Proof of a conjecture of Granath on optimal bounds of the Landau constants
aa r X i v : . [ m a t h . C A ] M a r Proof of a conjecture of Granath on optimal bounds of theLandau constants
Chun-Ru Zhao, Wen-Gao Long and Yu-Qiu Zhao ∗ Department of Mathematics, Sun Yat-sen University, GuangZhou 510275, China
Abstract
We study the asymptotic expansion for the Landau constants G n , πG n ∼ ln(16 N ) + γ + ∞ X k =1 α k N k as n → ∞ , where N = n + 1, and γ is Euler’s constant. We show that the signs of the coefficients α k demonstrate a periodic behavior such that ( − l ( l +1)2 α l +1 < l . We furtherprove a conjecture of Granath which states that ( − l ( l +1)2 ε l ( N ) < l = 0 , , , · · · and n = 0 , , , · · · , ε l ( N ) being the error due to truncation at the l -th order term. Consequently,we also obtain the sharp bounds up to arbitrary orders of the formln(16 N ) + γ + p X k =1 α k N k < πG n < ln(16 N ) + γ + q X k =1 α k N k for all n = 0 , , · · · , all p = 4 s + 1 , s + 2 and q = 4 m, m + 3, with s = 0 , , , · · · and m = 0 , , , · · · . MSC2010:
Keywords:
Landau constants; second-order linear difference equation; sharper bound; asymp-totic expansion; hypergeometric function
In 1913, Landau [11] proved that if f ( z ) is analytic in the unit disc, and | f ( z ) | < | z | < f ( z ) = a + a z + a z + · · · + a n z n + · · · , | z | < , then there exist constants G n such that | a + a + · · · + a n | ≤ G n , n = 0 , , , · · · , ∗ Corresponding author (
E-mail address: [email protected]). Investigation supported in part by theNational Natural Science Foundation of China under grant numbers 10871212 and 11571375. n , where G = 1, and G n = 1 + (cid:18) (cid:19) + (cid:18) · · (cid:19) + · · · + (cid:18) · · · · · (2 n − · · · · · (2 n ) (cid:19) for n = 1 , , · · · . (1.1)The constants G n are termed the Landau constants. The large- n behavior is known fromthe very beginning. Landau [11] derived that G n ∼ π ln n as n → ∞ ;see also Watson [18]. It is worth mentioning that there exist generating functions for theseconstants; cf. [7], possible q -versions of the constants; cf. [10], and an observation made byRamannujan (cf. [7]) that relates the Landau constants to the generalized hypergeometric func-tions. Useful integral representations for G n have been obtained from such relations; cf., e.g.,Watson [18]; see also Cvijovi´c and Srivastava [7].The approximation of G n has gone in two related directions. One is to obtain large- n asymptotic approximations for the constants, in a time period spanning from the early twentiethcentury [11, 18] to very recently [7, 12]. The other direction is to find sharper bounds of G n forall nonnegative integers n . Authors working on the sharper bounds includes Brutman [3] andFalaleev [8] (in terms of elementary functions), Alzer [2] and Cvijovi´c and Klinowski [6] (usingthe digamma function), Zhao [20], Mortici [14] and Granath [9] (involving higher order terms),and Chen and Choi [5] and Chen [4] (digamma function and higher order terms). The list is byno means complete. The reader is referred to [7, 12, 13] for a historic account. Attempts have been made to seek bounds in a sense optimal, and up to arbitrary accuracy.In 2012, Nemes [15] derived full asymptotic expansions. For 0 < h < /
2, he shows that theLandau constants G n have the asymptotic expansion G n ∼ π ln( n + h ) + 1 π ( γ + 4 ln 2) − X k ≥ g k ( h )( n + h ) k as n → + ∞ , (1.2)where γ = 0 . · · · is Euler’s constant. Earlier in 2011, the special cases h = and h = 1were established by Nemes and Nemes [16] using a formula in [6]. They also conjecture in [16] asymmetry property of the computable constant coefficients such that g k ( h ) = ( − k g k (3 / − h )for every k ≥
1. The conjecture has been proved by G. Nemes himself in [15]. A naturalconsequence is that for h = 3 /
4, all odd terms in the expansion vanish. In this importantspecial case, Nemes [15] has further proved that
