arXiv:1409.6411v1 [math.CA] 23 Sep 2014
SEVERAL COMPLETELY MONOTONE FUNCTIONS RELATED
TO DETEMPLE’S SEQUENCE
ZHEN-HANG YANG
Abstract. In this paper, we present the necessary and sufficient conditions
such that several functions involving R (x) = ψ (x + 1/2) − ln x with a parameter are completely monotone on (0, ∞), where ψ is the digamma function.
This generalizes some known results and verifies a conjecture posed by Chen.
1. Introduction
A function f is said to be completely monotonic on an interval I , if f has
derivatives of all orders on I and satisfies
(1.1)
(−1)n f (n) (x) ≥ 0 for all x ∈ I and n = 0, 1, 2, ....
If the inequality (1.1) is strict, then f is said to be strictly completely monotonic
on I. It is known (Bernstein’s Theorem) that f is completely monotonic on (0, ∞)
if and only if
f (x) =
Z
∞
e−xt dµ (t) ,
0
where µ is a nonnegative measure on [0, ∞) such that the integral converges for all
x > 0, see [1, p. 161].
For x > 0 the classical Euler’s gamma function Γ and psi (digamma) function ψ
are defined by
Z ∞
Γ′ (x)
,
(1.2)
Γ (x) =
tx−1 e−t dt,
ψ (x) =
Γ (x)
0
respectively. The derivatives ψ ′ , ψ ′′ , ψ ′′′ , ... are known as polygamma functions.
As the important role played in many branches, such as mathematical physics,
probability, statistics, engineering, the gamma and polygamma functions have attracted the attention of many scholars. In particular many authors published numerous interesting inequalities for the Euler-Mascheroni constant defined by
!
n
X
1
γ = lim Dn := lim
− ln n = 0.577215664...,
n→∞
n→∞
k
k=1
as well as digamma (psi) function ψ due to ψ (n + 1) = Hn − γ.
Date: July 15, 2014.
2010 Mathematics Subject Classification. 33B15, 26D15.
Key words and phrases. Psi function, completely monotone function, inequality.
This paper is in final form and no version of it will be submitted for publication elsewhere.
1
2
ZHEN-HANG YANG
It is known that the sequence Dn converges very slowly to its limit, due to
1
1
< Dn − γ <
2 (n + 1)
2n
(see [2], [3]). Some quicker approximations to the Euler-Mascheroni constant were
established in [4], [5], [6], [7], [8], [9], [10], [11], [12], [13], [14], [15], [16], [17], [18],
[19], [20], [21], [22], [23], [24], [25], [26], [27].
In 1993, DeTemple [7] defined a modified sequence
n
X
1
1
(1.3)
Rn =
− ln n +
k
2
k=1
to approximate to γ, which converges faster since
1
1
(1.4)
2 < Rn − γ < 24n2 ,
24 (n + 1)
and
(1.5)
1
1
7
7 1
< Rn − γ −
<
.
4
2
960 (n + 1)
24n
960 n4
Villarino [14, Theorem 1.7] proved that for n ∈ N, the following inequality
(1.6)
1
24(n+1/2)2 +21/5
< Rn − γ <
1
24(n+1/2)2 +1/(1−ln 3+ln 2−γ)−54
is valid with the best constants 21/5 and 1/ (1 − ln 3 + ln 2 + Psi (1))−54 ≈ 3.7393.
Chen [18] proved that
−2
1
1
−2
1
n+
,
(1.7)
< Rn − γ <
24 (n + λ)
24
2
where
1
− 1 ≈ 0.551 07
λ= p
2 6 (1 + Psi (1) − ln 3 + ln 2)
and 1/2 are the best constants.
Mortici [28, Theorem 2.1] gave new bounds as follows:
−2
−2
1
1
1
1
7
(1.8)
n+ +
n+
< Rn − γ <
.
24
2 80n
24
2
In [29], [30] (also see [31]), it was established that
γ + ln n +
which is equivalent to
1
2
<
n
X
1
≤ γ + ln n + e1−γ − 1 ,
k
k=1
0 < Rn − γ ≤ ln
n + e1−γ − 1
.
n + 1/2
On the other hand, Karatsuba [32] proved that for all integers n ≥ 1, H(n) <
H(n + 1), where H(n) is defined by
(1.9)
H(n) = (Rn − γ) n2 .
In view of ψ (n + 1) = Hn − γ, we see that
1
2
n2 .
H(n) = (Rn − γ) n = ψ (n + 1) − ln n +
2
SEVERAL COMPLETELY MONOTONE FUNCTIONS
3
As mentioned in [33], some computer experiments also seem to indicate that (1 +
1/n)2 H(n) is a decreasing convex function. Chen [15, Theorem 2] proved that
for all integers n ≥ 1, H(n) and ((n + 1/2) /n)2 H(n) are both strictly increasing
2
concave sequences, while [((n + 1)/n) H(n) is strictly decreasing convex sequence.
Also, he conjectured in [15, Cojecture 1] that
(i) the functions H(x) and ((x + 1/2) /x)2 H(x) are both so-called Bernstein
function on (0, ∞). That is,
n
H(x)
(n+1)
> 0, (−1) [H(x)]
> 0,
(n+1)
n
> 0, (−1) ((x + 1/2) /x)2 H(x)
>0
((x + 1/2) /x)2 H(x)
hold for x > 0 and n ∈ N.
2
(ii) The function ((x + 1)/x) H(x) is strictly completely monotonic on (0, ∞).
