New properties for the Ramanujan R-function

Chuan-Yu Cai; Lu Chen; Ti-Ren Huang; Yuming Chu

doi:10.1515/math-2022-0045

Open Access Published by De Gruyter Open Access September 1, 2022

New properties for the Ramanujan R-function

Chuan-Yu Cai , Lu Chen , Ti-Ren Huang and Yuming Chu

From the journal Open Mathematics

https://doi.org/10.1515/math-2022-0045

Abstract

In the article, we establish some monotonicity and convexity (concavity) properties for certain combinations of polynomials and the Ramanujan R-function by use of the monotone form of L’Hôpital’s rule and present serval new asymptotically sharp bounds for the Ramanujan R-function that improve some previously known results.

Keywords: Ramanujan R-function; Gaussian hypergeometric function; Riemann zeta function; generalized elliptic integral; monotonicity; convexity

MSC 2010: 33C05; 33B15; 26A51; 26D20

1 Introduction

Let Re ( a ) > 0 and Re ( b ) > 0 . Then, the classical gamma function Γ ( ⋅ ) [1,2, 3,4], psi function ψ ( ⋅ ) [5,6, 7,8], beta function B ( ⋅ , ⋅ ) [9,10], and Euler-Mascheroni constant γ [11,12,13, 14,15] are defined by

(1.1) Γ ( a ) = ∫ 0 ∞ t a − 1 e − t d t , ψ ( a ) = Γ ′ ( a ) Γ ( a ) , B ( a , b ) = Γ ( a ) Γ ( b ) Γ ( a + b )

and

γ = lim n → ∞ ∑ k = 1 n 1 k − log n = 0.57721566 … ,

respectively, and the Ramanujan R -function R ( ⋅ ) [16,17,18] is defined by

(1.2) R ( a ) = − 2 γ − ψ ( a ) − ψ ( 1 − a ) , a ∈ ( 0 , 1 ) ,

and it is the special case of the two parameter function

(1.3) R ( a , b ) = − 2 γ − ψ ( a ) − ψ ( b ) ,

and sometimes R ( a , b ) is also said to be the Ramanujan constant although it is a function of a and b .

For a ∈ ( 0 , 1 ) , we denote

(1.4) B ( a ) = B ( a , 1 − a ) = Γ ( a ) Γ ( 1 − a ) = π sin ( π a ) .

By the symmetry, in what follows, we assume that a ∈ ( 0 , 1 / 2 ] in (1.2) and (1.4).

For a , b , c ∈ R with c ≠ 0 , − 1 , − 2 , … , the Gaussian hypergeometric function [19,20,21, 22,23,24, 25,26,27, 28,29,30] is defined by

(1.5) F ( a , b ; c ; x ) = F 1 2 ( a , b ; c ; x ) = ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n x n n ! , x ∈ ( − 1 , 1 ) ,

where ( x ) n denotes the shifted factorial function defined by ( x ) n = x ( x + 1 ) ⋯ ( x + n − 1 ) for n ≥ 1 and ( x ) 0 = 1 for x ≠ 0 . It is well known that F ( a , b ; c ; x ) has wide applications in many branches of mathematics and physics. Many elementary and special functions in mathematical physics are the particular or limiting cases of the Gaussian hypergeometric function. F ( a , b ; c ; x ) is said to be zero-balanced if c = a + b . The asymptotic properties of F ( a , b ; a + b ; x ) are related to B ( a , b ) and R ( a , b ) as x → 1 [31,32,33, 34,35,36]. For example, F ( a , b ; a + b ; x ) satisfies the Ramanujan asymptotic identity [37]:

(1.6) B ( a , b ) F ( a , b ; a + b ; x ) + log ( 1 − x ) = R ( a , b ) + O ( ( 1 − x ) log ( 1 − x ) )

when x → 1 .

Let a , r ∈ ( 0 , 1 ) . Then, the generalized elliptic integral K a ( r ) [38,39,40, 41,42,43, 44,45,46, 47,48,49, 50,51,52] of the first kind is defined by

(1.7) K a ( r ) = π 2 F ( a , 1 − a ; 1 ; r 2 ) , K a ( 0 + ) = π 2 , K a ( 1 − ) = ∞ .

For the sake of simplicity and convenience, in what follows, we denote r ′ = 1 − r 2 for r ∈ [ 0 , 1 ] and K ′ ( r ) = K a ( r ′ ) . By the symmetry, we only need to discuss the case of a ∈ ( 0 , 1 / 2 ] . From (1.6), we obtain the asymptotic identity:

(1.8) B ( a ) K a ( r ) − π log e R ( a ) / 2 r ′ = O ( ( r ′ ) 2 log ( r ′ ) )

when r → 1 .

Recently, the properties and bounds involving K a ( r ) and R ( a ) have attracted the attention of many researchers. Wang et al. [53] proved that the double inequality

(1.9) π R ( a ) + [ B ( a ) − R ( a ) ] r 2 < K a ( r ) log ( e R ( a ) / 2 / r ′ ) < π B ( a ) [ 1 − a ( 1 − a ) + a ( 1 − a ) r 2 ]

holds for all a ∈ ( 0 , 1 / 2 ] and r ∈ ( 0 , 1 ) .

