Some results associated with Bernoulli and Euler numbers with applications

0.2 MB·

by  C. -P. Chen

SOME RESULTS ASSOCIATED WITH BERNOULLI AND EULER
NUMBERS WITH APPLICATIONS
6
1 CHAO-PINGCHEN∗ ANDRICHARDB.PARIS
0
2
n Abstract. Inthispaper,wepresentseriesrepresentationsoftheremaindersintheexpansions
a for2/(et+1),sechtandcotht. Forexample,weprovethatfort>0andN ∈N:={1,2,...},
J N−1
0 secht= X (E2j2)j!t2j+RN(t)
1 j=0
with
] (−1)N2t2N ∞ (−1)k
CA RN(t)= π2N−1 kX=0(k+ 21)2N−1(cid:16)t2+π2(k+ 12)2(cid:17),
and
.
h N−1
t secht= E2j t2j+Θ(t,N) E2N t2N
a X (2j)! (2N)!
m j=0
with a suitable 0 < Θ(t,N) < 1. Here En are the Euler numbers. By using the obtained
[
results, wededuce some inequalities andcompletely monotonic functions associated withthe
1 ratio of gamma functions. Furthermore, we give a (presumably new) quadratic recurrence
v relationfortheBernoullinumbers.
2
9
1
2 1. Introduction
0
1. The Bernoulli polynomials Bn(x) and Euler polynomials En(x) are defined, respectively, by
the generating functions:
0
6 text ∞ tn 2ext ∞ tn
1 = B (x) (t <2π) and = E (x) (t <π).
et 1 n n! | | et+1 n n! | |
:
v − nX=0 nX=0
i The numbers B = B (0) and E = 2nE (1), which are known to be rational numbers and
X n n n n 2
integers, respectively, are called Bernoulli and Euler numbers.
r
a It follows from [23, Chapter 4, Part I, Problem 154] that
2m 2m+1
B t t B
2j t2j < 1+ < 2j t2j (1.1)
(2j)! et 1 − 2 (2j)!
j=1 − j=1
X X
for t > 0 and m N := N 0 , N := 1,2,3,... . The inequality (1.1) can be also found
0
∈ ∪{ } { }
in [12,24]. It is also known [31, p. 64] that
n
t t B
1+ = 2j t2j +( 1)nt2n+2ν (t) (n N ), (1.2)
et 1 − 2 (2j)! − n ∈ 0
− j=1
X
*CorrespondingAuthor.
2010Mathematics SubjectClassification. Primary11B68;Secondary26A48,26D15.
Key words and phrases. Bernoulli polynomials and numbers; Euler polynomials and numbers; Completely
monotonicfunctions;Inequality.
1
2 C.-P.CHENANDR.B.PARIS
where
∞
2 1
ν (t)= . (1.3)
n (2π)2n k2n(t2+4π2k2)
k=1
X
Itiseasilyseenthat(1.2)implies(1.1). Koumandos[12]gavethefollowingintegralrepresentation
of ν (t):
n
( 1)n 1 1
ν (t)= − extB (x)dx. (1.4)
n (2n+1)!et 1 2n+1
− Z0
Remark 1.1. From (1.4), it is possible to deduce (1.3) by making use of the expansion [20, p.
592, Eq. (24.8.2)]
∞
( 1)n+12(2n+1)! sin(2kπx)
B (x)= − (n N, 0 x 1).
2n+1 (2π)2n+1 k2n+1 ∈ ≤ ≤
k=1
X
We then obtain from (1.4) that
∞ ∞
1 2 1 extsin(2kπx) 2 1
ν (t)= dx= .
n −et 1(2π)2n+1 k2n+1 (2π)2n k2n(t2+4π2k2)
− k=1Z0 k=1
X X
An alternative derivation of (1.2) and another integral representation of the remainder function
ν (t) are given in the appendix.
n
Binet’s first formula [30, p. 16] for the logarithm of Γ(x) states that
1 ∞ t t e−xt
lnΓ(x)= x lnx x+ln√2π+ 1+ dt (x>0). (1.5)
− 2 − et 1 − 2 t2
(cid:18) (cid:19) Z0 (cid:18) − (cid:19)
Combining(1.2)with(1.5),XuandHan[36]deducedin2009thatforeverym N ,thefunction
0
∈
m
1 B
R (x)=( 1)m lnΓ(x) x lnx+x ln√2π 2j (1.6)
m −  − − 2 − − 2j(2j 1)x2j−1
(cid:18) (cid:19) j=1 −
X
 
