2.1. Improvements Based on Computation of Mean Value
In this section, we consider the arithmetic mean value An, geometric mean value Gn, logarithmic mean value Ln, harmonic mean value Hn, and antilogarithmic mean Ωn of xn and yn, respectively; they are
(5)An=xn+yn2=(1+(1/n))n+(1+(1/n))n+12 =xn(1+12n),Gn=(xnyn)1/2=[(1+1n)n(1+1n)n+1]1/2 =xn(1+1n)1/2,Hn=2xnynxn+yn=xn(1+12n+1),Ln=yn-xnlnyn-lnxn =[(1+(1/n))n+1-(1+(1/n))nln(1+(1/n))n+1-ln(1+(1/n))n] =xnnln(1+(1/n)),Ωn=xnyn(lnyn-lnxn)yn-xn=xn(1+1n)lnxn =xn(n+1)ln(1+1n),
where Hn is a monotone increasing sequence, see [14] for a discussion of this issue, which means that the left-hand side of I. Schur inequality is valid. A lower bound for I. Schur inequality can be obtained via the monotone convergence theorem [27, pp. 87-88]. Moreover, it has been confirmed that An and Ωn are monotone decreasing in [28]: both An=xn(1+(1/2n)) and Ωn=xn(n+1)ln(1+(1/n)) are monotone decreasing sequences, and we obtained two inequalities as follows:
(6)e<xn(1+12n), e<xn(n+1)ln(1+1n), n=1,2,….
The first inequality has verified the rationality of the right-hand side of I. Schur inequality, and the second inequality has put forward a method for sharpening the upper bound of I. Schur inequality because of (1+(1/2n))>(n+1)ln(1+(1/n)), n∈N*. Besides, the monotonicity of the remaining mean value sequences Gn,Ln should also be checked, and the conclusions can be drawn as follows.
Proposition 2.
G
n
=
x
n
(
1
+
(
1
/
n
)
)
1
/
2
(
n
∈
N
*
)
is a monotone decreasing sequence.
Remark 3.
(
1
)
We can also continue to construct geometric mean value sequence Qn=((1+(1/n))2n(1+(1/n))1/2)1/2=xn(1+(1/n))1/4. By Lemma 1 and (1/4)=0.25<((2ln3-3ln2)/(2ln2-ln3))=0.409…, we can obtain that Qn=xn(1+(1/n))1/4 is a monotone decreasing sequence; thus xn(1+(1/n))1/4<e.
(
2
)
Consider m≥2, m∈N*, times of geometric mean value operation by combining with Lemma 1 and (1/2m)<((2ln3-3ln2)/(2ln2-ln3))=0.409…, and we have Qn=xn(1+(1/n))1/2m<e, which is a monotone increasing sequence for each m.
Theorem 4.
L
n
=
x
n
/
(
n
ln
(
1
+
(
1
/
n
)
)
)
is a monotone increasing sequence, and the following inequality
(7)Ln=xnnln(1+(1/n))>e
holds for n∈N*.
Proof.
We consider the monotonicity of f(x)=(x/lnx), (1<x<e), and
(8)f′(x)=lnx-1(lnx)2,
when x∈(1,e), f′(x)<0 and hence f(x) is monotone decreasing. It is known that xn=(1+(1/n))n is monotone increasing, and xn=(1+(1/n))n∈(1,e). According to monotonous property of the composite functions, Ln=xn/(nln(1+(1/n))) is a monotone decreasing sequence; thus
(9)f(x)=xlnx>f(e)=elne=e.
Take x=(1+(1/n))n∈(1,e); then
(10)Ln=xnnln(1+(1/n))>e.
Based on this conclusion, it is not hard to note that it is an improvement of upper bound for I. Schur inequality. There is another conclusion shown as follows (see [14]).
Proposition 5.
The double inequality
(11)(1+1n)n+α≤e≤(1+1n)n+β
for n≥1 is valid in the sense that the maximum α=(1/ln2)-1 and minimum β=1/2 in (11) are best possible.
Then we study the necessary and sufficient condition for the validness of the class of (1+(1/(2n+β)))xn≤e and obtain the following.
Proposition 6.
The inequality (1+(1/(2n+β)))xn≤e holds if and only if β≥(5/6), and β=5/6 is the best constant.