We remark that the functions H(x) is not Bernstein function on (0, ∞) due to
−H ′′ (0+ ) =
2 ln 2 − 2γ ≈ −0.231 86 < 0,
7
7
−H ′′ (1/2) = 2γ + 4 ln 2 + ζ (3) − π 2 + ≈ 0.014615 > 0.
2
4
The aim of this paper is to prove the complete monotonicity of certain functions
involving
(1.10)
R (x) = ψ (x + 1/2) − ln x.
In Section 2, some useful lemmas are given. Our main results are presented in
Section 3, in which Theorem 1 indicates that the second conjecture posed by Chen
is valid, while Theorems 2–4 show the necessary and sufficient conditions such
that three functions involving R (x) with a parameter are completely monotone
on (0, ∞). In the last section, some remarks, conjecture and open problem are
presented.
2. Lemmas
In order to prove our results, we need some lemmas.
Lemma 1. For t > 0, let the function Q defined by
(2.1)
Q (t) =
1
1
−
.
t
2 sinh 2t
Then
(2.2)
R (x) =
Z
∞
e−xt Q (t) dt,
0
(2.3)
xR (x) =
(2.4)
x2 R (x) =
(2.5)
x3 R (x) =
(2.6)
x4 R (x) =
Z
∞
e−xt Q′ (t) dt,
0
Z ∞
1
+
e−xt Q′′ (t) dt,
24
0
Z ∞
1
x+
e−xt Q′′′ (t) dt,
24
0
Z ∞
7
1 2
e−xt Q(4) (t) dt.
x −
+
24
960
0
4
ZHEN-HANG YANG
Proof. Using the integral representations [34, p. 259]
Z ∞ −t
Z ∞ −t
e − e−xt
e−xt
e
dt
and
ln
x
=
−
dt
ψ(x) =
t
1 − e−t
t
0
0
gives
R (x) =
=
Z ∞ −xt
e
e−(x+1/2)t
ψ(x + 21 ) − ln x =
dt
−
t
1 − e−t
0
Z ∞
Z ∞
1
1
dt
=
e−xt Q (t) dt.
e−xt
−
t
2 sinh 2t
0
0
An integration by part yields
Z ∞
Z ∞
−xt
Q (t) de−xt
e Q (t) dt = −
xR (x) = x
0
0
Z ∞
Z ∞
−xt ′
= −e−xt Q (t) |∞
+
e
Q
(t)
dt
=
e−xt Q′ (t) dt,
0
0
0
where the last equality holds due to limt→∞ (e−xt Q (t)) = limt→0 (e−xt Q (t)) = 0.
Integrations by parts again and noticing that
1
lim e−xt Q′ (t) = 0 and lim e−xt Q′ (t) = −
t→∞
t→0
24
lead to
Z ∞
Z ∞
e−xt Q′′ (t) dt
e−xt Q′ (t) dt = −e−xt Q′ (t) |∞
+
x2 R (x) = x
0
0
0
Z ∞
1
−xt ′′
+
e Q (t) dt.
=
24
0
Similarly, we have
x3 R (x)
=
=
Z ∞
Z ∞
1
1
x+x
e−xt Q′′ (t) dt =
x − e−xt Q′′ (t) |∞
+
e−xt Q′′′ (t) dt
0
24
24
0
Z ∞0
1
−xt ′′′
e Q (t) dt
x+
24
0
due to limt→∞ (e−xt Q′′ (t)) = limt→0 (e−xt Q′′ (t)) = 0 and
Z ∞
Z ∞
1 2
1 2
x4 R (x) =
x +x
e−xt Q′′′ (t) dt =
x − e−xt Q′′′ (t) |∞
+
e−xt Q(4) (t) dt
0
24
24
0
0
Z ∞
7
1 2
−xt (4)
x −
+
e Q (t) dt
=
24
960
0
due to
7
lim e−xt Q′′′ (t) = 0 and lim e−xt Q′′′ (t) = −
.
t→∞
t→0
960
This completes the proof.
In the same method as Lemma 5 in [35], we can prove the following statement.
Lemma 2 ([35, Lemma 5]). Let P (t) be a power series which is convergent on
(0, ∞) defined by
m
∞
X
X
i
a i ti ,
ai t −
P (t) =
i=m+1
i=0
SEVERAL COMPLETELY MONOTONE FUNCTIONS
5
where ai ≥ 0 for i ≥ m + 1 with maxi≥m+1 (ai ) > 0 and am > 0, ai ≥ 0 for
0 ≤ i ≤ m − 1. Then there is a unique number t0 ∈ (0, ∞) to satisfy P (t) = 0 such
that P (t) < 0 for t ∈ (0, t0 ) and P (t) > 0 for t ∈ (t0 , ∞).
Using Lemma 2 we can prove the following
Lemma 3. Let the function Q defined by (2.1). Then Q′ (t) ≥ c0 Q (t) for t > 0,
where
′ Q (t)
≈ −0.061875.
(2.7)
c0 = min
t>0
Q (t)
Proof. Differentiation yields
t
Q′ (t)
=
Q (t)
Q′ (t)
Q (t)
′
=
:
where
p (t)
1 cosh 2
1
4 sinh2 2t − t2
1
1
t − 2 sinh 2t
=
t2 cosh 2t − 4 sinh2
4t sinh2
t
2
t
2
− 2t2 sinh 2t
,
t
t
t
+ t4 cosh2 + 2t3 sinh3
2
2
2
t
t
4 t
2 t
2 t
2 t
2
3
4
+ 16 sinh
+ 8t cosh sinh
− 4t cosh
sinh )
−t sinh
2
2
2
2
2
2
p 2t
=
,
2
4t2 t − 2 sinh 2t sinh2 2t
1
2
4t2 (t−2 sinh 2t ) sinh2
t
2
(−16t sinh3
= 16(t4 cosh2 t + t3 sinh3 t − t4 sinh2 t + sinh4 t − 2t sinh3 t
+2t2 cosh t sinh2 t − 2t3 cosh2 t sinh t.