iscompletelymonotonicon(0, ). Recallthatafunctionf(x)issaidtobecompletelymonotonic
∞
on an interval I if it has derivatives of all orders on I and satisfies the following inequality:
( 1)nf(n)(x) 0 (x I, n N ). (1.7)
0
− ≥ ∈ ∈
For m=0,the complete monotonicitypropertyofR (x) wasprovedby Muldoon[19]. Alzer[2]
m
first proved in 1997 that R (x) is completely monotonic on (0, ). In 2006, Koumandos [12]
m
∞
proved the double inequality (1.1), and then used (1.1) and (1.5) to give a simpler proof of the
complete monotonicitypropertyofR (x). In2009,KoumandosandPedersen[13,Theorem2.1]
m
strengthened this result.
Chen and Paris [9, Lemma 1] presented an analogous result to (1.1) given by
2m+1(1 22j)B t2j−1 2 t 2m (1 22j)B t2j−1
2j 2j
− < 1+ < − (1.8)
j (2j 1)! et+1 − 2 j (2j 1)!
j=2 − j=2 −
X X
for t>0 and m N. The inequality (1.8) can also be written for t>0 and m N as
0
∈ ∈
2 m (1 22j)B t2j−1
( 1)m+1 1 − 2j >0. (1.9)
− et+1 − − j (2j 1)!
j=1 −
X
 
SOME RESULTS ASSOCIATED WITH BERNOULLI AND EULER NUMBERS WITH APPLICATIONS 3
Based on the inequality (1.9), Chen and Paris [9, Theorem 1] provedthat for every m N , the
0
∈
function
m
Γ(x+1) 1 1 B
F (x)=( 1)m ln lnx 1 2j (1.10)
m −  Γ(x+ 1) − 2 − − 22j j(2j 1)x2j−1
(cid:18) 2 (cid:19) j=1(cid:18) (cid:19) −
X
 
iscompletelymonotonicon(0, ). Thisresultissimilartothecompletemonotonicitypropertyof
∞
R (x) in (1.6). In analogywith (1.2), these authorsalso considered[9, Eq. (2.4)]the remainder
m
r (t) in the expansion
m
2 m (1 22j)B
=1+ − 2jt2j−1+r (t) (1.11)
et+1 j (2j 1)! m
j=1 · −
X
and gave an integral representation for r (t) when t>0.
m
Chen [6] proposed the following conjecture.
Conjecture 1.1. For t>0 and m N , let
0
∈
et/3 e2t/3 m 2B (1)
µ (t)= − 2j+1 3 t2j (1.12)
m et 1 − (2j+1)!
− j=0
X
and
et/4 e3t/4 m 2B (1)
ν (t)= − 2j+1 4 t2j, (1.13)
m et 1 − (2j+1)!
− j=0
X
where B (x) denotes the Bernoulli polynomials. Then, for t>0 and m N ,
n 0
∈
( 1)mµ (t)>0 (1.14)
m
−
and
( 1)mν (t)>0. (1.15)
m
−
Chen[6,Lemma 1]hasprovedthe statementsinConjecture 1.1form=0,1,2,and3. He has
also pointed out in [6] that, if Conjecture 1.1 is true, then it follows that the functions
Γ(x+ 2) m B (1) 1
U (x)=( 1)m ln 3 2j+1 3 (1.16)
m −  x1/3Γ(x+ 1) − j(2j+1)x2j
3 j=1
X
 