Proposition 1. (Nemes) The following asymptotic approximation holds: πG n ∼ ln( n + 3 /
4) + γ + 4 ln 2 + ∞ X s =1 β s ( n + 3 / s , n → ∞ , (1.3) where the coefficients ( − s +1 β s are positive rational numbers. The derivation of Nemes [15] is based on an integral representation of G n involving a Gausshypergeometric function in the integrand. An entirely different difference equation approach isapplied in Li et al. [12] to obtain full asymptotic expansions with coefficients iteratively given.2hat is more, in a follow-up paper [13], it is shown that the error due to truncation of (1.3)is bounded in absolutely value by, and of the same sign as, the first neglected term for all n = 0 , , , · · · . An immediate corollary is Proposition 2. (Li, Liu, Xu and Zhao) For N = n + 3 / , it holds ln N + γ + 4 ln 2 + m X s =1 β s N s < πG n < ln N + γ + 4 ln 2 + k − X s =1 β s N s (1.4) for all n = 0 , , , · · · , m = 0 , , , · · · , and k = 1 , , · · · . In a sense, the formulas (1.3) and (1.4) in the above propositions seem to have ended ajourney since one has thus obtained optimal bounds up to arbitrary orders. Yet there is aninteresting observation worth mentioning, as presented in the 2012 paper [9] of Granath; seealso [13].Granath derives an asymptotic expansion πG n ∼ ln(16 N ) + γ + ∞ X k =1 α k N k , n → ∞ , (1.5)where α k are effectively computable constants but not explicitly given, except for the first few.Here and hereafter we use the notation N = n + 1.Denoting the truncation of (1.5) A l ( N ) = ln(16 N ) + γ + l X k =1 α k N k , (1.6)then one of the main results in Zhao [20] reads A ( N ) < πG n < A ( N ) for n = 0 , , , · · · .Mortici [14] have actually proved that A ( N ) < πG n < A ( N ) for all non-negative n .In [9], Granath proves that A ( N ) < πG n < A ( N ) and states that A ( N ) < πG n < A ( N ),for all non-negative n . Based on these formulas and numerical evidences, Granath proposes aconjecture. Conjecture 1. (Granath) It holds ( − l ( l +1)2 ( πG n − A l ( N )) < for all n = 0 , , , · · · and l = 0 , , , · · · . We will show that the conjecture is true. To do so, we will make use of the second orderdifference equation for G n employed in [12], and some estimating techniques used in [13].First we denote the error term ε l ( N ) = πG n − A l ( N ) = πG n − ( ln(16 N ) + γ + l X k =1 α k N k ) ; (1.8)cf. (1.6), where N = n + 1. It is readily seen that ε l ( N ) ∼ α l +1 /N l +1 as N → ∞ . Hence wemay start by showing that (1.7) holds for large n . To this aim, we have3 heorem 1. The coefficients of the asymptotic expansion (1.5) satisfy ( − l ( l +1)2 α l +1 < , l = 0 , , , · · · . (1.9)Next, we will prove the conjecture for all non-negative n . Theorem 2.
For N = n + 1 , it holds ( − l ( l +1)2 ε l ( N ) < for l = 0 , , , · · · and n = 0 , , , · · · . As a straightforward application of Theorem 2, we obtain the following sharp bounds up toarbitrary orders.
Corollary 1.
For N = n + 1 , it holds A p ( N ) < πG n < A q ( N ) , that is, ln(16 N ) + γ + p X k =1 α k N k < πG n < ln(16 N ) + γ + q X k =1 α k N k (1.11) for all n = 0 , , , · · · and for all p = 4 s + 1 , s + 2 and q = 4 m, m + 3 , with s = 0 , , , · · · and m = 0 , , , · · · . In view of Theorem 1, we see that the bounds in (1.11) are optimal as n → ∞ .Theorem 2 can actually be understood as an estimate of the error term, such that the errordue to truncation is bounded in absolute value by, and of the same sign as, the first one or twoneglected terms. Indeed, since ε l ( N ) = α l +1 N l +1 + ε l +1 ( N ) and ε l ( N ) = α l +1 N l +1 + α l +2 N l +2 + ε l +2 ( N ) , taking into account the signs in Theorems 1 and 2, we have0 < ε k +1 ( N ) < α k +2 N k +2 + α k +3 N k +3 and 0 < ε k +2 ( N ) < α k +3 N k +3 for all non-negative integers n and k , and α k +1 N k +1 < ε k ( N ) < α k +4 N k +4 + α k +5 N k +5 < ε k +3 ( N ) < n and k .As a by-product of the proof of Theorem 2, we have approximations of the asymptoticcoefficients, follows respectively from (4.10) and (4.15): Corollary 2.