If we prove that there is a unique number t0 > 0 such that Q′ /Q is decreasing on
(0, t0 ) and increasing on (t0 , ∞), then we have Q′ (t) /Q (t) ≥ Q′ (t0 ) /Q (t0 ) for all
t > 0. For this end, we use the ”product into sum” formulas and Taylor expansion
to get
1
p (t)
2
= cosh 4t + 4t2 cosh 3t − 2t3 sinh 3t − 4t sinh 3t − 4 cosh 2t
−4t2 cosh t − 10t3 sinh t + 12t sinh t + 8t4 + 3
∞
∞
∞
∞
X
X
X
X
42n t2n
32n−2 t2n
32n−3 t2n
32n−1 t2n
=
+4
−2
−4
(2n)!
(2n − 2)!
(2n − 3)!
(2n − 1)!
n=0
n=1
n=2
n=1
−4
∞
∞
∞
X
X
X
22n t2n
t2n
t2n
−4
− 10
(2n)!
(2n − 2)!
(2n − 3)!
n=0
n=1
n=2
+12
:
where
∞
X
t2n
+ 8t4 + 3
(2n
−
1)!
n=1
∞
X
un 2n
=
t ,
(2n)!
n=3
un = 42n − 8n 2n2 − 9n + 13 32n−3 − 22n+2 − 8n 10n2 − 13n + 1 .
6
ZHEN-HANG YANG
A direct verification gives u3 = 0, un < 0 for 4 ≤ n ≤ 10, and we now show that
un ≥ 0 for n ≥ 11. It is easy to check that un satisfies the recursive relation
un+1 − 16un = 8 14n3 − 117n2 + 199n − 54 32n−3 + 48 × 22n
+1200n3 − 1800n2 + 88n + 16
>
0 for n ≥ 11,
which together with u11 = 1636 643 754 240 > 0 indicates that un ≥ 0 for n ≥ 11.
By Lemma 2 we see that there is a unique t′0 > 0 such that p (t) < 0 for t ∈ (0, t′0 )
and p (t) < 0 for t ∈ (t′0 , ∞), which implies that Q′ /Q is decreasing on (0, 2t′0 ) and
increasing on (2t′0 , ∞). Numeric calculation shows that t0 = 2t′0 = 15.40151..., and
so Q′ (t0 ) /Q (t0 ) ≈ −0.061875.
This completes the proof.
To simplify the proofs of some crucial inequalities, we need to the following
lemma.
Lemma 4. For t > 0, the following inequalities hold:
sinh t
t
sinh t
t
sinh t
t
(2.8)
(2.9)
(2.10)
>
>
<
2 cosh t + 3
,
cosh t + 14
2 cosh2 t + 10 cosh t + 9
,
15
2 cosh2 t + 101 cosh t + 212
18 cosh2 t + 160 cosh t + 179
15
cosh t.
1159 cosh2 t + 4192 cosh t + 4
3
Proof. The inequality (2.8) has been proven in [36, Theorem 18].
To prove (2.9), it suffices to show that for t > 0
p1 (t) :=
2 cosh2 t + 101 cosh t + 212
sinh t − 15t > 0.
2 cosh2 t + 10 cosh t + 9
Differentiating and factoring give
p′1 (t)
=
2 cosh2 t+101 cosh t+212
2 cosh2 t+10 cosh t+9
2
cosh t+116 cosh t+173
cosh t − 7 26
sinh2 t − 15
(2 cosh2 t+10 cosh t+9)2
5
=
4 (cosh t − 1)
2 cosh2 t + 10 cosh t + 9
2 > 0,
which yields p1 (t) > p1 (0) = 0.
Similarly, inequality (2.10) is equivalent to
p2 (t) :=
1159 cosh2 t + 4192 cosh t + 4
sinh t − 15t < 0.
18 cosh2 t + 160 cosh t + 179 cosh t
Differentiating and factoring lead us to
p′2 (t) =
=
1159 cosh2 t+4192 cosh t+4
(18 cosh2 t+160 cosh t+179) cosh t
−
4 (1215x + 179) (x − 1)5
2
x2 (18x2 + 160x + 179)
cosh t + sinh2 t
< 0,
d 1159x2 +4192x+4 − 15
2
dx (18x +160x+179)x
where x = cosh t > 1. This implies that p2 (t) < p2 (0) = 0.
SEVERAL COMPLETELY MONOTONE FUNCTIONS
7
Lemma 5. The function Q defined by (2.1) satisfies
q1 (t) := Q′′ (t) +
7
Q (t) > 0
40
for all t > 0.
Proof. Differentiation yields
2 t
2
t
2
1 cosh
1
−
4 sinh3
8 sinh 2t
Q′′ (t) =
(2.11)
2
,
t3
+
and then
q1 (t) = −
2 t
2
1 20t3 cosh
80
− 14t2 sinh3
t
2
− 3t3 sinh2
t
2
− 160 sinh3
t3 sinh3 2t
t
2
.