In fact, the proof of this theorem can be introduced via the conclusions in [14] as follows:
(12)6e12x+11<e-(1+1x)x<7e14x+12 (x≥1).
Use the left-hand side of (12) and change x into n, n∈N*; then
(13)(1+1n)n<e-6e12n+11=12n+512n+11e, n∈N*
and we can also obtain (1+(1/(2n+(5/6))))xn≤e. The conclusion on the best optimality of β can be found in [13]. It needs to be mentioned that the authors have made a mistake when they cited this inequality in [14]; for instance, let x=(1/10); then e-(1+(1/x))x=1.447300213…, but (7e/(14x+12))=1.419997970…. It means that the right-hand side of the inequality may not be valid. It is found that they mistake x≥1 for x>0 after checking the original paper [8]. In fact, we can prove e-(1+(1/x))x<(e/(2x+α)), x>0, and the constant α=e/(e-1) is the best possible.
Theorem 7.
The sequence
(14)Sn=pn+qn2=xn8n2+8n+18n2+4n
is monotone decreasing, and the inequality xn(1+(4n+1)/(8n2+4n))>e holds for n∈N*.
Proof.
We consider g(x)=ln(8x2+8x+1)-ln(8x2+4x)+xln(x+1)-xlnx, x>0; then
(15)g′(x)=16x+88x2+8x+1-16x+48x2+4x +ln(x+1)-lnx-1x+1,g′′(x)=168x2+8x+1-(16x+8)2(8x2+8x+1)2 -168x2+4x+(16x+4)2(8x2+4x)2-1x(x+1)2=(128x6+384x5+488x4+336x3 +125x2+21x+1) ×((8x2+8x+1)2(2x+1)2(x+1)2x2)-1>0,
where x>0. By limx→∞g′(x)=0, we obtain g′(x)<0 (x>0). Therefore, Sn is monotone decreasing, and we can straightforwardly prove the desired inequality of the theorem with limn→∞Sn=e.
Theorem 8.
The sequence
(16)Tn=2pnqnpn+qn=xn8n2+12n+48n2+8n+1
is monotone decreasing, and the following inequality
(17)xn8n2+12n+48n2+8n+1=xn(1+4n+38n2+8n+1)>e
holds for n∈N*.
Proof.
We consider h(x)=ln(8x2+12x+4)-ln(8x2+8x+1)+xln(x+1)-xlnx, x≥1, then h′(x)=((16x+12)/(8x2+12x+4))-((16x+8)/(8x2+8x+1))+ln(x+1)+(x/(x+1))-lnx-1, and
(18)h′′(x)=168x2+12x+4-(16x+12)2(8x2+12x+4)2-168x2+8x+1 +(16x+8)2(8x2+8x+1)2+2x+1-x(x+1)2-1x=128x4+256x3+152x2+24x-1x(x+1)(8x2+8x+1)2(2x+1)2>0,
where x≥1. According to limx→∞h′(x)=0, we obtain h′(x)<0 (x≥1). Therefore, Tn is monotone decreasing, and we can straightforwardly prove the desired inequality with limn→∞Tn=e.
In fact, the previous theorems can be viewed as the different improvements of original I. Schur inequality. The above improved inequalities follow the motivation of researching mean value sequences of the upper and lower bounds for original inequalities; the authors study the monotonicity of the sequences constructed by the mean value of upper and lower bounds for I. Schur inequality and its relationship with e. The arithmetic mean value should be replaced with the “weighted” mean value Wn=(xn+λyn)/(1+λ)=[1+λ/(λ+1)n]xn, λ∈ℝ; then we have the following.
Theorem 9.
For λ∈[(-1-3)/4,(-1+3)/4] and n∈N*, the sequence Wn is monotone decreasing, and the inequality xn[1+λ/(λ+1)n]<e holds.
Proof.
We consider d(x)=ln(x+λx+λ)-ln(x+λx)+xln(x+1)-xlnx (x>0); then
(19)d′(x)=1+λx+λx+λ-1+λx+λx+ln(x+1) +xx+1-lnx-1,(20)d′′(x)=-(1+λ)2(x+λx+λ)2+(1+λ)2(x+λx)2+2x+1 -x(x+1)2-1x=(2λx2+3λ2x2+3λ2x +x3λ2-x3+λ2+2λx) ×((x+1)2x2(x+λx+λ)2)-1.