Assume that α k are the coefficients in the asymptotic expansion (1.5) . Thenwe have α k = ( − k +1 k − π ) k (cid:18) O (cid:18) k (cid:19)(cid:19) (1.12) and α k +1 = ( − k +1 k )! ln(2 k + 1)(2 π ) k +2 (cid:18) O (cid:18) k (cid:19)(cid:19) (1.13) as k → ∞ . The asymptotic coefficients and the proof of Theorem 1
From the representation (1.1) one obtains the recurrence relation G n +1 − G n = (cid:18) n + 12 n + 2 (cid:19) ( G n − G n − ) . Set N = n + 1, we may rewrite it as a standard second-order difference equation w ( N + 1) − (cid:18) − N + 14 N (cid:19) w ( N ) + (cid:18) − N (cid:19) w ( N −
1) = 0 , (2.1)where w ( N ) = πG n . An interesting fact is that the formal solution to (2.1) is an asymptoticsolution; cf. Li and Wong [19]; see also [12]. Hence the asymptotic series (1.5) furnishes a formalsolution of (2.1). Therefore, one way to determine the coefficients α k is to substitute (1.5) into(2.1) and equalizing the coefficients of the same powers of x = 1 /N . We include some detailsas follows.ln(1 + x ) + ∞ X k =1 α k x k (1 + x ) k − (cid:18) − x + x (cid:19) ∞ X k =1 α k x k + (cid:16) − x (cid:17) " ln(1 − x ) + ∞ X k =1 α k x k (1 − x ) k = 0 . Using the Maclaurin series expansions, we have − ∞ X s =3 d ,s x s + ∞ X k =1 α k x k ∞ X j =2 d k,j + k x j = ∞ X s =3 s − X k =1 d k,s α k − d ,s ! x s = 0 . Accordingly, coefficients α k are determined by d s − ,s α s − + d s − ,s α s − + · · · + d ,s α − d ,s = 0 , s = 3 , , · · · , (2.2)where the coefficients d s − ,s = ( s − for s = 3 , , · · · , d ,s = ( − s + 1 s − s − s −
2) for s = 3 , , · · · , and (2.3) d k,s = (cid:0) ( − s − k + 1 (cid:1) ( k ) s − k ( s − k )! − ( k ) s − k − ( s − k − k ) s − k − s − k − k = 1 , , · · · , s − s = k + 3 , k + 4 , · · · .Appealing to (2.2)-(2.4), the first few coefficients α k can be evaluated as α = − , α = , α = , α = − ,α = − , α = , α = , α = − ,α = − , α = , α = , α = − . One readily sees a periodic phenomenon of the signs of the coefficients, which agrees withTheorem 1. To give a full proof of the theorem, we may connect the coefficients with those in(1.3), and eventually with a certain hypergeometric function. Indeed, re-expanding the formula(1.3) in descending powers of N = n + 1 yields the expansion (1.5). Hence we have α k = 4 − k − k + k X j =1 ( k − j β j ( j − k − j )! , k = 1 , , · · · ; (2.5)5f. [13, (4.4)], where β j vanish for odd integers j . We also note that the coefficients β k possessa generating function, that is, u ( x ) = ∞ X k =0 ρ k x k = x x F (cid:18) ,
12 ; 1; sin x (cid:19) := x x F (cid:16) sin x (cid:17) , (2.6)where ρ = 1 and ρ s = ( − s +1 β s (2 s − , s = 1 , , · · · are the positive constants defined in [13, Sec. 3.1].It is shown in [13] that the generating function u solves a second-order differential equation,and consequently the hypergeometric function F (cid:0) , ; 1; t (cid:1) is brought in. It is worth notingthat the function also furnishes a generating relation for the Landau constants, namely F ( x )1 − x = P ∞ n =0 G n x n for small x ; see [15]. Here and hereafter we denote for short the hypergeometricfunction as F ( t ) = F (cid:0) , ; 1; t (cid:1) . Proof of Theorem 1 . From (2.5) we have α k (2 k − k X s =0 ( − s +1 ρ s (2 k − s )! (cid:18) (cid:19) k − s for k = 1 , , · · · . (2.7)Here use has been made of the fact that β s − = 0 for s = 1 , , · · · . From (2.7) we further have1 + ∞ X k =1 ( − k +1 α k (2 k − x k = ( ∞ X s =0 ρ s x s ) ( ∞ X s =0 s )! (cid:18) − x (cid:19) s ) = u ( x ) cos x . (2.8)Combining (2.6) with (2.8), and applying a quadratic transformation formula, we have1 + ∞ X k =1 ( − k +1 α k (2 k − x k = x sin x F (cid:16) sin x (cid:17) = x sin x x F (cid:16) tan x (cid:17) ;see [1, (15.3.17)]. Each factor on the right-hand side possesses a Maclaurin expansion withpositive coefficients; see Nemes [15, pp. 842-843]. Hence we conclude that( − k +1 α k > k = 1 , , · · · . (2.9)Similarly, we may write ∞ X k =0 ( − k +1 α k +1 (2 k )! x k = 14 ( ∞ X s =0 ρ s x s ) ( ∞ X s =0 s + 1)! (cid:18) − x (cid:19) s ) = u ( x ) sin x x . (2.10)Taking (2.6) into account, we can write the right-hand side term as14 cos x F (cid:16) sin x (cid:17) = 14 cos x x F (cid:16) tan x (cid:17) , which again has a Maclaurin expansion with all positive coefficients. Here we have used theformula1cos t = ∞ X k =0 ( − k E k (2 k )! t k , where E k are the Euler numbers such that ( − k E k > k = 0 , , , · · · ; see [17, (24.2.6)-(24.2.7)]. Accordingly we have( − k +1 α k +1 > k = 0 , , , · · · . (2.11)A combination of (2.9) and (2.11) then gives (1.9).6 Proof of Theorem 2
To give a rigorous proof of Theorem 2, we introduce R l ( N ) = ε l ( N + 1) − (cid:18) − N + 14 N (cid:19) ε l ( N ) + (cid:18) − N (cid:19) ε l ( N −
1) (3.1)for l = 0 , , · · · and N = n + 1 = 1 , , , · · · , where ε l is the remainder term given in (1.8).Similar to the derivation of (2.2), substituting (1.8) into (3.1), and again denoting x = 1 /N , wesee that R l ( N ) is an analytic function of x at the origin, with the Maclaurin expansion R l ( N ) = ∞ X k =3 d ,k x k − l X k =1 α k x k ∞ X j =2 d k,j + k x j = ∞ X s = l +3 r l,s x s , (3.2)where, for s = l + 3 , l + 4 , · · · , and l = 0 , , , · · · , the coefficients in (3.2) are r l,s = − ( d l,s α l + d l − ,s α l − + · · · + d ,s α − d ,s ) . (3.3)To justify Theorem 2, we state a lemma as follows, leaving the proof of it to later sections. Lemma 1.