In [54], the authors established the following two-sided inequality:

sin ( π a ) 1 − b + b r 2 log e R ( a ) / 2 r ′ − C ≤ K a ( r ) ≤ sin ( π a ) 1 − b + b r 2 log e R ( a ) / 2 r ′ − C r ′ 2 ,

for all a ∈ ( 0 , 1 / 2 ] and r ∈ ( 0 , 1 ) , where C = [ R ( a ) − ( 1 − b ) B ( a ) ] / 2 and b = a ( 1 − a ) .

Qiu et al. [17] presented the power series expansions of the function R ( a ) at a = 0 and a = 1 / 2 , and provided the monotonicity and convexity properties of certain combinations defined in terms of polynomials and the difference between the Ramanujan R -function R ( a ) and the beta function B ( a ) , and proved that

(1.10) 2 ( 2 a 2 n + α 2 n − 1 ) x 2 n + 1 ≤ B ( x ) − R ( x ) + 2 ∑ k = 1 2 n a k x k ≤ ( 2 a 2 n + α 2 n − 1 ) x 2 n

and

(1.11) ( 2 a 2 n + 1 + α 2 n ) x 2 n + 1 ≤ B ( x ) − R ( x ) + 2 ∑ k = 1 2 n + 1 a k x k ≤ 2 ( 2 a 2 n + 1 + α 2 n ) x 2 ( n + 1 )

for n ∈ N and x ∈ ( 0 , 1 / 2 ] , where

a n = { ( − 1 ) n + [ 1 + ( − 1 ) n + 1 ] 2 − n − 1 } ζ ( n + 1 ) ,

α n = 2 n + 1 π − log 16 + 2 ∑ k = 1 n 2 − k a k

and

(1.12) ζ ( s ) = ∑ k = 1 ∞ 1 k s ( Re ( s ) > 1 )

is the Riemann zeta function [55,56].

Huang et al. [54] established several monotonicity, concavity, and convexity properties for certain combinations defined in terms of R ( a ) and b = a ( 1 − a ) , and proved that

(1.13) ( 7 log 2 − 3 ) b ≤ b ( 2 − b ) R ( a ) + b − 1 ,

(1.14) A 2 b ≤ 1 − b 2 − b ( 1 − b ) R ( a ) < 2 b ,

(1.15) log 2 + D 1 ( 1 − 2 a ) 2 − D 2 ( 1 − 2 a ) 4 − D 3 ( 1 − 2 a ) 5 ≤ b R ( a ) ≤ log 2 + ( 1 − log 2 ) ( 1 − 2 a ) ,

where D 1 = λ ( 3 ) − log 2 , D 2 = λ ( 3 ) − λ ( 5 ) , D 3 = λ ( 5 ) − 1 , A 2 = 15 / 4 − log 8 and

(1.16) λ ( n + 1 ) = ∑ k = 0 ∞ 1 ( 2 k + 1 ) n + 1 .

From [31], one has

(1.17) λ ( n + 1 ) = ( 1 − 2 − n − 1 ) ζ ( n + 1 ) .

We also denote

(1.18) β ( n ) = ∑ k = 0 ∞ ( − 1 ) k 1 ( 2 k + 1 )

in the full article.

The Ramanujan R -function often appears in the studies of the properties of the generalized Grötzsch ring function μ a ( r ) [57,58], the generalized Hersch-Pfluger distortion function φ K a ( r ) [59,60,61], and the generalized η K -distortion function η K a ( t ) [62]. Therefore, the function R ( a ) is indispensable for studying the properties of K a ( r ) and many other special functions [63].

The main purpose of this article is to establish new asymptotically sharp lower and upper bounds for the Ramanujan R -function R ( a ) and to improve inequalities (1.14) and (1.15).

2 Lemmas

To prove our main results, we need several technical lemmas, which we present in this section.

Lemma 2.1

(See [54, Lemma 2.1]) The following two statements are true:

The function λ ( n ) defined by (1.17) is strictly decreasing with respect to n ∈ N \ { 1 } with λ ( 2 ) = π 2 / 8 and lim n → ∞ λ ( n ) = 1 , and the function β ( n ) defined by (1.18) is strictly increasing with respect to n ∈ N with β ( 1 ) = π / 4 and lim n → ∞ β ( n ) = 1 .
The following identities
(2.1) R ( a ) = 1 a + 2 ∑ n = 1 ∞ ζ ( 2 n + 1 ) a 2 n = log 16 + 4 ∑ n = 1 ∞ λ ( 2 n + 1 ) ( 1 − 2 a ) 2 n = log 16 + 4 ∑ n = 1 ∞ ( 1 − 2 − 2 n − 1 ) ζ ( 2 n + 1 ) ( 1 − 2 a ) 2 n = 1 b − 4 ( 1 − log 2 ) + 4 ∑ n = 1 ∞ [ λ ( 2 n + 1 ) − 1 ] ( 1 − 2 a ) 2 n
and
(2.2) B ( a ) = 1 a + 2 ∑ n = 1 ∞ ( 1 − 2 1 − 2 n ) ζ ( 2 n ) a 2 n − 1 = 4 ∑ n = 1 ∞ β ( 2 n + 1 ) ( 1 − 2 a ) 2 n
hold for all a ∈ ( 0 , 1 / 2 ] .