To prove the desired inequality, we write q1 (2t) in the form of
3
sinh t
sinh t
80 sinh3 t q1 (2t) = 20
+ 7 sinh2 t
+ 3 sinh2 t − 20 cosh2 t.
t
t
Employing inequality (2.8), we get
3
2 cosh t + 3
2 cosh t + 3
+ 7 cosh2 t − 1 3
80 sinh3 t q1 (2t) > 20 3
cosh t + 14
cosh t + 14
2
2
+3 cosh t − 1 − 20 cosh t
= 25 cosh2 t + 24 cosh t + 240
which proves the desired inequality.
(cosh t − 1)3
3
(cosh t + 14)
> 0,
Lemma 6. The function Q defined by (2.1) satisfies
q2 (t) := Q(4) (t) −
31
Q (t) < 0
336
for all t > 0.
Proof. Differentiation yields
(2.12)
Q(4) (t) =
2
4
7 cosh 12 t 3 cosh 12 t
5
24
−
−
+ 5,
8 sinh3 21 t
4 sinh5 12 t
t
32 sinh 21 t
and then
1
q2 (t) = − 336
252t5 cosh4
4
t
2 +31t
sinh5
5
t
2 +37t
sinh4 2t −8064 sinh5
t5 sinh5 2t
5
t
2 −294t
cosh2
t
2
sinh2
t
2
.
A simple identical transformation gives
5
sinh t
sinh t
+ 31 sinh4 t
− 672 sinh5 t q2 (2t) = −504
t
t
+504 cosh4 t − 588 cosh2 t sinh2 t + 74 sinh4 t.
In order to prove the desired inequality, we define
U (y) = −504y 5 + 31(sinh4 t)y + 504 cosh4 t − 588 cosh2 t sinh2 t + 74 sinh4 t,
and it suffices to prove that U ((sinh t) /t) > 0 for t > 0.
8
ZHEN-HANG YANG
′
An easy computation
= 31 sinh4 t − 2520y 4, which reveals that U is
p yields U (y) p
increasing for y ∈ 1, 4 31/2520 sinh t and decreasing for y ∈ [ 4 31/2520 sinh t, ∞).
We now distinguish two cases to prove U ((sinh t) /t) > 0 for t > 0.
p
In the case when t ∈ 4 2520/31, ∞ , by inequality (2.8), we have
r
sinh t
2 cosh t + 3
31
4
<
<
sinh t,
1<3
cosh t + 14
t
2520
that is,
2 cosh t + 3 sinh t p
3
,
∈ 1, 4 31/2520 sinh t ,
cosh t + 14
t
and so
2 cosh t + 3
sinh t
U
> U 3
= 504 cosh4 t − 588 cosh2 t sinh2 t + 74 sinh4 t
t
cosh t + 14
5
2 cosh t + 3
2 cosh t + 3
4
+31 sinh t × 3
.
− 504 3
cosh t + 14
cosh t + 14
Putting cosh t = x, then sinh2 t = x2 − 1, and factoring yield
3
(x − 1)
sinh t
>
U
5 U1 (x)
t
(x + 14)
where
U1 (x) =
176x6 + 10 523x5 + 245 869x4 + 2810 864x3
+12 467 224x2 + 12 511 688x − 20 756 344.
It is easy to verify that U1 (x) > U1 (1)
= 7290 000 > 0, which implies that
p
4
2520/31, ∞ .
p
In the case of when t ∈ (0, 4 2520/31], from inequality (2.10) we see that
r
31
sinh t
18 cosh2 t + 160 cosh t + 179
4
cosh t >
∞ > 15
>
sinh t.
2
t
2520
1159 cosh t + 4192 cosh t + 4
p
Using the decreasing property of U for u ∈ 4 31/2520 sinh t, ∞ , we have
18 cosh2 t + 160 cosh t + 179
sinh t
cosh
t
.
> U 15
U
t
1159 cosh2 t + 4192 cosh t + 4
Letting cosh t = x and factoring indicate that
(x − 1)4
18x2 + 160x + 179
x
=
U 15
5 U2 (x) ,
1159x2 + 4192x + 4
(1159x2 + 4192x + 4)
U ((sinh t) /t) > 0 for t ∈
where
U2 (x)
=
14 379 675 269 523 570x11 + 357 214 567 270 415 330x10
+3604 910 878 299 956 955x9 + 19 027 526 850 473 930 600x8
+55 570 610 110 726 848 080x7 + 85 295 682 448 077 545 696x6
+54 079 668 524 631 977 864x5 + 560 130 320 580 220 160x4
+1016 873 963 329 280x3 + 923 378 178 560x2 + 418 677 504x + 75 776.
p
Evidently, U2 (x) > 0, and so U ((sinh t) /t) > 0 for t ∈ (0, 4 2520/31].
SEVERAL COMPLETELY MONOTONE FUNCTIONS
This proof is completed.
9
Lemma 7. The function Q defined by (2.1) satisfies
199 849
11 165 ′′
Q (t) +
Q (t) > 0
q3 (t) := Q(4) (t) +
8284
1391 712
for all t > 0.
Proof. From (2.1), (2.11) and (2.12) it is obtained that
t
t
t
q3 (t) = 2783 424t15 sinh5 t (−2087 568t5 cosh4 + 1497 636t5 cosh2 sinh2
2
2
2
2
t
t
−165 829t5 sinh4 + 399 698t4 sinh5
2
2
t
t
+7502 880t2 sinh5 + 66 802 176 sinh5 ).
