝑘 𝐵 𝑘 𝑈 ∧ 4 𝑘 − 6 = 𝐴 ∨ 𝑘 − 1 𝐻 𝑘 − 1 𝑈 ∧ 4 𝑘 − 6 = 𝐴 ∨ 𝑘 − 1 𝑈 ∨ 𝑘 − 1 = 𝑈 ∨ 𝑘 . ( 2 . 4 2 ) So, the equality (2.41) holds for any 𝑘 ∈ ℕ . In view of (2.41),

𝑃 ∨ 𝑘 = ( 2 ( 𝑘 + 1 ) ! ) − 1 𝑃 ∧ 4 𝑘 − 2 , 𝑄 ∨ 𝑘 = ( 2 ( 𝑘 + 1 ) ! ) − 1 𝑄 ∧ 4 𝑘 − 2 ( 2 . 4 3 ) for 𝑘 ∈ [ 2 , + ∞ ) ∩ ℤ . Since

𝑃 ∨ 𝜈 = ( 𝜈 ! ) 3 𝑣 𝜈 , 𝑄 ∨ 𝜈 = ( 𝜈 ! ) 3 𝑢 𝜈 ( 2 . 4 4 ) for 𝑣 𝜈 and 𝑢 𝜈 in (1.6) and 𝜈 ∈ ℕ 0 , it follows from (2.43) and (2.44), that

𝑃 ∧ 4 𝑘 − 2 = 2 ( 𝑘 + 1 ) ( 𝑘 ! ) 4 𝑣 𝑘 , 𝑄 ∧ 4 𝑘 − 2 = 2 ( 𝑘 + 1 ) ( 𝑘 ! ) 4 𝑢 𝑘 . ( 2 . 4 5 ) As it is well known, for any 𝜀 > 0 there exist 𝐶 1 ( 𝜀 ) > 0 and 𝐶 2 ( 𝜀 ) > 0 such that

𝐶 1 √ ( 𝜀 ) 1 + 2 4 𝑘 ( 1 − 𝜀 ) < | | 𝑢 𝑘 | | < 𝐶 2 √ ( 𝜀 ) 1 + 2 4 𝑘 ( 1 + 𝜀 ) , 𝐶 ( 2 . 4 6 ) 1 √ ( 𝜀 ) 1 + 2 4 𝑘 ( 1 − 𝜀 ) < | | 𝑣 𝑘 | | < 𝐶 2 √ ( 𝜀 ) 1 + 2 4 𝑘 ( 1 + 𝜀 ) 𝐶 , ( 2 . 4 7 ) 1 ( 𝜀 ) √ 1 + 2 8 𝑘 ( 1 + 𝜀 ) < | | | | 𝑣 𝜁 ( 3 ) − 𝑘 𝑢 𝑘 | | | | < 𝐶 2 ( 𝜀 ) √ 1 + 2 8 𝑘 ( 1 − 𝜀 ) . ( 2 . 4 8 ) 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 ≤ 4 𝑘 − 2 + 𝜂 𝜅 = 1 𝑎 𝜅 ≤ 4 𝑘 + 1 𝜅 = 1 𝑎 𝜅 ≤ 𝑎 4 𝑘 − 1 𝑎 4 𝑘 𝑎 4 𝑘 + 1 × 𝑘 3 ( 𝑘 + 1 ) 3 4 𝑘 − 2 𝜅 = 1 𝑎 𝜅 = 4 𝑘 3 ( 𝑘 + 1 ) 3 𝑘 𝜅 = 2 𝑎 4 𝜅 − 5 𝑎 4 𝜅 − 4 𝑎 4 𝜅 − 3 𝑎 4 𝜅 − 2 = 4 𝑘 3 ( 𝑘 + 1 ) 3 𝑘 𝜅 = 2 ( 𝜅 − 1 ) 2 𝜅 2 ( 𝜅 − 1 ) 𝜅 𝜅 ( 𝜅 + 1 ) = 2 ( 𝑘 ! ) 8 ( 𝑘 + 1 ) 4 , ( 2 . 4 9 ) 4 ( 𝑘 + 1 ) 2 ( 𝑘 ! ) 8 𝑢 2 𝑘 = 𝑄 4 𝑘 − 2 2 < 𝑄 4 𝑘 − 3 + 𝜂 𝑄 4 𝑘 − 2 + 𝜂 . ( 2 . 5 0 ) In view of (2.23), (2.50), and (2.49), if 𝜃 = 1 , 2 , 3

| | | | 𝑃 4 𝑘 − 2 𝑄 4 𝑘 − 2 − 𝑃 4 𝑘 − 2 + 𝜃 𝑄 4 𝑘 − 1 + 𝜃 | | | | ≤ 𝜃 𝜂 = 1 | | | | 𝑃 4 𝑘 − 3 + 𝜂 𝑄 4 𝑘 − 3 + 𝜂 − 𝑃 4 𝑘 − 2 + 𝜂 𝑄 4 𝑘 − 2 + 𝜂 | | | | ≤ 3 𝜂 = 1 | | | | 𝑃 4 𝑘 − 3 + 𝜂 𝑄 4 𝑘 − 3 + 𝜂 − 𝑃 4 𝑘 − 2 + 𝜂 𝑄 4 𝑘 − 2 + 𝜂 | | | | ( ≤ 3 𝑘 + 1 ) 2 2 𝑢 2 𝑘 ≤ √ 1 + 2 8 𝑘 ( − 1 + 𝑜 ( 1 ) ) , ( 2 . 5 1 ) when 𝑘 → + ∞ . In view of (2.45), (2.48), and (2.51), there exist 𝐶 3 ( 𝜀 ) > 0 and 𝐶 4 ( 𝜀 ) > 0 such that

𝐶 3 ( 𝜀 ) √ 1 + 2 8 𝑘 ( 1 + 𝜀 ) < | | | | 𝜁 𝑃 ( 3 ) − ∧ 4 𝑘 − 2 + 𝜃 𝑄 ∧ 4 𝑘 − 2 | | | | < 𝐶 4 ( 𝜀 ) √ 1 + 2 8 𝑘 ( 1 − 𝜀 ) , ( 2 . 5 2 ) where 𝜃 = 0 , 1 , 2 , 3 . So, the equality (2.1) is proved. In view of (2.23),

𝑃 𝜁 ( 3 ) − ∧ 0 𝑄 ∧ 0 = ∞ 𝜈 = 1 ( − 1 ) 𝜈 − 1 𝑑 𝜈 , ( 2 . 5 3 ) where

0 < 𝑑 𝜈 = 𝜈 ∏ 𝑘 = 1 𝑎 ∧ 𝑘 ( 𝑄 ∧ 𝜈 𝑄 𝜈 − 1 ) ∧ . ( 2 . 5 4 )

Further, we have

𝑑 𝜈 + 1 𝑑 𝜈 = 𝑎 𝜈 + 1 ∧ 𝑄 𝜈 − 1 ∧ 𝑏 ∧ 𝜈 + 1 𝑄 ∧ 𝜈 + 𝑎 ∧ 𝜈 + 1 𝑄 ∧ 𝜈 − 1 < 1 . ( 2 . 5 5 ) Hence, the series (2.53) is the series of Leibnitz type. Therefore, 𝑃 ⋀ 2 𝑘 − 1 / 𝑄 ⋀ 2 𝑘 − 1 decreases, when 𝑘 increases in ℕ , and 𝑃 ⋀ 2 𝑘 / 𝑄 ⋀ 2 𝑘 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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