Lemma 2.2

The function f ( s ) = − 2 ζ ( 2 s + 3 ) + ζ ( 2 s + 5 ) + 9 ζ ( 2 s + 7 ) / 8 is strictly increasing from [ 1 , ∞ ) onto [ f ( 1 ) , 1 / 8 ) , where f ( 1 ) = − 2 ζ ( 5 ) + ζ ( 7 ) + 9 ζ ( 9 ) / 8 = 0.0617 … .

Proof

It follows from (1.12) that

f ( s ) = − 2 ζ ( 2 s + 3 ) + ζ ( 2 s + 5 ) + 9 8 ζ ( 2 s + 7 ) = ∑ k = 1 ∞ − 2 k − ( 2 s + 3 ) + k − ( 2 s + 5 ) + 9 8 k − ( 2 s + 7 ) = ∑ k = 1 ∞ − 2 k − 3 + k − 5 + 9 8 k − 7 k − 2 s = ∑ k = 1 ∞ − 16 k 4 + 8 k 2 + 9 8 k 7 k − 2 s .

Differentiating f ( s ) leads to

f ′ ( s ) = ∑ k = 1 ∞ ( − 2 log k ) − 16 k 4 + 8 k 2 + 9 8 k 7 k − 2 s = ∑ k = 2 ∞ 2 log k 16 k 4 − 8 k 2 − 9 8 k 7 k − 2 s .

Therefore, f ′ ( s ) > 0 for all s > 0 and f is strictly increasing due to 16 k 4 − 8 k 2 − 9 > 0 and log k > 0 for k ≥ 2 . Note that f ( 1 ) = − 2 ζ ( 5 ) + ζ ( 7 ) + 9 ζ ( 9 ) / 8 = 0.06175 … and f ( ∞ ) = 1 / 8 .□

The following monotone form of L’Hôpita’s rule can be found in the literature [[32], Theorem 1.25].

Lemma 2.3

Let − ∞ < a < b < ∞ , f , g : [ a , b ] → R be continuous on [ a , b ] and be differentiable on ( a , b ) , and g ′ ( x ) ≠ 0 on ( a , b ) . If f ′ ( x ) / g ′ ( x ) is increasing (decreasing) on ( a , b ) , then so are the functions

[ f ( x ) − f ( a ) ] / [ g ( x ) − g ( a ) ] , [ f ( x ) − f ( b ) ] / [ g ( x ) − g ( b ) ] .

If f ′ ( x ) / g ′ ( x ) is strictly monotone, then the monotonicity in the conclusion is also strict.

3 Main results

Throughout this paper, we always assume that a ∈ ( 0 , 1 / 2 ] and b = a ( 1 − a ) . Let

(3.1) A 1 ( 2 ) = log 2 − λ ( 3 ) , A 1 ( 2 k ) = λ ( 2 k − 1 ) − λ ( 2 k + 1 ) ( k ≥ 2 ) , B 1 ( 1 ) = 1 , B 1 ( 2 ) = 0 , B 1 ( 2 k − 1 ) = − 2 ζ ( 2 k − 1 ) , B 1 ( 2 k ) = 2 ζ ( 2 k − 1 ) ( k ≥ 2 ) .

Theorem 3.1

Let n ∈ N and f 1 ( a ) = − b R ( a ) = − a ( 1 − a ) R ( a ) . Then one has

f 1 has the power series formula
(3.2) f 1 ( a ) = − log 2 + ∑ n = 1 ∞ A 1 ( 2 n ) ( 1 − 2 a ) 2 n = − 1 + ∑ n = 1 ∞ B 1 ( n ) a n .
f 1 is increasing, concave, and ( − 1 ) k f 1 ( k ) ( a ) > 0 on ( 0 , 1 / 2 ) for k ≥ 3 .
Let
(3.3) F 1 , n ( a ) = f 1 ( a ) − P 1 , n ( a ) ( 1 − 2 a ) 2 ( n + 1 ) ,
where P 1 , n ( a ) = − log 2 + ∑ k = 1 n A 1 ( 2 k ) ( 1 − 2 a ) 2 k and A 1 ( 2 k ) ( k ≥ 1 ) is defined by (3.1). Then, F 1 , n is strictly decreasing and convex from ( 0 , 1 / 2 ) onto ( A 1 ( 2 n + 2 ) , λ ( 2 n + 1 ) − 1 ) . In particular, the double inequality
(3.4) 0 ≤ − b R ( a ) − P 1 , n ( a ) − A 1 ( 2 n + 2 ) ( 1 − 2 a ) 2 n + 2 ≤ [ λ ( 2 n + 1 ) − 1 − A 1 ( 2 n + 2 ) ] ( 1 − 2 a ) 2 n + 3
holds for all n ∈ N and a ∈ ( 0 , 1 / 2 ] , with equality in each instance if and only if a = 1 / 2 .
Let
(3.5) G 1 , n ( a ) = f 1 ( a ) − R 1 , n ( a ) a n + 1 ,
where R 1 , n ( a ) = − 1 + ∑ k = 1 n B 1 ( k ) a k and B 1 ( k ) ( k ≥ 1 ) is defined by (3.1). Then, G 1 , 2 n ( G 1 , ( 2 n + 1 ) ) is strictly increasing (decreasing) and concave (convex) on ( 0 , 1 / 2 ] with G 1 , n ( 0 + ) = B 1 ( n + 1 ) . In particular, the inequalities
(3.6) 2 ( G 1 , 2 n ( 1 / 2 ) + 2 ζ ( 2 n + 1 ) ) a 2 n + 2 ≤ − b R ( a ) − R 1 , 2 n ( a ) + 2 ζ ( 2 n + 1 ) a 2 n + 1 ≤ ( G 1 , 2 n ( 1 / 2 ) + 2 ζ ( 2 n + 1 ) ) a 2 n + 1
and
(3.7) ( G 1 , ( 2 n + 1 ) ( 1 / 2 ) − 2 ζ ( 2 n + 1 ) ) a 2 n + 2 ≤ − b R ( a ) − R 1 , ( 2 n + 1 ) ( a ) − 2 ζ ( 2 n + 1 ) a 2 n + 2 ≤ 2 ( G 1 , ( 2 n + 1 ) ( 1 / 2 ) − 2 ζ ( 2 n + 1 ) ) a 2 n + 3
hold for all n ∈ N and a ∈ ( 0 , 1 / 2 ] , with equality in each instance if and only if a = 1 / 2 .