2
2
We write q3 (2t) as
2783 424 sinh5 t q3 (2t)
= − 2087 568 cosh4 t + 165 829 sinh4 t − 1497 636 cosh2 t sinh2 t
t
t 3
t 5
+199 849 sinh4 t sinh
+ 937 860 sinh2 t sinh
+ 2087 568 sinh
.
t
t
t
Utilizing inequality (2.9) and putting cosh t = x > 1 and then factoring yield
2
2783 424 sinh5 t q3 (2t) > − 2087 568x4 + 165 829 x2 − 1 − 1497 636x2 x2 − 1
2
2x2 + 10x + 9
15 2
+199 849 x2 − 1
2x + 101x + 212
3
2x2 + 10x + 9
2
+937 860 x − 1 15 2
2x + 101x + 212
5
2
2x + 10x + 9
+2087 568 15 2
2x + 101x + 212
5
=
7 (x − 1)
5 q4
(2x2 + 101x + 212)
where the last inequality holds due to
q4 (x)
(x) > 0,
= 10 249 024x9 + 2015 594 800x8 + 163 876 520 192x7 + 6681 271 280 040x6
+136 012 433 414 956x5 + 1069 481 086 377 851x4 + 4121 483 475 973 500x3
+8450 810 874 059 188x2 + 8899 895 239 232 240x + 3802 278 457 617 584.
This completes the proof.
3. Main Results
Now we state and prove the first result, which shows that the second conjecture
posed by Chen is valid.
Theorem 1. Let the function R be defined on (0, ∞) by (1.10). Then the function
ha (x) = (x + a)2 R (x)
is strictly completely monotonic on (0, ∞) if a ≥ a0 =
where c0 ≈ −0.061875 is defined by (2.7).
p
c20 + 7/40 − c0 ≈ 0.48476,
10
ZHEN-HANG YANG
Proof. Using the relations (2.2)–(2.4) we get that
ha (x) =
=
=
(x + a)2 R (x) = x2 R (x) + 2axR (x) + a2 R (x)
Z ∞
Z ∞
Z ∞
1
e−xt Q′ (t) dt + a2
e−xt Q′′ (t) dt + 2a
e−xt Q (t) dt
+
24
0
0
0
Z ∞
1
+
e−xt Q′′ (t) + 2aQ′ (t) + a2 Q (t) dt
24
0
Z ∞
△ 1
=
+
e−xt Q (t) δ a (t) dt.
24
0
Evidently, for t > 0, Q (t) > 0 and by Lemmas 3 and 5,
δ a (t) =
=
if a ≥
Q′ (t)
7
Q′′ (t)
+ 2a
+ a2 ≥ a2 + 2ac0 −
Q (t)
Q (t)
40
!
!
r
r
7
7
a + c0 + c20 +
a + c0 − c20 +
≥0
40
40
p
c20 + 7/40 − c0 , which proves the desired result.
Taking a = 1/2 and replacing x by (x + 1/2) in the above theorem, we have
Corollary 1. The function ((x + 1)/x)2 H(x) is strictly completely monotonic on
(−1/2, ∞).
Remark 1. Evidently, Theorem 1 reveals that the second conjecture posed by Chen
in [15, Theorem 2] is true.
Our second result states that
Theorem 2. Let the function R be defined on (0, ∞) by (1.10). Then the function
x 7→ Fa (x) = 24 x2 + a R (x) − 1
is strictly completely monotonic on (0, ∞) if and only if a ≥ a1 = 17/40.
Proof. The necessity follows from
24 x2 + a (ψ (x + 1/2) − ln x) − 1
7
Fa (x)
= lim
=a−
≥ 0.
lim
−2
x→∞
x→∞ x−2
x
40
Using the relations (2.2) and (2.4) we obtain that
Fa (x) = 24 x2 + a R (x) − 1 = 24x2 R (x) + 24aR (x) − 1
Z ∞
Z ∞
1
−xt ′′
= 24
+
e Q (t) dt + 24a
e−xt Q (t) dt − 1
24
0
0
Z ∞
e−xt (Q′′ (t) + aQ (t)) dt.
= 24
0
If a ≥ 7/40, then by Lemma 5 it follows that
Q′′ (t) + aQ (t) ≥ Q′′ (t) +
for t > 0, which proves the sufficiency.
7
Q (t) > 0
40
SEVERAL COMPLETELY MONOTONE FUNCTIONS
11
Noting that
1
= 24 (n + 1/2)2 + 7/40 Rn − 1
F7/40 n +
2
and the facts that
286 291 3 291
−
ln −
γ ≈ 0.007979 and F7/40 (∞) = 0,
5
5
2
5
by Theorem 3 we get immediately
F7/40 (3/2) =
Corollary 2. Let Rn be defined by (1.3). Then the double inequality
(3.1)
1
1 + λ1
< Rn − γ <
2
2
24 (n + 1/2) + 7/40
24 (n + 1/2) + 7/40
holds for n ∈ N, where λ1 = f7/40 (3/2) ≈ 0.007979 is the best constant.
Our third result is contained in the following theorem.
Theorem 3. Let the function R be defined on (0, ∞) by (1.10). Then the function
7
x 7→ fa (x) = −24 x4 + a R (x) + x2 − ,
40
is strictly completely monotonic on (0, ∞) is and only if a ≤ a2 = −31/336.
Proof. The necessity can be deduced by
7
−24 x4 + a (ψ (x + 1/2) − ln x) + x2 − 40
31
fa (x)
=
lim
= −a −
≥ 0.
lim
−2
x→∞
x→∞ x−2
x
336
By the relations (2.2), (2.4) and (2.6), we obtain that
7
fa (x) = −24x4 R (x) − 24aR (x) + x2 −
40
Z ∞
Z ∞
7
7
1 2
−xt (4)
x −
+
e Q (t) dt − 24a
e−xt Q (t) dt + x2 −
= −24
24
960
40
0
0
Z ∞
= 24
e−xt −Q(4) (t) − aQ (t) dt.