Now rewrite numerator of (20) and denote Ix(λ)=(3x2+3x+x3+1)λ2+(2x+2x2)λ-x3. This is a quadratic function of λ, and its discriminant is
(21)Δ=(2x+2x2)2+4(3x2+3x+x3+1)x3>0.
We note that two roots of Ix(λ) are
(22)λ1=-(1+1+x2+x)x(x+1)2,λ2=(-1+1+x2+x)x(x+1)2.
It is not hard to prove that {-((1+1+x2+x)x/(x+1)2)}x=1∞ is strictly monotone decreasing, and
(23)-1<-(1+1+x2+x)x(x+1)2≤-1+34=-0.6830127…,
and {(-1+1+x2+x)x/(x+1)2}x=1∞ is strictly monotone increasing; we also have
(24)-1+34=0.1830127… ≤(-1+1+x2+x)x(x+1)2<1.
Hence, the inequality Ix(λ)≤0 always holds for x≥1 and λ∈[(-1-3)/4,(-1+3)/4]. By using limx→∞d′(x)=0, thus d′(x)≥0. It shows that sequence Wn=[1+λ/(λ+1)n]xn is monotone increasing under this condition, and the inequality in this theorem can be proved by using limn→∞Wn=e.
2.2. Further Discussions Based on Introducing the Parameters
Here we study the monotonicity of two new sequences and their relationships with e by introducing real parameter λ. Two sequences are, respectively, in=(1+(λ/n))n and jn=(1+(λ/n))n+1, where λ>0, and it is easy to obtain limn→∞in=limn→∞jn=eλ. Next, we consider their arithmetic mean value sequence, geometric mean value sequence, and harmonic mean value sequence, respectively; consider
(25)A1(n,λ)=in+jn2=(1+λ2n)(1+λn)n,G1(n,λ)=(injn)1/2=(1+λn)1/2(1+λn)n,H1(n,λ)=2injnin+jn=(1+λ2n+λ)(1+λn)n.
Theorem 10.
For 0<λ≤1 and n∈N*, the sequences A1(n,λ)=(1+(λ/2n))(1+(λ/n))n are monotone decreasing, and the inequality eλ<(1+(λ/2n))(1+(λ/n))n holds.
Proof.
We consider M(x)=xln(x+λ)-xlnx+ln(2x+λ)-ln(2x), x>0; then
(26)M′(x)=ln(x+λ)+xx+λ-lnx-1+22x+λ-1x,M′′(x)=21x+λ-x(x+λ)2-1x-4(2x+λ)2+1x2=-(λ(4x3λ+4x2λ2+xλ3-4x3-9x2λ -6xλ2-λ3))×(x2(x+λ)2(2x+λ)2)-1.
Denote the numerator of M′′(x) in (26) as
(27)N(x)≔-λ(4x3λ+4x2λ2+xλ3-4x3-9x2λ -6xλ2-λ3)≔λY(x),
where Y(x):=4x3-4x3λ+9x2λ-4x2λ2+6xλ2-xλ3+λ3. Take ∀p≥0; then λ:=1/(p+1)∈(0,1] and
(28)Y(x)∶=4x3-4x3p+1+9x2p+1-4x2(p+1)2+6x(p+1)2 -x(p+1)3+1(p+1)3=(4x3p3+8x3p2+4x3p+9x2p2 +14x2p+5x2+6xp+5x+1)×((p+1)3)-1.
It is found that if 0<λ≤1 and x>0, then Y(x)≥0. Furthermore, it implies that
(29)N(x)=λY(x)≥0;
then M′(x) is monotone increasing, and limx→∞M′(x)=0 has been verified, such that M′(x)<0, and A1(n,λ)=(1+(λ/2n))(1+(λ/n))n are monotone decreasing; hence eλ<(1+(λ/2n))(1+(λ/n))n can be proved because of limn→∞A1(n,λ)=eλ.
Theorem 11.
For 0<λ≤1 and n∈N*, the sequences G1(n,λ)=(1+(λ/n))1/2(1+(λ/n))n are monotone decreasing, and the inequality (1+(λ/n))1/2(1+(λ/n))n>eλ holds.
Proof.