For N = n + 1 , it holds ˜ R l ( N ) := ( − l +1 R l ( N ) > , n = 1 , , , · · · , l = 0 , , , · · · . (3.4)Now we prove the theorem, assuming that Lemma 1 holds true. Proof of Theorem 2 . For fixed l , l = 0 , , · · · , first we show that˜ ε l ( N ) := ( − l +1 ε l ( N ) > n = 1 , , · · · , where N = n + 1. To this aim, we note that˜ ε l ( N ) = ( − l +1 ε l ( N ) = ( − l +1 α l +1 N l +1 (cid:26) O (cid:18) N (cid:19)(cid:27) = | α l +1 | N l +1 (cid:26) O (cid:18) N (cid:19)(cid:27) > N large enough; cf. (1.5), (1.8) and (2.11). Now assume that ˜ ε l ( N ) > N . Then there exists a finite positive M such that M = max { N = n + 1 : n ∈ N and ˜ ε l ( N ) ≤ } . Thus for the positive integer M , we have ˜ ε l ( M ) ≤
0, while ˜ ε l ( M + 1) , ˜ ε l ( M + 2) , · · · > b ( N ) = (cid:0) − N (cid:1) for simplicity, from (3.1) we have˜ ε l ( M + 2) = (1 + b ( M + 1))˜ ε l ( M + 1) + b ( M + 1)( − ˜ ε l ( M )) + ˜ R l ( M + 1) . The later terms on the right-hand side are nonnegative (where M + 1 ≥ ε l ( M + 2) ≥ (1 + b ( M + 1))˜ ε l ( M + 1) > ˜ ε l ( M + 1) . Moreover, from (3.4) we further have˜ ε l ( M + 3) ≥ (1 + b ( M + 2))˜ ε l ( M + 2) + b ( M + 2)( − ˜ ε l ( M + 1)) > ˜ ε l ( M + 2) . ε l ( M + k + 1) > ˜ ε l ( M + k ) , k = 1 , , · · · . By induction we conclude˜ ε l ( M + 1) < ˜ ε l ( M + k ) (3.7)for k ≥
2. Recalling that ˜ ε l ( N ) = O (cid:0) N − l − (cid:1) for N → ∞ ; cf. (3.6), letting k → ∞ in (3.7)gives ˜ ε l ( M + 1) ≤
0. This contradicts the fact that ˜ ε l ( M + 1) >
0. Hence (3.5) holds.Now from (1.8), (2.9) and (3.5), we have( − l ε l − ( N ) = ( − l α l N l + ( − l ε l ( N ) < l = 1 , , , · · · , and n = 1 , , · · · , (3.8)where N = n + 1.A combination of (3.5) and (3.8) gives (1.10). Thus completes the proof of the theorem. α k To prove Lemma 1, first we estimate the coefficients α k , or, more precisely, the quantities˜ α k = ( − k +1 α k (2 k − and ˜ α k +1 = ( − k +1 α k +1 (2 k )! for k = 1 , , · · · . These are positive constants; cf.(2.9) and (2.11). As a preparation, we give a brief account of the analytic continuation of thehypergeometric function. The reader is referred to [13, Sec. 3.2] for full details. We denote ϕ ( z ) = F (cid:16) sin z (cid:17) = F (cid:18) ,
12 ; 1; sin z (cid:19) for Re z ∈ ( − π, π ) ∪ (2 π, π ) . (4.1)Then the piecewise-defined function v ( z ) = (cid:26) ϕ ( z ) , ≤ Re z < π,ϕ ( z ) ± iϕ ( z − π ) , π < Re z < π and ± Im z > ϕ ( z ) in (4.1) from the strip Re z ∈ [0 , π ) to the cutstrip 0 ≤ Re z < π and z [2 π, + ∞ ). What is more, we have the connection formula(see [13, (3.17)]) v ( z ) = v A ( z ) − π ϕ ( z − π ) ln (2 π − z ) (4.3)for 0 < Re z < π , with v A ( z ) being analytic in the strip, and the branch of the logarithm beingchosen as arg(2 π − z ) ∈ ( − π, π ). We proceed to show that Lemma 2.
It holds . k − π ) k ≤ ˜ α k ≤ . k − π ) k , k = 9 , , , · · · (4.4) and k + 1) + 0 . π (2 π ) k +1 ≤ ˜ α k +1 ≤ k + 1) + 2 . π (2 π ) k +1 , k = 9 , , , · · · . (4.5)8 Γ−3π 3π−2π 2π .. . z− planeO .. Figure 1: The deformed contour Γ: the oriented curve (see [13, Fig. 2]).