Proof

(1) It follows from Lemma 2.1(2) and b = a ( 1 − a ) = [ 1 − ( 1 − 2 a ) 2 ] / 4 that

(3.8) f 1 ( a ) = − a ( 1 − a ) R ( a ) = − 1 − ( 1 − 2 a ) 2 4 log 16 + 4 ∑ n = 1 ∞ λ ( 2 n + 1 ) ( 1 − 2 a ) 2 n = − log 2 + ∑ n = 1 ∞ A 1 ( 2 n ) ( 1 − 2 a ) 2 n .

On the other hand, from Lemma 2.1(2), f 1 ( a ) can be expressed by

(3.9) f 1 ( a ) = − a ( 1 − a ) R ( a ) = − a ( 1 − a ) 1 a + 2 ∑ n = 1 ∞ ζ ( 2 n + 1 ) a 2 n = − 1 + ∑ n = 1 ∞ B 1 ( n ) a n .

(2) From Lemma 2.1(1), we know that A 1 ( 2 n ) > 0 for all n ≥ 2 . Differentiating f 1 gives

(3.10) f 1 ( k ) ( a ) = ∑ n = [ ( k + 1 ) / 2 ] ∞ ( − 1 ) k 2 k ( 2 n ) ! ( 2 n − k ) ! A 1 ( 2 n ) ( 1 − 2 a ) 2 n − k ,

where and in what follows [ ⋅ ] denotes the greatest integral function. Therefore, ( − 1 ) k f 1 ( k ) ( a ) ≥ 0 for all a ∈ ( 0 , 1 / 2 ] and k = 3 , 4 , … and f 1 ″ is decreasing on ( 0 , 1 / 2 ] due to f 1 ‴ ( a ) ≤ 0 . It follows from (3.9) that

f 1 ″ ( a ) = ∑ n = 2 ∞ n ( n − 1 ) B 1 ( n ) a n − 2

with f 1 ″ ( 0 + ) = 0 . Therefore, f 1 ″ ( a ) < 0 for a ∈ ( 0 , 1 / 2 ] , so that f 1 ′ ( a ) is strictly decreasing on ( 0 , 1 / 2 ] with f 1 ′ ( 1 / 2 ) = 0 , which leads to the desired monotonicity and concavity of f 1 .

(3) Differentiating

F 1 , n ( a ) = f 1 ( a ) − P 1 , n ( a ) ( 1 − 2 a ) 2 ( n + 1 ) = ∑ k = n + 1 ∞ A 1 ( 2 k ) ( 1 − 2 a ) 2 ( k − n − 1 ) = ∑ k = 0 ∞ A 1 ( 2 ( k + n + 1 ) ) ( 1 − 2 a ) 2 k

gives

(3.11) F 1 , n ( l ) ( a ) = ∑ k = [ ( l + 1 ) / 2 ] ∞ ( − 1 ) l 2 k ( 2 k ) ! ( 2 k − l ) ! A 1 ( 2 ( k + n + 1 ) ) ( 1 − 2 a ) 2 k − l ,

which implies that

F 1 , n ′ ( a ) = − 4 ∑ k = 1 ∞ k A 1 ( 2 ( k + n + 1 ) ) ( 1 − 2 a ) 2 k − 1

and

F 1 , n ″ ( a ) = 8 ∑ k = 1 ∞ k ( 2 k − 1 ) A 1 ( 2 ( k + n + 1 ) ) ( 1 − 2 a ) 2 k − 2 .

Therefore, the monotonicity and concavity properties of F 1 , n follow easily from A 1 ( 2 n ) > 0 for all n ≥ 2 . Clearly, F 1 , n ( 0 + ) = λ ( 2 n + 1 ) − 1 , F 1 , n ( 1 / 2 − ) = A 1 ( 2 n + 2 ) , the double inequality follows from the monotonicity and convexity properties of F 1 , n ( a ) .

(4) Let G 1 , n ( a ) = h 1 ( a ) / h 2 ( a ) , where h 2 ( a ) = a n + 1 and

(3.12) h 1 ( a ) = f 1 ( a ) − R 1 , n ( a ) .

Then h 1 ( m ) ( 0 + ) = h 2 ( m ) ( 0 + ) = 0 for all m ∈ N ∪ { 0 } with 0 ≤ m ≤ n and

(3.13) h 1 ( n + 1 ) ( a ) h 2 ( n + 1 ) ( a ) = f 1 ( n + 1 ) ( a ) ( n + 1 ) ! .