We consider f1=xln(1+(λ/x))+(1/2)ln(1+(λ/x)), x≥1; then
(30)f1′(x)=ln(1+λx)-λx(1+(λ/x))-λ2x2(1+(λ/x)),(31)f1′′(x)=-λ2x3(1+(1/x))2+λx3(1+(λ/x)) -λ22x4(1+(λ/x))2=-λ(2λx-2x-λ)2x2(x+λ)2=(-2x+1)λ2+2λx2x2(x+λ)2.
Denote the numerator of (31) as δ(λ)=(-2x+1)λ2+2xλ, which is a quadratic function of λ. And its discriminant is Δ1=4x2>0. Then two roots of δ(λ) are
(32)λ1=0, λ2=2x2x-1.
We note that {2x/(2x-1)}x=1∞ is strict monotone decreasing and 1<(2x/(2x-1))≤2, x≥1. According to the characteristic of parabola curve, δ(λ)=(-2x+1)λ2+2xλ>0 always holds for 0<λ≤1 and x≥1; then from limx→∞f1′(x)=0, f1′(x)≤0. It shows that the sequences G1(n,λ) are monotone decreasing.
Theorem 12.
If λ≥1, then H1(n,λ)=(2injn)/(in+jn)=(1+λ/(2n+λ))(1+(λ/n))n are monotone increasing, and the following inequality
(33)(1+λ2n+λ)(1+λn)n<eλ
holds for n∈N*.
Proof.
We consider φ(x)=ln(2x+2λ)-ln(2x+λ)+xln(x+λ)-xlnx, x>0; then
(34)φ′(x)=22x+2λ-22x+λ+ln(x+λ)+xx+λ-lnx-1,(35)φ′′(x)=-4(2x+2λ)2+4(2x+λ)2+2x+λ-x(x+λ)2 -1x=-(λ(-4x2-3λx+4λx2+4xλ2+λ3)) ×((x+λ)2(2x+λ)2x)-1.
Denote numerator of (35) as
(36)γ(x)=-λ(-4x2-3λx+4λx2+4xλ2+λ3)≔-λZ(x),
where Z(x)∶=-4x2-3λx+4λx2+4xλ2+λ3. Since τ+1∶=λ≥1, that is, τ≥0, we have
(37)Z(x)∶=-4x2-3x(τ+1)+4x2(τ+1) +4x(τ+1)2+(τ+1)3=4x2τ+(τ+1)(4τ+1)x+(τ+1)3≥0. (x>0, τ≥0)
It implies that the inequality γ(x)=-λZ(x)≤0 always holds for x>0,λ≥1. Then φ′(x) is monotone decreasing, and according to limx→∞φ′(x)=0, we have φ′(x)>0. Hence, H1(n,λ)=(1+λ/(2n+λ))(1+(λ/n))n are monotone increasing sequences, and we prove that
(38)eλ>(1+λ2n+λ)(1+λn)n.
A remaining issue of Theorem 12 can be described as follows: when λ<1, are the H1(n,λ) monotone increasing or monotone decreasing? With the help of the Bottema (available at the site of http://old.irgoc.org/Soft/ShowSoft.asp?SoftID=15, and the practical implementing Maple codes also are available by the authors’ email) (developed by Yang and Xia and based on the Maple programme; see [29] and references therein), we can find that the inequality γ(x)≥0 holds for 0<λ≤(4/5), x≥1. Moreover, it can be also shown in Algorithm 1.
Then we can conclude that H1(n,λ), n=1,2,…, is monotone decreasing under the previous discussion in the proof of the Theorem 12. The further conclusions need to be studied and proven by analytical methods in future work.
Algorithm 1
Input:
(
i)
read “D:/Program Files/bottema2009”;
(ii)
ineq:= -lambda*(-4*x
∧
2-3*x*lambda+4*x
∧
2*lambda+4*x*lambda
∧
2+lambda
∧
3);
(iii) xprove(ineq >= 0, [lambda <= 4/5, x >= 1]);
Output:
Found border curves…
(-4x2-3xλ+4x2λ+4xλ2+λ3)(5λ-4)(x-1)xλ
Start to project curves…, 3.307
[x,λ]
do 1-th partition…
Start to find the sample points., 3.307
in 1-dimensional space…
finished in 1-dimensional space.
in 2-dimensional space…
finished in 2-dimensional space.
number(s) of sample points:, 1, 3.322
[x,λ]
[[2,12]]
−1
The inequality holds.
true