0
If a ≤ a2 = −31/336, then by Lemma 6 it follows that
31
Q (t) > 0,
−Q(4) (t) − aQ (t) ≥ −Q(4) (t) +
336
which proves the sufficiency.
Utilizing the decreasing property of fa2 and noting the facts that
3
835
835 3 32 819
f a2
=
γ+
ln −
≈ 0.0090636 and fa2 (∞) = 0,
2
7
7
2
280
we have
Corollary 3. Let Rn be defined by (1.3). Then the double inequality
2
2
7
7
− λ2
1 n + 21 − 40
1 n + 21 − 40
<
R
−
γ
<
(3.2)
n
4
24
24 n + 1 4 − 31
n + 1 − 31
2
336
2
336
holds for n ∈ N, where λ2 = fa2 (3/2) ≈ 0.0090636 is the best constant.
12
ZHEN-HANG YANG
Remark 2. The upper bound for Rn − γ given in (3.2) is better than the one given
in (1.5), because
2
7
1 n + 21 − 40
7 1
1
+
−
24 n + 1 4 − 31
24n2 960 n4
2
336
6720n5 + 10 752n4 + 5712n3 + 1564n2 + 588n − 35
= −
< 0.
960n4 (168n4 + 336n3 + 252n2 + 84n − 5)
The last result is the following theorem
Theorem 4. Let the function R be defined on (0, ∞) by (1.10). Then the function
7
7
31
2
4
2
R (x) − x −
(3.3)
x 7→ Ga (x) = 24 x + ax + a −
+a
40
336
40
is strictly completely monotonic on (0, ∞) if and only if a ≥ a3 = 11165/8284.
Proof. The necessity can be deduced by
Ga (x)
x→∞ x−4
=
lim
=
7
7
31
24(x4 +ax2 + 40
a− 336
+a)
)(Psi(x+1/2)−ln x)−(x2 − 40
−4
x
x→∞
2071
11 165
≥ 0.
33 600 a − 8284
lim
By the relations (2.2), (2.4) and (2.6) we obtain that
31
7
7
2
4
2
a−
+a
R (x) − x −
Ga (x) = 24x R (x) + 24ax R (x) + 24
40
336
40
Z ∞
Z ∞
7
1
1 2
−xt (4)
−xt ′′
e Q (t) dt + 24a
e Q (t) dt
x −
+
+
= 24
24
960
24
Z 0∞
0
7
31
7
+24
a−
+a
e−xt Q (t) dt − x2 −
40
336
40
0
Z ∞
7
31
(4)
′′
−xt
Q (t) + aQ (t) +
e
= 24
Q (t) dt
a−
40
336
0
Z ∞
△
= 24
e−xt ga (t) dt.
0
If a ≥ a3 = 11165/8284, then it follows from Lemmas 5 and 7 that
31
7
(4)
′′
ga (t) = Q (t) −
Q (t) + a Q (t) + Q (t)
336
40
7
31
′′
(4)
11 165
Q (t) + 8284 Q (t) + Q (t)
≥ Q (t) −
336
40
199 849
(4)
11 165 ′′
= Q (t) + 8284 Q (t) +
Q (t) > 0,
1391 712
which proves the sufficiency.
Application of the decreasing property of Ga3 and notice the facts that
112 672 809 11 465 761 3 11 465 761
3
=
−
ln −
γ ≈ 0.0016903 and Ga3 (∞) = 0,
Ga3
2
579 880
57 988
2
57 988
we have
SEVERAL COMPLETELY MONOTONE FUNCTIONS
13
Corollary 4. Let Rn be defined by (1.3). Then the double inequality
(3.4)
2
2
153
153
1
1
(n+ 12 ) + 97
(n+ 21 ) + 97
82 840
82 840 +λ3
< Rn − γ <
2
1 2
11 165
199 849
11 165
199 849
1 4
1 4
24 (n+ 2 ) + 8284 (n+ 2 ) + 1391 712
24 (n+ 2 ) + 8284 (n+ 12 ) + 1391
712
holds for n ∈ N, where λ3 = Ga3 (3/2) ≈ 0.0016903 is the best constant.
4. Remarks
Remark 3. The function Ga defined by (3.3) can be written as
F−31/336 (x)
(4.1) Ga (x) = a × f7/40 (x) − F−31/336 (x) = f7/40 (x) × a −
.
f7/40 (x)
Theorem 4 tell us that
7
31
F−31/336 (x)
F−31/336 (x)
11 165
−24(x4 − 336
)R(x)+x2 − 40
≤
(4.2)
≤ lim
= lim
.
7
2
24(x + 40 )R(x)−1
x→∞
x→∞
f7/40 (x)
f7/40 (x)
8284
On the other hand, we can prove that
F−31/336 (x)
F−31/336 (x)
(4.3)
≥ lim
= lim
+
f7/40 (x)
f7/40 (x)
x→0
x→0+
7
31
−24(x4 − 336
)R(x)+x2 − 40
7
24(x2 + 40
)R(x)−1
≥
155
.
294
It suffices to prove the function
x 7→ V (x) = ψ (x + 1/2) − ln x −
is increasing on (0, ∞). Differentiation gives
V ′ (x) = ψ ′ (x + 1/2) −
2071
1 x2 + 5880
2
2
24 x x + 155
294
1 1728 720x4 + 1217 748x2 + 321 005
.