From Lemma 2.3, we know that the function G 1 , 2 n ( G 1 , ( 2 n + 1 ) ) has the same monotonicity with the function f ( 2 n + 1 ) ( f ( 2 n + 2 ) ) . It follows from ( − 1 ) n f 1 ( n ) ( a ) ≥ 0 for a ∈ ( 0 , 1 / 2 ] and n ≥ 3 that G 1 , 2 n ( G 1 , ( 2 n + 1 ) ) is increasing (decreasing) on ( 0 , 1 / 2 ] . Differentiating G 1 , n gives

( G 1 , n ( a ) ) ′ = h 1 ( a ) h 2 ( a ) ′ = a ( f 1 ( a ) − R 1 , n ( a ) ) ′ − ( n + 1 ) ( f 1 ( a ) − R 1 , n ( a ) ) a n + 2 .

Let

h 3 ( a ) = a ( f 1 ( a ) − R 1 , n ( a ) ) ′ − ( n + 1 ) ( f 1 ( a ) − R 1 , n ( a ) ) ,

and h 4 ( a ) = a n + 2 . Then from (3.9), we know that h 3 ( a ) can be rewritten as follows:

h 3 ( a ) = ∑ k = n + 1 ∞ ( k − n − 1 ) B 1 ( k ) a k ,

which leads to h 3 ( m ) ( 0 + ) = h 4 ( m ) ( 0 + ) = 0 for all m ∈ N ∪ { 0 } with 0 ≤ m ≤ n and

(3.14) h 3 ( n + 1 ) ( a ) h 4 ( n + 1 ) ( a ) = f 1 ( n + 2 ) ( a ) ( n + 2 ) ! .

From Lemma 2.3, we know that the function ( G 1 , n ) ′ has the same monotonicity with the function f ( n + 2 ) . Therefore, the desired monotonicity of ( G 1 , 2 n ) ′ and ( G 1 , ( 2 n + 1 ) ) ′ is obtained, the concavity (convexity) property of G 1 , 2 n ( G 1 , ( 2 n + 1 ) ) follows from part (2), and the double inequality (3.6) (3.7) and its equality case follow from the monotonicity and concavity (convexity) properties of G 1 , 2 n ( G 1 , ( 2 n + 1 ) ) .□

A function f is said to be strictly completely monotonic on an interval I ⊂ R if ( − 1 ) n f ( n ) ( x ) > 0 for all x ∈ I and n = 0 , 1 , 2 , 3 … . If ( − 1 ) n f ( n ) ( x ) ≥ 0 for all x ∈ I and n = 0 , 1 , 2 , 3 … , then f is called completely monotonic on I . The completely monotonic functions have important applications in areas of numerical analysis [64,65], probability theory [66], physics [67], and so on.

Corollary 3.2

For each n ∈ N , the function F 1 , n is completely monotonic on ( 0 , 1 / 2 ] .

Proof

It follows from (3.11) that ( − 1 ) l F 1 , n ( l ) ( a ) ≥ 0 for all a ∈ ( 0 , 1 / 2 ] , and l = 0 , 1 , 2 , 3 … . Therefore, F 1 , n is completely monotonic on ( 0 , 1 / 2 ] .□

Let

(3.15) A 2 ( 2 ) = [ 9 λ ( 3 ) − 10 log 2 ] / 4 , A 2 ( 4 ) = [ 9 λ ( 5 ) − 10 λ ( 3 ) + log 2 ] / 4 , A 2 ( 2 k ) = [ 9 λ ( 2 k + 1 ) − 10 λ ( 2 k − 1 ) + λ ( 2 k − 3 ) ] / 4 ( k ≥ 3 ) , B 2 ( 1 ) = − 1 , B 2 ( 2 ) = − 2 , B 2 ( 3 ) = 4 ζ ( 3 ) + 1 , B 2 ( 4 ) = − 2 ζ ( 3 ) , B 2 ( 2 k + 1 ) = − 4 ( ζ ( 2 k − 1 ) − ζ ( 2 k + 1 ) ) ( k ≥ 2 ) , B 2 ( 2 k + 2 ) = 2 ( ζ ( 2 k − 1 ) − ζ ( 2 k + 1 ) ) ( k ≥ 2 ) .

Theorem 3.3

Let f 2 ( a ) = b ( 2 + b ) R ( a ) . Then, the following statements are true:

f 2 has the power series formula:
(3.16) f 2 ( a ) = 9 log 2 4 + ∑ n = 1 ∞ A 2 ( 2 n ) ( 1 − 2 a ) 2 n = 2 + ∑ n = 1 ∞ B 2 ( n ) a n .
The function f 2 is strictly decreasing on ( 0 , 1 / 2 ] , and f 2 is neither convex nor concave on ( 0 , 1 / 2 ] and ( − 1 ) k f 2 ( k ) ( a ) > 0 for ( 0 , 1 / 2 ] and k ≥ 5 .
Let
(3.17) F 2 , n ( a ) = f 2 ( a ) − P 2 , n ( a ) ( 1 − 2 a ) 2 ( n + 1 ) ,
where P 2 , n ( a ) = ( 9 log 2 ) / 4 + ∑ k = 1 n A 2 ( 2 k ) ( 1 − 2 a ) 2 k and A 2 ( 2 k ) ( k ≥ 1 ) is given in (3.15). Then, F 2 , n is strictly decreasing and convex from ( 0 , 1 / 2 ) onto ( A 2 ( n + 1 ) , F 2 , n ( 0 + ) ) . In particular, the double inequality
(3.18) 0 ≤ b ( 2 + b ) R ( a ) − P 2 , n ( a ) − A 2 ( n + 1 ) ( 1 − 2 a ) 2 ( n + 1 ) ≤ [ F 2 , n ( 0 + ) − A 2 ( n + 1 ) ] ( 1 − 2 a ) 2 ( n + 1 ) + 1
holds for all n ∈ N and a ∈ ( 0 , 1 / 2 ] , with equality in each instance if and only if a = 1 / 2 .
Let
(3.19) G 2 , n ( a ) = f 2 ( a ) − R 2 , n ( a ) a n + 1 ,
where R 2 , n ( a ) = 2 + ∑ k = 1 n B 2 ( k ) a k and B 2 ( k ) ( k ≥ 1 ) is defined by (3.15). Then, for n ≥ 2 , G 2 , 2 n ( G 2 , ( 2 n − 1 ) ) is strictly increasing (decreasing) and concave (convex) on ( 0 , 1 / 2 ] with G 2 , n ( 0 + ) = B 2 ( n + 1 ) . In particular, inequalities
(3.20) 2 ( G 2 , 2 n ( 1 / 2 ) − B 2 ( 2 n + 1 ) ) a 2 n + 2 ≤ b ( 2 + b ) R ( a ) − R 2 , 2 n ( a ) − B 2 ( 2 n + 1 ) a 2 n + 1 ≤ ( G 2 , 2 n ( 1 / 2 ) − B 2 ( 2 n + 1 ) ) a 2 n + 1
and
(3.21) ( G 2 , ( 2 n − 1 ) ( 1 / 2 ) − B 2 ( 2 n ) ) a 2 n + 2 ≤ b ( 2 + b ) R ( a ) − R 2 , 2 n ( a ) − B 2 ( 2 n ) a 2 n + 2 ≤ 2 ( G 2 , ( 2 n − 1 ) ( 1 / 2 ) − B 2 ( 2 n ) ) a 2 n + 3
hold for all n ∈ N and a ∈ ( 0 , 1 / 2 ] , with equality in each instance if and only if a = 1 / 2 .

Proof

(1) It follows from Lemma 2.1(2) and b = a ( 1 − a ) = [ 1 − ( 1 − 2 a ) 2 ] / 4 that

(3.22) f 2 ( a ) = b ( 2 + b ) R ( a ) = 9 − 10 ( 1 − 2 a ) 2 + ( 1 − 2 a ) 4 16 log 16 + 4 ∑ n = 1 ∞ λ ( 2 n + 1 ) ( 1 − 2 a ) 2 n = 9 log 2 4 + ∑ n = 1 ∞ A 2 ( 2 n ) ( 1 − 2 a ) 2 n ,

where A 2 ( 2 n ) ( n ≥ 1 ) is defined by (3.15). Again use Lemma 2.1(2), we obtain

(3.23) f 2 ( a ) = b ( 2 + b ) R ( a ) = b ( 2 + b ) 1 a + 2 ∑ n = 1 ∞ ζ ( 2 n + 1 ) a 2 n = 2 − a − 2 a 2 + ( 4 ζ ( 3 ) + 1 ) a 3 − 2 ζ ( 3 ) a 4 + ∑ n = 2 ∞ [ − 4 ζ ( 2 n − 1 ) + 4 ζ ( 2 n + 1 ) ] a 2 n + 1 + [ 2 ζ ( 2 n − 1 ) − 2 ζ ( 2 n + 1 ) ] a 2 n + 2 = 2 + ∑ n = 1 ∞ B 2 ( n ) a n .

(2) It follows from (1.16) that A 2 ( 2 n ) = [ 9 λ ( 2 n + 1 ) − 10 λ ( 2 n − 1 ) + λ ( 2 n − 3 ) ] / 4 and

A 2 ( 2 n ) = 1 4 ∑ k = 1 ∞ 9 − 10 ( 2 k + 1 ) 2 + ( 2 k + 1 ) 4 ( 2 k + 1 ) 2 n + 1 = 4 ∑ k = 2 n ( k 2 + 2 k ) ( k 2 − 1 ) ( 2 k + 1 ) 2 n + 1 > 0

for n > 2 . Differentiating f 2 gives

(3.24) f 2 ( k ) ( a ) = ∑ n = [ ( k + 1 ) / 2 ] ∞ ( − 1 ) k 2 k ( 2 n ) ! ( 2 n − k ) ! A 2 ( 2 n ) ( 1 − 2 a ) 2 n − k ,

(3.25) f 2 ( k ) ( a ) = ∑ n = [ ( k + 1 ) / 2 ] ∞ ( n ) ! ( n − k ) ! B 2 ( n ) a n − k .

From (3.24), we know that ( − 1 ) k f 2 ( k ) ( a ) ≥ 0 for a ∈ ( 0 , 1 / 2 ] and k = 5 , 6 , … ; and f 2 ( 5 ) ( a ) is increasing on ( 0 , 1 / 2 ] . By the series formula of f 2 ( a ) , we obtain f 2 ( 5 ) ( 1 / 2 ) = 0 ; therefore, f 2 ( 5 ) ( a ) ≤ 0 for a ∈ ( 0 , 1 / 2 ] , so that f 2 ( 4 ) is strictly decreasing on ( 0 , 1 / 2 ] .