Proof . We understand (2.8) as a generating relation for ˜ α k . Using the Cauchy integral formula,and in view of (2.6), we have˜ α k = 12 πi I u ( z ) cos( z/ z k +1 dz = 18 πi Z Γ v ( z )sin( z/ dzz k , where initially the integration path Γ is a loop encircling the origin anti-clockwise, and is thendeformed to the oriented curve illustrated in Figure 1; see also [13, Fig. 2], and v ( z ) is thefunction defined in (4.2). From (4.3), paying attention to the symmetric properties of v ( z ) andΓ, we have˜ α k = 14 πi Z Γ v v ( z )sin( z/ dzz k − πi Z Γ l π ϕ ( z − π ) ln(2 π − z )sin( z/ dzz k := I v + I l , (4.6)where Γ v is the vertical part Re z = 3 π , and Γ l is the remaining right-half part of Γ, consistingof a circular part around z = 2 π , and a pair of horizontal line segments, respectively along theupper and lower edges of (2 π, π ), joining the circle with the vertical line; see Figure 1.First, straightforward calculation gives | I v | ≤ M v π π ) k Z ∞−∞ dy (cid:12)(cid:12) sin π + iy (cid:12)(cid:12) = M v π (3 π ) k B (cid:18) , (cid:19) ≈ . · · · (3 π ) k , (4.7)where | v (3 π + iy ) | ≤ √ y ∈ R | ϕ ( π + iy ) | ≤ M v = 2 . · · · ; cf. [13, p.297], and B (cid:0) , (cid:1) isthe Beta function.Now we turn to the dominant part I l . It is readily seen that I l = 1 π Z π π ϕ ( x − π )sin( x/ dxx k = 1 π Z π π dxx k + 1 π Z π π (cid:26) g ( x ) − x − π (cid:27) ( x − π ) x k dx, (4.8)where g ( x ) = ϕ ( x − π )sin( x/ such that g (2 π ) = 1. One can see that g ( x ) − x − π is positive and monotoneincreasing for x ∈ (2 π, π ] since g ( x ) − x − π = (cid:26) sin( t/ t cos( t/ (cid:27) (cid:26) ϕ ( t ) − t/ (cid:27) + (cid:26) t (cid:18) t/ − (cid:19)(cid:27) , t = x − π, t ∈ (0 , π ]; see [13, p.299] for the monotonicity of ϕ ( t ) − t/ . Therefore, we have for x ∈ (2 π, π ],0 ≤ g ( x ) − x − π ≤ g (3 π ) − π := M g = 1 π (cid:20) √ F (cid:18) ,
12 ; 1; 12 (cid:19) − (cid:21) = 0 . · · · . Substituting it into (4.8), we have I l = 22 k − π ) k + δ l,k (2 k − π ) k (4.9)with − (cid:0) (cid:1) k < δ l,k ≤ πM g k − . Further substituting (4.7) and (4.9) into (4.6) gives˜ α k = 22 k − π ) k + δ k (2 k − π ) k (4.10)with − { . k −
1) + 3 } (cid:18) (cid:19) k < δ k < . k − (cid:18) (cid:19) k + 2 πM g ( k − . Hence for k ≥
9, we obtain the inequalities in (4.4).Now we turn to the inequality (4.5) for the odd terms. From (2.6) and (2.10) we have˜ α k +1 = 12 πi I u ( z ) sin( z/ dzz k +2 = 18 πi Z Γ v ( z ) dz cos( z/ z k +1 , where Γ is the same path illustrated in Figure 1. Then, in view of the connection formula (4.3),we may write˜ α k +1 = 14 πi Z Γ v v ( z )cos z dzz k +1 + 14 πi Z Γ l v A ( z )cos z dzz k +1 − πi Z Γ l π ϕ ( z − π ) ln(2 π − z )cos z dzz k +1 := J v + J a + J l , (4.11)where the integration paths Γ v and Γ l are the same as in (4.6); see Figure 1. We note that theprocedure in [13, Sec.3.3] applies here, with minor modifications. Case by case estimating gives | J v | ≤ (cid:26) M v π B (cid:18) , (cid:19)(cid:27) π ) k +1 ≈ . · · · (3 π ) k +1 ; (4.12)see (4.7). Also, picking up the residue at z = 2 π yields J a = 2 v A (2 π )(2 π ) k +1 = 16 ln 2 π (2 π ) k +1 , (4.13)where v A (2 π ) = π ; see [13, (3.21)]. The dominant contribution comes from the last integral J l . We follow the steps in [13, pp.299-301], and eventually obtain J l = 4 ln(2 k + 1) − (4 γ + 4 ln(2 π )) + δ l,k π (2 π ) k +1 , (4.14)10here | δ l,k | < M ϕ + πk ˜ M f + k + e − k − for positive integers k with M ϕ = e − e ; see [13, (3.23)],and such that0 < ϕ ( x − π )sin x − π − x − π = (cid:26) ϕ ( t ) − t (cid:27) + (cid:26) t − t (cid:27) ≤ ˜ M f = √ F (cid:18) ,
12 ; 1; 12 (cid:19) − π for x ∈ (2 π, π ], or, t ∈ (0 , π ] for t = x − π . Here use has been made of the fact that both termsin the curly braces are monotone increasing positive functions for t ∈ (0 , π ]; cf. the derivationof (4.9). Now substituting (4.12), (4.13) and (4.14) into (4.11) yields˜ α k +1 = 4 ln(2 k + 1) + (16 ln 2 − γ − π )) + δ k π (2 π ) k +1 , (4.15)where | δ k | ≤ . π (cid:0) (cid:1) k +1 +2 M ϕ + πk ˜ M f + k + e − k − for positive integers k , which is monotonedecreasing in k . Straightforward calculation from (4.15) yields (4.5) for k ≥ Corollary 3.