+
2
x
10x3 (294x2 + 155)
Utilizing ψ ′ (x + 1) − ψ ′ (x) = −1/x2 yields
V ′ (x + 1) − V ′ (x)
=
1
− (x+1/2)
2 −
+ x1 −
=
1
x+1
+
1728 720(x+1)4 +1217 748(x+1)2 +321 005
2
10(x+1)3 (294(x+1)2 +155)
1728 720x4 +1217 748x2 +321 005
10x3 (294x2 +155)2
(x)
,
− 10x3 (2x+1)2 (294x2 +155)V21 (x+1)
3
(294x2 +588x+449)2
where
V1 (x)
=
1718 371 882 080x12 + 10 310 231 292 480x11 + 29 399 355 669 600x10
+52 486 324 833 600x9 + 66 690 983 696 400x8 + 65 258 530 001 280x7
+51 909 045 513 612x6 + 34 352 301 620 196x5 + 18 881 999 450 054x4
+8378 736 976 048x3 + 2808 871 359 013x2 + 622 502 847 155x + 64 714 929 005.
It is evident that V1 (x) > 0 for x > 0, and so V ′ (x + 1) − V ′ (x) < 0 for x > 0.
This leads us to
V ′ (x) > V ′ (x + 1) > V ′ (x + 2) > · · · > lim V ′ (x + n) = 0,
n→∞
which reveals that V is increasing on (0, ∞).
Meanwhile, (4.1) implies that an necessary condition such that the function −Ga
is completely monotone on (0, ∞) is
7
31
F−31/336 (x)
155
−24(x4 − 336
)R(x)+x2 − 40
= lim
.
=
7
2
24(x + 40 )R(x)−1
x→0+
x→0+
f7/40 (x)
294
a ≤ lim
14
ZHEN-HANG YANG
This together with inequalities (4.2) and (4.3) yield two conjectures.
Conjecture 1. Let the function R be defined on (0, ∞) by (1.10). Then
(i) the function
7
31
R (x) + x2 − 40
−24 x4 − 336
x 7→
7
R (x) − 1
24 x2 + 40
is increasing on (0, ∞);
(ii) the function −Ga is completely monotone on (0, ∞) if and only if a ≤
155/294.
Remark 4. In addition, using the increasing property of the function V proved in
Remark 3, and noting that
3
866 519
3
V
=
− ln − γ ≈ −0.000032387 and V (∞) = 0
2
881 820
2
we have
2071
1
1
(n+21 )2 + 5880
(n+21 )2 + 2071
5880
+ λ4 < Rn − γ <
,
2
1 2
155
155
1 2
1 2
24 (n+ 2 ) (n+ 2 ) + 294
24 (n+ 2 ) (n+ 12 ) + 294
where λ4 ≈ −0.000032387 is the best possible.
Clearly, it is an improvement of (3.2), since
2 2071
2
7
n + 12 + 5880
1
1 n + 21 − 40
−
4
24 n + 1 2 n + 1 2 + 155
24 n + 1 − 31
2
336
2
2
294
= −
64 201
2
120 (2n + 1)
(588n2
+ 588n + 457) (168n4 + 336n3 + 252n2 + 84n − 5)
< 0.
Remark 5. Theorems 3 and 4 and Remark 3 offer in fact three new sequences
convergent to γ with increasingly higher speed. Denote by
2
n
7
X
1 n + 12 − 40
1
− ln (n + 1/2) −
,
wn =
4
k
24 n + 1 − 31
k=1
yn
=
zn
=
2
336
n
2
X
153
1
1
(n+ 12 ) + 97
82 840
,
− ln (n + 1/2) −
4
165
199 849
(n+ 12 )2 + 1391
k
24 (n+ 12 ) + 118284
712
k=1
2 2071
n
X
n + 12 + 5880
1
1
− ln (n + 1/2) −
k
24 n + 1 2 n + 1 2 + 155
k=1
2
2
294
Then by Theorems 3 and 4 and Remark 4 we have
wn < zn < γ < yn .
And, a simple check yields
319
,
92 160
627 404 761
lim n10 (yn − γ) =
,
n→∞
246 900 842 496
199 849
lim n8 (zn − γ) = −
n→∞
94 832 640
These show that the sequences (wn ), (yn ) and (zn ) converge to γ as n−8 , n−10 and
n−8 , respectively.
lim n8 (wn − γ) = −
n→∞
SEVERAL COMPLETELY MONOTONE FUNCTIONS
15
Lastly, inspired by Theorem 2–4, we post the following open problem.
Problem 1. Determine the best constants ak and bk such that the function
!
!
n
n+1
X
X
2k
2k
R (x) −
x 7→
bk x
ak x
k=0
k=0
is completely monotone on (0, ∞), and satisfies that
P
Pn
n+1
2k
2k
R (x) −
a
x
k
k=0 bk x
k=0
= c 6= 0, ±∞.
lim
x→∞
x−2n−4
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
[13]
[14]
[15]
[16]
[17]
[18]
[19]
[20]
[21]
[22]
D. V. Widder, The Laplace Transform, Princeton Univ. Press, Princeton, 1941.
P. J. Rippon, Convergence with pictures, Amer. Math. Monthly 93 (1986), 476–478.
R. M. Young, Euler’s Constant, Math. Gaz. 75 (1991), 187–190.
S. R. Tims and J. A. Tyrrell, Approximate evaluation of Euler’s constant, Math. Gaz. 55
(1971), 65–67.