and
Γ(x+ 3) m B (1) 1
V (x)=( 1)m ln 4 2j+1 4 (1.17)
m −  x1/2Γ(x+ 1) − j(2j+1)x2j
4 j=1
X
form N arecompletelymonotonicon(0, ). ThecompletemonotonicitypropertiesofU (x)
0 m
∈ ∞
and V (x) are similar to the complete monotonicity property of F (x) in (1.10).
m m
In this paper, we obtain the following results: (i) a series representation of the remainder
r (t) in (1.11) (Theorem 2.1); (ii) a series representation of the remainder in the expansion of
m
secht involving the Euler numbers (Theorem 2.2), together with the double inequality for t>0
and m N ,
0
∈
2m+1 2m
E E
2j t2j <secht< 2j t2j; (1.18)
(2j)! (2j)!
j=0 j=0
X X
4 C.-P.CHENANDR.B.PARIS
(iii) the proof of the inequality (1.15) for all m N , and a demonstration that the function
0
∈
V (x) in (1.17) is completely monotonic on (0, ) (Remark 2.4); (iv) a series representation of
m
∞
the remainder in the expansion for cotht (Theorem 2.3); and finally, (v) a quadratic recurrence
relation for the Bernoulli numbers (Theorem 3.1).
2. Main results
Theorem 2.1. For t>0 and m N,
∈
2 m (1 22j)B
=1+ − 2jt2j−1+( 1)m+1t2m+1s (t), (2.1)
et+1 j (2j 1)! − m
j=1 · −
X
where s (t) is given by
m
∞
4 1
s (t)= . (2.2)
m π2m (2k+1)2m t2+π2(2k+1)2
k=0
X
Proof. Boole’s summation formula ((cid:0)see [31, p. 17, Th(cid:1)eorem 1.4]) for a function f(t) defined on
[0,1] with k continuous derivatives states that, for k N,
∈
1k−1E (1) 1 1
f(1)= j f(j)(1)+f(j)(0) + f(k)(x)Ek−1(x)dx. (2.3)
2 j! 2(k 1)!
Xj=0 (cid:16) (cid:17) − Z0
Noting [20, p. 590] that
2(2n+1 1)
E (1)= − B (n N), (2.4)
n n+1
n+1 ∈
we see that
(22j 1)B
E2j−1(1)= − 2j and E2j(1)=0 (j N).
j ∈
The choice1 k =2m+1 in (2.3) yields
m (22j 1)B
f(1) f(0)= − 2j f(2j−1)(1)+f(2j−1)(0)
− j (2j 1)!
Xj=1 · − (cid:16) (cid:17)
1 1
+ f(2m+1)(x)E (x)dx. (2.5)
2m
(2m)!
Z0
Application of the above formula to f(x)=ext then produces
2 m (1 22j)B
et+1 =1+ j −(2j 1)2!jt2j−1+rm(t), (2.6)
j=1 · −
X
where
1 t2m+1 1
r (t)= extE (x)dx. (2.7)
m −et+1(2m)! 2m
Z0
1Itisalsopossibletochoosek=2min(2.3)andtousetheFourierexpansion forE2m+1(x)in[31,p.16]to
obtainthesameresult.
SOME RESULTS ASSOCIATED WITH BERNOULLI AND EULER NUMBERS WITH APPLICATIONS 5
Using the following formula (see [31, p. 16]):
∞
4(2m)! sin[(2k+1)πx]
E (x)=( 1)m (m N, 0 x 1), (2.8)
2m − π2m+1 (2k+1)2m+1 ∈ ≤ ≤
k=0
X
we obtain
( 1)m+14t2m+1 ∞ 1 sin[(2k+1)πx]
r (t)= − ext dx
m et+1 π2m+1 (2k+1)2m+1
k=0Z0
X
∞
4t2m+1 1
=( 1)m+1 .
− π2m+1 (2k+1)2m t2+π2(2k+1)2
k=0
X
This completes the proof of Theorem 2.1. (cid:0) (cid:1) (cid:3)
Remark 2.1. From (2.1) we retrieve (1.9).
Remark 2.2. From [20, p. 592, Eq. (24.7.9)] and [32, p. 43, Ex. 12(i)] we have
∞ 4t2ncosh(πt)
E (x)=( 1)nsin(πx) dt (0<x<1, n N ),
2n 0
− cosh(2πt) cos(2πx) ∈
Z0 −
from which it follows that
E (x)>0 and E (x)<0 (0<x<1, m N ).
4m 4m+2 0
∈
By combining these inequalities with (2.6) and (2.7) we immediately obtain (1.8).
Corollary 2.1. For t>0 and m N,
∈
2et m (22j 1)B
( 1)m − 2jt2j−2 >0. (2.9)
− (et+1)2 − j (2j 2)! 
j=1 · −
X
 