Assume that ˜ α k are the positive constants in Lemma 2. Then we have ˜ α k ˜ α k +2 < . for k = 5 , , · · · , (4.16)˜ α k +1 ˜ α k +3 < for k = 0 , , , · · · , (4.17) and (2 k + 1) ˜ α k +2 ˜ α k +1 < . and (2 k + 1) ˜ α k +2 ˜ α k +3 < . for k = 0 , , , · · · . (4.18)The results follow accordingly from Lemma 2 and Table 1. To obtain (4.17) one may haveto evaluate the ratio ˜ α k +1 ˜ α k +3 up to k = 13, such that ˜ α k +1 ˜ α k +3 = 38 . , . , . , . , . k = 9 , , , , k ˜ α k ˜ α k +2 .
305 60 .
184 57 .
345 53 .
150 49 .
797 47 .
533 46 .
044 45 .
031 44 . ˜ α k − ˜ α k +1 .
333 30 .
720 34 .
632 36 .
358 37 .
227 37 .
730 38 .
055 38 .
282 38 . (2 k − α k ˜ α k − . . . . . . . . . (2 k − α k ˜ α k +1 . . . . . . . . . Proof of Lemma 1
Now that we have proved Lemma 2, we turn to the proof of Lemma 1.
Proof of Lemma 1 . To prove (3.4), the idea is as follows: First we show that( − l r l, j +2 > j = l + 1 , l + 2 , · · · , l = 0 , , , · · · , (5.1)and ( − l +1 (cid:18) r l, j +1 + 12 r l, j +2 (cid:19) > j = l + 1 , l + 2 , · · · , l = 0 , , , · · · . (5.2)Then (3.4) follows immediately from (5.1) and (5.2) since x = 1 /N ∈ (0 , /
2] for N ≥
2, and˜ R l ( N ) = ∞ X j = l +1 ( − l +1 ( r l, j +1 + r l, j +2 x ) x j +1 ≥ ∞ X j = l +1 ( − l +1 (cid:18) r l, j +1 + 12 r l, j +2 (cid:19) x j +1 > N = n + 1 = 2 , , · · · , and l = 0 , , , · · · .The above idea is simple, yet the verification of (5.1) and (5.2) is quite complicated. Webegin with (5.1). First, a combination of (2.2) and (3.3) gives( − l r l, l +4 = d l +2 , l +4 n ( − l α l +2 o + {− d l +1 , l +4 } n ( − l +1 α l +1 o > l = 0 , , , · · · . Here use has been made of (2.9), (2.11), and the facts that d l +2 , l +4 > d l +1 , l +4 <
0. Hence (5.1) is true for j = l + 1. Therefore, we need only to prove (5.1) for j = l + 2 , l + 3 , · · · . In view of (3.3), it suffices to show, by an induction argument, that r El,j := ( − l +1 l X k =0 d k, j +2 α k > r Ol,j := ( − l +1 l − X k =0 d k +1 , j +2 α k +1 ≥ j = l + 2 , l + 3 , · · · and l = 0 , , , · · · , where α = − Proving r El,j > for j = l + 2 , l + 3 , · · · and l = 0 , , , · · · :Straightforward verification shows that the first inequality in (5.3) holds for l = 0: We seefrom (3.3) and (2.3) that r E ,j = r , j +2 = (cid:18) j + 1 − j + 1 (cid:19) + 18 j > j = 2 , , · · · . Similarly, from (2.2) and (3.3) we have r E ,j = α d , j +2 + α d , j +2 . Hence r E ,j = 5192 (cid:20) j + 2 + 2 j − (cid:21) − (cid:20) j + 1 − j + 1 + 18 j (cid:21) > j + 1)96 − j + 1) > j = 3 , , · · · . Thus the first inequality in (5.3) is true for l = 1.12ow assume (5.3) for a non-negative integer l , then, replacing l with l + 2, we have r El +2 ,j = r El,j + ˜ α l +4 (cid:20) (2 l + 3)! d l +4 , j +2 − ˜ α l +2 ˜ α l +4 (2 l + 1)! d l +2 , j +2 (cid:21) > j = l + 4 , l + 5 , l + 6 , · · · . Indeed, if we write(2 l + 1)! d l +2 , j +2 = (2 j + 2 l + 2)(2 j )!(2 j − l )! + (2 j − j − l − A l + B l , Then, noting that for l ≥ j − l ≥
5, in view of (4.16) and Table 1, we have A l +1 + B l +1 − ˜ α l +2 ˜ α l +4 ( A l + B l ) ≥ (2 j − l − j − l ) A l + (2 j − l − j − l − B l −
61 ( A l + B l ) ≥ A l + 56 B l −
61 ( A l + B l ) > , since A l > B l by straightforward verification. Alternatively, applying (4.16), for l ≥ j − l ≥
4, we can modify the above inequalities to give A l +1 + B l +1 − ˜ α l +2 ˜ α l +4 ( A l + B l ) ≥ A l + 30 B l − A l + B l ) > j = l + 4 with l = 0 , , ,