L. Tóth, Problem E3432, Amer. Math. Monthly 98 (1991), 264.
L. Tóth, Problem E3432 (Solution), Amer. Math. Monthly 99 (1992), 684–685.
D. W. DeTemple, A quicker convergence to Euler’s constant, Amer. Math. Monthly 100
(1993), 468–470.
T. Negoi, A faster convergence to the constant of Euler, Gazeta Matematicǎ, seriac A 15
(1997), 111–113 (in Romanian).
A. Vernescu, A new accelerate convergence to the constant of Euler, Gaz. Mat. Ser. A (17)
96 (1999), 273-278 (in Romanian).
F. Qi, R.-Q. Cui, Ch.-P. Chen, and B.-N. Guo, Some completely monotonic functions involving polygamma functions and an application, Journal of Mathematical Analysis and Applications 310 (2005), no. 1, 303–308; Available online at
http://dx.doi.org/10.1016/j.jmaa.2005.02.016.
A. Sintǎmǎrian, A generalization of Euler’s constant, Numer. Algorithms 46 (2007), 141–151.
Ch.-P. Chen and F. Qi, The best bounds of the n-th harmonic number, Global Journal of
Applied Mathematics and Mathematical Sciences 1 (2008), no. 1, 41–49.
A.
Sintǎmǎrian,
Some
inequalities
regarding
a
generalization
of
Euler’s
constant,
J. Inequal. Pure Appl. Math. 9 (2008),
no. 2,
Article 46.
http://www.emis.de/journals/JIPAM/images/352 07 JIPAM/352 07.pdf.
M. B. Villarino, Ramanujan’s harmonic number expansion into negative powers of triangular
number, J. Inequal. Pure Appl. Math. 9 (2008), no. 3, Article 89, 12 pp. Available online at
http://www.emis.de/journals/JIPAM/images/245 07 JIPAM/245 07.pdf.
C.-P. Chen, Inequalities and monotonicity properties for some special functions, J. Math.
Inequal. 3 (2009), no. 1, 79–91.
C.-P. Chen, The best bounds in Vernescu’s inequalities for the Euler’s Constant, RGMIA Res. Rep. Coll. 12 (2009), no.3, Article 11. Available online at
http://ajmaa.org/RGMIA/v12n3.php.
C.-P. Chen, Monotonicity properties of functions related to the psi function, Appl. Math.
Comput. 217 (2010), 2905–2911.
C.-P. Chen, Inequalities for the Euler-Mascheroni constant, Appl. Math. Lett. 23 (2010),
161–164.
C. Mortici, On new sequences converging towards the Euler-Mascheroni constant, Comput.
Math. Appl. 59 (2010), 2610–2614.
C. Mortici, Improved convergence towards generalized Euler-Mascheroni constant, Appl.
Math. Comput. 215 (2010), 3443–3448.
C. Mortici, Fast convergences towards Euler-Mascheroni constant, Comput. Appl. Math. 29
(2010) 479–491.
B.-N. Guo and F. Qi, Sharp bounds for harmonic numbers, Applied Mathematics and Computation 218 (2011), no. 3, 991–995;
Available online at
http://dx.doi.org/10.1016/j.amc.2011.01.089.
16
ZHEN-HANG YANG
[23] C.P. Chen, Sharpness of Negoi’s inequality for the Euler–Mascheroni constant, Bull. Math.
Anal. Appl. 3 (2011) 134–141.
[24] C. Mortici, A new Stirling series as continued fraction, Numer. Algorithms 56 (2011), 17–26.
[25] C.P. Chen and C. Mortici, New sequence converging towards the Euler–Mascheroni constant,
Comput. Math. Appl. 64 (2012), 391–398.
[26] I. Gavrea and M. Ivan, A solution to an open problem on the Euler–Mascheroni constant,
Applied Mathematics and Computation 224 (2013) 54–57.
[27] D. Lu, Some new convergent sequences and inequalities of Euler’s constant, J. Math. Anal.
Appl. 419 (2014), 541–552.
[28] C. Mortici, New bounds for a convergence by DeTemple, Journal of Science and Arts 2
(2010), no. 13, 239-242.
[29] N. Batir, Inequalities for the gamma function, Arch. Math. 91 (2008) 554–563.
[30] F. Qi, B.-N. Guo, Sharp inequalities for the psi function and harmonic numbers, Available
online at http://arxiv.org/abs/0902.2524.
[31] B.-N. Guo and F. Qi, Sharp inequalities for the psi function and harmonic numbers, Analysis
(Berlin) 34 (2014), 1–10. Available online at http://dx.doi.org/10.1515/anly-2014-0001.
[32] E. A. Karatsuba, On the computation of the Euler constant γ, Numer. Algorithms 24 (2000),
83–97.
[33] G. D. Anderson, M. K. Vamanamurthy and M. Vuorinen, Topics in special functions, Papers
on analysis: Report. Univ. Jyväskylä 83 (2001), 5–26.
[34] Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, M.
Abramowitz and I. A. Stegun (Eds.), National Bureau of Standards, Applied Mathematics
Series 55, 4th printing, with corrections, Washington, 1965.
[35] Zh.-H. Yang, Y.-M. Chu and X.-J. Tao, A double inequality for the tri-gamma function and
its applications, Abstr. Appl. Anal. 2014, Art. ID 702718.
[36] Zh.-H. Yang and Y.-M. Chu, A note on Jordan, Adamović-Mitrinović, and Cusa inequalities,
Abstr. Appl. Anal. 2014, Art. ID 364076.
Power Supply Service Center, ZPEPC Electric Power Research Institute, Hangzhou,
Zhejiang, China, 310009
E-mail address: [email protected]