Proof. Differentiating the expression in (2.1), we find
2 et = m (22j −1)B2jt2j−2+( 1)m+1 t2m+1s (t) ′. (2.10)
−(et+1)2 − j (2j 2)! − m
j=1 · −
X (cid:0) (cid:1)
It is easy to see that
∞
4 1 4
t2sm(t)+sm−1(t)= π2m (2k+1)2m = π2m(1−2−2m)ζ(2m),
k=0
X
where ζ(z) is the Riemann zeta function. This last expression can be written as
4
t2sm(t)= π2m(1−2−2m)ζ(2m)−sm−1(t). (2.11)
Then, since s (t) is strictly decreasing for t > 0, we deduce from (2.11) that t2s (t) is strictly
m m
increasing for t>0. Hence, t2m+1s (t) is strictly increasing for t>0, and we then obtain from
m
(2.10) that
( 1)m 2et m (22j −1)B2jt2j−2 = t2m+1s (t) ′ >0
− (et+1)2 − j (2j 2)!  m
j=1 · −
X (cid:0) (cid:1)
for t>0 and m N. The proof is complete.  (cid:3)
∈
6 C.-P.CHENANDR.B.PARIS
Theorem 2.2. For t>0 and N N, we have
∈
N−1
E
secht= 2j t2j +R (t) (2.12)
N
(2j)!
j=0
X
with
∞
( 1)N2t2N ( 1)k
R (t)= − − , (2.13)
N π2N−1 (k+ 1)2N−1 t2+π2(k+ 1)2
Xk=0 2 2
(cid:16) (cid:17)
and
N−1
E E
secht= 2j t2j +Θ(t,N) 2N t2N (2.14)
(2j)! (2N)!
j=0
X
with a suitable 0<Θ(t,N)<1.
Proof. It follows from [34, p. 136] (see also [5, p. 458, Eq. (27.3)]) that
∞
π ( 1)k(2k+1)
= − ,
4cosh πx (2k+1)2+x2
2 k=0
X
which can be w(cid:0)ritt(cid:1)en as
∞
4 ( 1)k
secht= − . (2.15)
π 2
k=0(2k+1) 1+ 2t
X π(2k+1)
(cid:18) (cid:16) (cid:17) (cid:19)
Substitution of x= 1 in (2.8) leads to
2
∞
( 1)k ( 1)jπ2j+1
− = − E . (2.16)
(2k+1)2j+1 22j+2(2j)! 2j
k=0
X
Using the identity
1 N−1 qN
= ( 1)jqj +( 1)N (q = 1) (2.17)
1+q − − 1+q 6 −
j=0
X
and (2.16), we obtain from (2.15) that
2N
secht= 4 ∞ (−1)k N−1( 1)j 2t 2j +( 1)N π(22kt+1)
π (2k+1) − π(2k+1) − (cid:16) (cid:17) 2
k=0 j=0 (cid:18) (cid:19) 1+ 2t
X X π(2k+1)
 
N−1  (cid:16) (cid:17) 
E
= 2j t2j +R (t),
N
(2j)!
j=0
X
with
∞
2 ( 1)N+k t2N
R (t)= − .
N π2N−1 (k+ 1)2N−1 t2+π2(k+ 1)2
k=0 2 2
X
Noting that (2.16) holds, we find that R(cid:0) (t) can be wr(cid:1)itten as
N
E t2N F(t)
2N
R (t)=Θ(t,N) , Θ(t,N):= ,
N
(2N)! F(0)
SOME RESULTS ASSOCIATED WITH BERNOULLI AND EULER NUMBERS WITH APPLICATIONS 7
where
∞
1 1
F(t):= ( 1)kα , α := .
− k k (k+ 1)2N−1 t2+π2(k+ 1)2
k=0 2 2
X
Then it is easily seen that α > α for k N , t > 0 and N N; thus F(t) > 0 for t > 0.
2k 2k+1 0
∈ ∈
Differentiation yields
∞
( 1)kα
′ k
F (t)= 2t −
− t2+π2(k+ 1)2
k=0 2
X
and a similar reasoning shows that F′(t)<0 for t>0. Hence, for all t>0 and N N, we have
0<F(t)<F(0) and thus 0<Θ(t,N)<1. The proof of Theorem 2.2 is complete.∈ (cid:3)
Remark 2.3. Recalling that
E >0 and E <0 (m N ),
4m 4m+2 0
∈
we can deduce (1.18) from (2.14). Note that the inequality (1.18) can also be written as
m
E
( 1)m+1 secht 2j t2j >0 (t>0, m N ). (2.18)
0
−  − (2j)!  ∈
j=0
X
 