3, can be justified by direct calculation:The values of(2 l + 3)! d l +4 , l +10 − ˜ α l +2 ˜ α l +4 (2 l + 1)! d l +2 , l +10 = 62 . , . , . × , . × , respectively for l = 0 , , ,
3. Summarizing all above, we see the validity of (5.4). Therefore, thefirst inequality in (5.3) is true for j = l + 2 , l + 3 , · · · and l = 0 , , , · · · . Proving r Ol,j ≥ for j = l + 2 , l + 3 , · · · and l = 0 , , , · · · :The analysis of r Ol,j is similar to, and simpler than, that of the even terms r El,j . First, for l = 0, the sum in (5.3) is empty and thus we understand that r O ,j = 0 for all j . Also, it isreadily seen that r O ,j = d , j +2 α ≡ for j = 3 , , · · · ; cf. (2.4). Hence, the equality for r Ol,j in(5.3) also holds for l = 1.Now assume that r Ol,j ≥ l and j = l + 2 , l + 3 , · · · . From (3.3)we may write r Ol +2 ,j = r Ol,j − ˜ α l +3 (2 l + 2)! d l +3 , j +2 + ˜ α l +1 (2 l )! d l +1 , j +2 := r Ol,j + c + ∆ l,j (5.5)with a positive constant c + = ˜ α l +3 (2 l )! | d l +1 , j +2 | , and try to prove that∆ l,j := (2 l + 2)! d l +3 , j +2 (2 l )! d l +1 , j +2 − ˜ α l +1 ˜ α l +3 = 6 j + 2 l + 26 j + 2 l (2 j − l − j − l ) − ˜ α l +1 ˜ α l +3 > j = l + 4 , l + 5 , · · · . Here the last inequality comes from (4.17). Therefore, from (5.5) and byinduction, we have justified the validity of both inequalities in (5.3) for all j ≥ l + 2 and l ≥ j ≥ l + 1 and l ≥
0, noting that the only exceptionalcase j = l + 1 has been discussed earlier in this section.13 roving (5.2):In what follows we proceed to prove (5.2). First, taking into account the formulas (2.2) and(3.3), we see that r l, l +3 + r l, l +4 can be represented as a linear combination of α l +1 and α l +2 . More precisely, substituting in the coefficients d k,s ; see (2.3) and (2.4), we have( − l +1 (cid:20) r l, l +3 + 12 r l, l +4 (cid:21) = (2 l )! ˜ α l +1 (cid:20) (2 l + 1)(12 l + 5)8 − ˜ α l +2 ˜ α l +1 (2 l + 2) (2 l + 1)2 (cid:21) , which is positive for all l since ˜ α l +2 ˜ α l +1 < l +1) for l ≥
0; cf. (4.18). Thus (5.2) is true for j = l + 1, allowing us to just prove (5.2) for j = l + 2 , l + 3 , · · · and l = 0 , , , · · · .For l = 0, it is readily verified from (3.3) and (2.3) that − r , j +1 − r , j +2 = (cid:18) j − j + 2 (cid:19) + (cid:18) j + 1) − j − − j (cid:19) > j = 2 , , · · · . Here the right-hand side is the sum of positive numbers when j ≥
3, and equalsto when j = 2. Hence (5.2) holds for l = 0.For l = 1, recalling that r ,s = − d ,s + d ,s + d ,s ; cf. (3.3), from (2.3) and (2.4) we maywrite r , j +1 + 12 r , j +2 = 5 j
768 + 2811536 − (cid:18) j − j + 2 (cid:19) − (cid:18) j + 1) − j − (cid:19) + 116 j . Using the facts that j − j +2 ≤ j for j ≥
3, and j +1) − j − < j for j ≥
1, we have r , j +1 + r , j +2 > j + − j > j = 3 , , , · · · . Hence (5.2) holds for l = 1.Now assume (5.2) for a non-negative integer l , then, from (3.3) we have( − l +3 (cid:18) r l +4 , j +1 + 12 r l +4 , j +2 (cid:19) = ( − l +1 (cid:18) r l, j +1 + 12 r l, j +2 (cid:19) + O l + E l . It suffices to show that O l + E l > j = l + 4 , l + 5 , · · · , where for l = 0 , , , · · · , O l := ( − l (cid:20) α l +3 (cid:18) d l +3 , j +1 + 12 d l +3 , j +2 (cid:19) + α l +1 (cid:18) d l +1 , j +1 + 12 d l +1 , j +2 (cid:19)(cid:21) (5.6)and E l := ( − l (cid:20) α l +4 (cid:18) d l +4 , j +1 + 12 d l +4 , j +2 (cid:19) + α l +2 (cid:18) d l +2 , j +1 + 12 d l +2 , j +2 (cid:19)(cid:21) . (5.7)We may write O l = ˜ α l +3 ( ˜ A l +1 + ˜ B l +1 ) − ˜ α l +1 ( ˜ A l + ˜ B l ) = ˜ α l +3 (cid:20) ˜ A l +1 + ˜ B l +1 − ˜ α l +1 ˜ α l +3 ( ˜ A l + ˜ B l ) (cid:21) (5.8)with ˜ A l = (2 j − j +7 l )4(2 j − l )! and ˜ B l = (2 j − j − l − . Observing that ˜ A l +1 > (2 j − l )(2 j − l −