It follows from (3.25) that f 2 ( 4 ) ( 0 + ) = − 48 ζ ( 3 ) < 0 , so f 2 ( 4 ) ( a ) < 0 for a ∈ ( 0 , 1 / 2 ] . Hence, f 2 ‴ ( a ) are strictly decreasing on ( 0 , 1 / 2 ] with f 2 ‴ ( 1 / 2 ) = 0 , which implies that f 2 ‴ ( a ) > 0 and f 2 ″ are strictly increasing on ( 0 , 1 / 2 ] .

By (3.24) and (3.25), we obtain f 2 ″ ( 0 + ) = − 4 and f 2 ″ ( 1 / 2 ) = − 20 log 2 + 63 ζ ( 3 ) / 4 = 5.070 … > 0 . Therefore, there exists a ∗ ∈ ( 0 , 1 / 2 ] such that f 2 ″ ( a ∗ ) = 0 , and f 2 ′ is decreasing on ( 0 , a ∗ ] and f 2 ′ is increasing on on ( a ∗ , 1 / 2 ] . Since f 2 ′ ( 0 + ) = − 1 and f 2 ′ ( 1 / 2 ) = 0 , so f 2 ′ ( a ) < 0 and f 2 is strictly decreasing on ( 0 , 1 / 2 ] with f 2 ( 0 + ) = 2 and f 2 ( 1 / 2 ) = 9 log 2 / 4 .

(3) Differentiating

F 2 , n ( a ) = f 2 ( a ) − P 2 , n ( a ) ( 1 − 2 a ) 2 ( n + 1 ) = ∑ k = n + 1 ∞ A 2 ( 2 k ) ( 1 − 2 a ) 2 ( k − n − 1 ) = ∑ k = 0 ∞ A 2 ( 2 ( k + n + 1 ) ) ( 1 − 2 a ) 2 k

leads to

(3.26) F 2 , n ( l ) ( a ) = ∑ k = [ ( l + 1 / 2 ) ] ∞ ( − 1 ) l 2 k ( 2 k ) ! ( 2 k − l ) ! A 2 ( 2 ( k + n + 1 ) ) ( 1 − 2 a ) 2 k − l ,

which implies that

F 2 , n ′ ( a ) = − 4 ∑ k = 1 ∞ k A 2 ( 2 ( k + n + 1 ) ) ( 1 − 2 a ) 2 k − 1

and

F 2 , n ″ ( a ) = 8 ∑ k = 1 ∞ k ( 2 k − 1 ) A 2 ( 2 ( k + n + 1 ) ) ( 1 − 2 a ) 2 k − 2 .

Therefore, the monotonicity and concavity properties of F 2 , n follow from A 2 ( 2 n ) > 0 for all n > 2 , and the desired inequalities follow from the monotonicity and convexity properties of F 2 , n together with F 2 , n ( ( 1 / 2 ) − ) = A 2 ( 2 n + 2 ) .

(4) Let G 2 , n ( a ) = h 5 ( a ) / h 6 ( a ) , where h 6 ( a ) = a n + 1 and

(3.27) h 5 ( a ) = f 2 ( a ) − R 2 n ( a ) .

Then it is easy to verify that h 5 ( m ) ( 0 + ) = h 6 ( m ) ( 0 + ) = 0 for all m ∈ N ∪ { 0 } with 0 ≤ m ≤ n . Simple computations lead to

(3.28) h 5 ( n + 1 ) ( a ) h 6 ( n + 1 ) ( a ) = f 2 ( n + 1 ) ( a ) ( n + 1 ) ! .

From Lemma 2.3, we know that G 2 , 2 n ( G 2 , ( 2 n − 1 ) ) has the same monotonicity with the function f 2 ( 2 n + 1 ) ( f 2 ( 2 n ) ) . Therefore, G 2 , 2 n ( G 2 , ( 2 n − 1 ) ) is increasing (decreasing) on ( 0 , 1 / 2 ] due to ( − 1 ) n f 2 ( n ) ( a ) ≥ 0 for a ∈ ( 0 , 1 / 2 ] and n ≥ 5 .

Differentiating G 2 , n gives

( G 2 , n ( a ) ) ′ = h 5 ( a ) h 6 ( a ) ′ = a ( f 2 ( a ) − R 2 , n ( a ) ) ′ − ( n + 1 ) ( f 2 ( a ) − R 2 , n ( a ) ) a n + 2 .

Let h 8 ( a ) = a n + 2 and

h 7 ( a ) = a ( f 2 ( a ) − R 2 , n ( a ) ) ′ − ( n + 1 ) ( f 2 ( a ) − R 2 , n ( a ) ) .

Then from (3.23), we know that h 7 ( a ) can be rewritten as follows:

h 7 ( a ) = ∑ k = n + 1 ∞ ( k − n − 1 ) B 2 ( k ) a k .