Remark 2.4. It was shown in [6] that (1.13) can be written as
m 2j
1 E t
2j
ν (t)= + (2.19)
m −2cosh(t) 2(2j)! 4
4 j=0 (cid:18) (cid:19)
X
and (1.15) is equivalent to (2.18). Hence, for t>0 and m N , (1.15) holds true.
0
∈
It was also shown in [6] that
∞ et/4 e3t/4 1 e−xt m 2B (1) ∞
V (x)=( 1)m − + dt 2j+1 4 t2j−1e−xtdt
m − "Z0 (cid:18) et−1 2(cid:19) t −j=1 (2j+1)! Z0 #
X
∞ e−xt
= ( 1)mν (t) dt. (2.20)
m
− t
Z0
We obtain from (2.20) that for all m N ,
0
∈
∞
( 1)nV(n)(x)= ( 1)mν (t)tn−1e−xtdt>0
− m − m
Z0
for x>0 and n N . Hence, the function V (x), defined by (1.17), is completely monotonic on
0 m
∈
(0, ).
∞
Sondow and Hadjicostas [29] introduced and studied the generalized-Euler-constant function
γ(z), defined by
∞
1 n+1
γ(z)= zn−1 ln , (2.21)
n − n
n=1 (cid:18) (cid:19)
X
where the series converges when z 1. Pilehrood and Pilehrood [22] considered the function
| | ≤
zγ(z) (z 1). The function γ(z) generalizes both Euler’s constant γ(1) and the alternating
| | ≤
Euler constant ln 4 = γ( 1) [27,28]. An interesting comparison by Sondow [27] is the double
π −
integral and alternating series
4 1 1 x 1 ∞ 1 n+1
ln = − dxdy = ( 1)n−1 ln . (2.22)
π (1+xy)ln(xy) − n − n
Z0 Z0 n=1 (cid:18) (cid:19)
X
8 C.-P.CHENANDR.B.PARIS
The formula (2.20) can provide integral representations for the constant π. For example, the
choice (x,m)=(1/4,0) in (2.20) yields
∞ et/4 e3t/4 1 2e−t/4 4
− + dt=ln , (2.23)
et 1 2 t π
Z0 (cid:18) − (cid:19)
which provides a new integralrepresentationfor the alternating Euler constantln 4. The choice
π
(x,m)=(3/4,0) in (2.20) yields
∞ et/4 e3t/4 1 2e−3t/4 π
− + dt=ln . (2.24)
et 1 2 t 3
Z0 (cid:18) − (cid:19)
Many formulas exist for the representation of π, and a collection of these formulas is listed
in [25,26]. For more history of π see [3,4,10].
Noting [6, Eq. (3.26)] that B (1) can be expressed in terms of the Euler numbers
2n+1 4
(2n+1)E
B (1)= 2n (n N ), (2.25)
2n+1 4 − 42n+1 ∈ 0
we find that (1.17) can be written as
Γ(x+ 3) m E 1
V (x)=( 1)m ln 4 + 2j . (2.26)
m −  x1/2Γ(x+ 1) j 42j+1x2j
4 j=1 ·
X
 
From the inequalities V (x)>0 for x>0, we obtain the following
m
Corollary 2.2. For x>0,
2m E 1 Γ(x+ 3) 2m+1 E 1
x1/2exp 2j < 4 <x1/2exp 2j . (2.27)
− j 42j+1x2j Γ(x+ 1) − j 42j+1x2j
j=1 · 4 j=1 ·
X X
   
TheproblemoffindingnewandsharpinequalitiesforthegammafunctionΓand,inparticular,
for the Wallis ratio
(2n 1)!! 1 Γ(n+ 1)
− = 2 (2.28)
(2n)!! √π Γ(n+1)
hasattractedtheattentionofmanyresearchers(see[8,9,14–16,18]andreferencestherein). Here,
we employ the special double factorial notation as follows:
(2n)!!=2 4 6 (2n)=2nn!, 0!!=1, ( 1)!!=1,
· · ··· −
1
(2n 1)!!=1 3 5 (2n 1)=π−1/22nΓ n+ ;
− · · ··· − 2
(cid:18) (cid:19)
see [1, p. 258]. For example, Chen and Qi [8] proved that for n N,
∈
1 (2n 1)!! 1
− < , (2.29)
π n+ 4 1 ≤ (2n)!! π n+ 1
π − 4
q q
where the co(cid:0)nstants 4(cid:1) 1 and 1 are the be(cid:0)st poss(cid:1)ible. This inequality is a consequence of the
π − 4
complete monotonicity on (0, ) of the function (see [7])
∞
Γ(x+1)
V(x)= . (2.30)
x+ 1Γ(x+ 1)
4 2
q
SOME RESULTS ASSOCIATED WITH BERNOULLI AND EULER NUMBERS WITH APPLICATIONS 9
If we write (2.27) as
1 2m+1 E 1 Γ(x+ 1) 1 2m E 1
exp 2j < 4 < exp 2j
√x  j 42j+1x2j Γ(x+ 3) √x  j 42j+1x2j
j=1 · 4 j=1 ·
X X
   