1) ˜ A l ≥
90 ˜ A l and ˜ B l +1 = (2 j − l − j − l −
3) ˜ B l ≥
56 ˜ B l for j ≥ l + 5 and l ≥
0, and recalling that ˜ α l +1 ˜ α l +3 <
43 for l ≥
0, we have O l ≥ ˜ α l +3 (cid:20) ˜ A l +1 (cid:18) − (cid:19) + ˜ B l +1 (cid:18) − (cid:19)(cid:21) ≥ α l +3 ˜ A l +1 , j ≥ l + 5 , l ≥ . E l . Similar to the discussion of O l , we may also write E l = ˜ α l +2 ( ˜ C l − ˜ D l ) − ˜ α l +4 ( ˜ C l +1 − ˜ D l +1 ) , where ˜ C l = (2 j +1)!(2 j − l )! − (2 j − j − l − = (4 l +2)(2 j − l )(2 j − j − l )! , and ˜ D l =
12 (2 j )!(2 j − l − −
18 (2 j − j − l − −
14 (2 j − j − l − . It is readily verified that both constants are positive, and such that D l < ˜ D l < D l for j ≥ l + 4 and l ≥ D l = (2 j )!(2 j − l − . Therefore, we have for j ≥ l + 5 and l ≥ O l + E l > α l +3 ˜ A l +1 −
12 ˜ α l +2 D l − ˜ α l +4 ˜ C l +1 := (2 j − α l +3 (2 j − l − , (5.9)whereΩ = (2 j − l − (cid:20) j + 7 l + 7) − l + 3) ˜ α l +4 ˜ α l +3 (2 j − l − (cid:21) − (2 l + 1) ˜ α l +2 ˜ α l +3 j l + 1 ≥ (cid:20) j + 7 l + 7) − . j − l − (cid:21) − . j = 193600 j + 2077200 l + 2077200 , and thus is positive for j ≥ l + 5 and l ≥
1. Here use has been made of (4.18). For the specialcase l = 0 and j ≥
5, taking Table 1 into account, again we have the positivity of Ω:Ω ≥ (2 j − (cid:20) j + 7) − . j − (cid:21) − . j = 701100 + 60491800 ( j −
5) + 311900 ( j − What remains is the case when j = l + 4 with l = 0 , , , · · · . Still we have (5.8). Since˜ A l +1 = (2 j − l +27)4 · >
56 ˜ A l and ˜ B l +1 = (2 j − · = 30 ˜ B l , in view of (4.17) we have O l ≥ ˜ α l +3 (cid:20) ˜ A l +1 (cid:18) − (cid:19) − ˜ B l +1 (cid:18) − (cid:19)(cid:21) ≥ (cid:20) − l + 7)(12 l + 27) (cid:21) ˜ α l +3 ˜ A l +1 , from which we see that O l > ˜ α l +3 ˜ A l +1 for j = l + 4 with l ≥
2. As a results, we have amodified version of (5.9) as j = l + 4, O l + E l >
15 ˜ α l +3 ˜ A l +1 −
12 ˜ α l +2 D l − ˜ α l +4 ˜ C l +1 := (2 j − α l +3
6! Ω , whereΩ = 12 l + 2720 − (2 l + 3) ˜ α l +4 ˜ α l +3 (2 l + 14) − (2 l + 1) ˜ α l +2 ˜ α l +3 l + 47(2 l + 1) . From (4.18) we readily see thatΩ ≥ l + 2720 − . l + 14) − . ×
17 = 31140 + 925 ( l − , and is positive for l ≥ O l + E l = 3 . , . , . l = 0 , ,
2, with j = l + 4. Thus we complete the proof of (5.2), and hence ofLemma 1. 15 Discussion
We have proved the conjecture of Granath [9], as stated in Theorem 2 and Corollary 1, ofwhich the results of Zhao [20], Mortici [14] and Granath [9] are special cases. The asymptoticexpansion involved, namely (1.5), corresponds to the special case of Nemes’ expansion (1.2) indescending powers of n + h , with h = 1.Earlier in [13], Li et al. consider the case h = 3 /
4; cf. (1.3) and (1.4). According to a resultin [13], the error due to truncation is bounded in absolute value by, and of the same sign as,the first neglected term for all nonnegative n . As an application, we obtain optimal upper andlower bounds up to all orders, holding for all integers n ≥ Question 1. (Li, Liu, Xu and Zhao) Considering the general expansion in (1.2) , for what h do we have the “best” approximation in the sense of [13, Theorem 1] (or, (1.4) in the presentpaper), or in the sense of Theorem 2 and Corollary 1? It is worth noting that the coefficients of the expansion (1.2) possess a symmetric property,namely, g k ( h ) = ( − k g k (3 / − h ). Hence, if we take h = 1 /
2, write N = n + 1 /
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