Therefore, h 7 ( m ) ( 0 + ) = h 8 ( m ) ( 0 + ) = 0 for all m ∈ N ∪ { 0 } with 0 ≤ m ≤ n and

(3.29) h 7 ( n + 1 ) ( a ) h 8 ( n + 1 ) ( a ) = f 2 ( n + 2 ) ( a ) ( n + 2 ) ! ,

which shows that G 2 , n ′ has the same monotonicity with the function f 2 ( n + 2 ) by Lemma 2.3. Therefore, the desired monotonicity of G 2 , 2 n ′ and G 2 , ( 2 n − 1 ) ′ , the concavity (convexity) of G 2 , 2 n ( G 2 , ( 2 n − 1 ) ) , are obtained, and the inequalities (3.20) and (3.21) follow easily from the monotonicity and concavity (convexity) properties of G 2 , 2 n ( G 2 , ( 2 n + 1 ) ) .□

Corollary 3.4

The function F 2 , n ( n > 2 ) is completely monotonic on ( 0 , 1 / 2 ] .

Proof

From (3.26), we clearly see that A 2 ( 2 n ) > 0 for all n > 2 . Therefore, ( − 1 ) l F 2 , n ( l ) ( a ) ≥ 0 for all a ∈ ( 0 , 1 / 2 ] and l = 0 , 1 , 2 , 3 … , and F 2 , n is completely monotonic on ( 0 , 1 / 2 ] .□

Corollary 3.5

G 2 , 1 is increasing and concave, and G 2 , 2 is decreasing and concave on ( 0 , 1 / 2 ] .

Proof

It follows from (3.28) and Lemma 2.3 that G 2 , 2 ( G 2 , 1 ) has the same monotonicity with the function f 2 ‴ ( f 2 ″ ) . Since f 2 ‴ ( f 2 ″ ) is strictly decreasing (increasing) on ( 0 , 1 / 2 ] , we can obtain the desired monotonicity of G 2 , 2 and G 2 , 1 . It follows from (3.29) and Lemma 2.3 that G 2 , 2 ′ ( G 2 , 1 ′ ) has the same monotonicity with the function f 2 ( 4 ) ( f 2 ‴ ) . Since f 2 ‴ and f 2 ( 4 ) is strictly decreasing on ( 0 , 1 / 2 ] , the desired monotonicity of G 2 , 2 ′ ( a ) and G 2 , 1 ′ ( a ) , and the concavity (convexity) of G 2 , 2 ( a ) and G 2 , 1 ( a ) are obtained.□

Let

(3.30) A 3 ( 0 ) = 7 log 2 − 4 = 0.8520 … , A 3 ( 2 ) = 7 λ ( 3 ) + log 2 − 4 = 4.055745 … , A 3 ( 2 k ) = 7 λ ( 2 k + 1 ) + λ ( 2 k − 1 ) − 4 ( k ≥ 2 ) , B 3 ( 1 ) = 0 , B 3 ( 2 ) = 4 ζ ( 3 ) − 1 , B 3 ( 2 k − 1 ) = − 1 − 2 ζ ( 2 k − 1 ) ( k ≥ 2 ) , B 3 ( 2 k ) = − 1 + 2 ζ ( 2 k − 1 ) + 4 ζ ( 2 k + 1 ) ( k ≥ 2 ) .

Theorem 3.6

Let f 3 ( a ) = ( 2 − b ) R ( a ) − 1 / b . Then the following statements are true:

f 3 has the power series formula:
(3.31) f 3 ( a ) = 7 log 2 − 4 + ∑ n = 1 ∞ A 3 ( 2 n ) ( 1 − 2 a ) 2 n = 1 a − 2 + ∑ n = 1 ∞ B 3 ( n ) a n .
f 3 ( a ) is completely monotonic on ( 0 , 1 / 2 ] .
Let
(3.32) F 3 , n ( a ) = f 3 ( a ) − P 3 , n ( a ) ( 1 − 2 a ) 2 ( n + 1 ) ,
where P 3 , n ( a ) = ∑ k = 0 n A 3 ( 2 k ) ( 1 − 2 a ) 2 k and A 3 ( k ) ( k ≥ 0 ) is given in (3.30). Then, F 3 , n is strictly decreasing and convex from ( 0 , 1 / 2 ] onto ( A 3 ( n + 1 ) , ∞ ) . In particular, the inequality
(3.33) A 3 ( n + 1 ) ( 1 − 2 a ) 2 ( n + 1 ) < ( 2 − b ) R − 1 / b − P 3 , n ( a )
holds for all a ∈ ( 0 , 1 / 2 ] and n ∈ N , with equality in each instance if and only if a = 1 / 2 .
Let
(3.34) G 3 , n ( a ) = f 3 ( a ) − R 3 , 2 n ( a ) a 2 n + 1 ,
where R 3 , n ( a ) = 1 / a − 2 + ∑ k = 1 n B 3 ( k ) a k and B 3 ( k ) ( k ≥ 1 ) is defined in (3.30). Then, G 3 , n is strictly increasing from ( 0 , 1 / 2 ] onto ( B 3 ( 2 n − 1 ) , G 3 , n ( 1 / 2 ) ] . In particular, the inequality
(3.35) B 3 ( 2 n − 1 ) a 2 n − 1 < ( 2 − b ) R ( a ) − 1 / b − R 3 , n ( a ) ≤ G 3 , n 1 2 a 2 n − 1 ,
holds for all a ∈ ( 0 , 1 / 2 ] and n ∈ N , with equality in each instance if and only if a = 1 / 2 .

Let

(3.36) H 3 , n ( a ) = f 3 ( a ) − R 3 , ( 2 n − 1 ) ( a ) a 2 n ,

where R 3 , n ( a ) =