and replace x by x+ 1, we find
4
1 2m+1 E 1 Γ(x+ 1)
exp 2j < 2
x+ 41  Xj=1 j·42j+1(x+ 14)2j Γ(x+1)
q  
2m
1 E 1
2j
< exp . (2.31)
x+ 14 Xj=1 j·42j+1(x+ 14)2j
q  
Noting that (2.28) holds, we then deduce from (2.31) that
2m+1
1 E 1 (2n 1)!!
2j
exp < −
π(x+ 14)  Xj=1 j·42j+1(x+ 14)2j (2n)!!
q  
2m
1 E 1
2j
< exp , (2.32)
π(x+ 14) Xj=1 j·42j+1(x+ 14)2j
q  
which generalizes a recently published result of Chen [6, Eq. (3.40)], who proved the inequality
(2.32) for m=1.
Theorem 2.3. For t>0 and N N , we have
0
∈
N 22jB
cotht= 2jt2j−1+σ (t), (2.33)
N
(2j)!
j=0
X
where
∞
( 1)Nt2N+1 2
σ (t)= − , (2.34)
N π2N k2N(t2+π2k2)
k=1
X
and
N 22jB 22N+2B
cotht= 2jt2j−1+θ(t,N) 2N+2t2N+1 (2.35)
(2j)! (2N +2)!
j=0
X
with a suitable 0<θ(t,N)<1.
Proof. It follows from [20, p. 126, Eq. (4.36.3)] that
∞ ∞
1 1 1 2t 1
cotht= +2t = + . (2.36)
t π2k2+t2 t π2 k2 1+( t )2
k=1 k=1 πk
X X
It is well known that (cid:0) (cid:1)
∞ 1 ( 1)j−1(2π)2jB
2j
= − . (2.37)
k2j 2(2j)!
k=1
X
10 C.-P.CHENANDR.B.PARIS
Using (2.17) and (2.37), we obtain from (2.36) that
1 ∞ 1 N−1 t 2j t 2N
cotht= +2t ( 1)j +( 1)N kπ
t Xk=1k2π2 Xj=0 − (cid:18)kπ(cid:19) − 1(cid:0)+ (cid:1)ktπ 2
1 N−122j+2B (cid:0) (cid:1) 
= + 2j+2t2j+1+σ (t)
N
t (2j+2)!
j=0
X
N 22jB
= 2jt2j−1+σ (t)
N
(2j)!
j=0
X
with
∞
2( 1)N t2N+1
σ (t)= − .
N π2N k2N(t2+π2k2)
k=1
X
Noting that (2.37) holds, we can rewrite σ (t) as
N
22N+2B
σ (t)=θ(t,N) 2N+2 t2N+1,
N
(2N +2)!
where
∞
f(t) 1
θ(t,N):= , f(t):= .
f(0) k2N(t2+π2k2)
k=1
X
Obviously, f(t)>0 and is strictly decreasing for t>0. Hence, for all t>0, 0<f(t)<f(0)and
thus 0<θ(t,N)<1. The proof of Theorem 2.3 is complete. (cid:3)
The following expansion for Barnes G-function was established by Ferreira and Lo´pez [11,
Theorem 1]. For arg(z) <π,
| |
1 1 1 1
lnG(z+1)= z2+zlnΓ(z+1) z2+ z+ lnz lnA
4 − 2 2 12 −
(cid:18) (cid:19)
N−1
B
+ 2k+2 + (z) (N N),
2k(2k+1)(2k+2)z2k RN ∈
k=1
X
where B are the Bernoulli numbers and A is the Glaisher–Kinkelin constant defined by
2k+2
n n2 n 1 n2
lnA= lim ln kk + + lnn+ , (2.38)
n→∞( k=1 !−(cid:18) 2 2 12(cid:19) 4 )
Y
the numerical value of A being 1.282427.... For (z)>0, the remainder (z) is given by
N
ℜ R
∞ t 2N B e−zt
(z)= ktk dt. (2.39)
RN Z0 et−1 −k=0 k! ! t3
X
Estimatesfor (z) werealsoobtainedbyFerreiraandLo´pez[11],showingthattheexpansion
N
|R |
is indeed an asymptotic expansion of lnG(z + 1) in sectors of the complex plane cut along
the negative axis. Pedersen [21, Theorem 1.1] proved that for any N 1, the function x
≥ 7→
( 1)N (x) is completely monotonic on (0, ).
N
− R ∞