# Refined Estimates and Generalizations of Inequalities Related to the Arctangent Function and Shafer’s Inequality

Authors: Branko Malešević; Marija Rašajski; Tatjana Lutovac
Journal: Mathematical Problems in Engineering
 (2018)

---

## Abstract

We give some sharper refinements and generalizations of inequalities related to Shafer’s inequality for the arctangent function, stated inTheorems  1, 2,  and  4 in Mortici and Srivastava, 2014, by C. Mortici and H.M. Srivastava.


---

## Body

## 1. Introduction

Inverse trigonometric functions play an important role and have many applications in engineering [1–4]. In particular, the arctangent function and various related inequalities have been studied and effectively applied to problems in fundamental sciences and many areas of engineering, such as electronics, mechanics, and aeronautics [3, 5, 6]; see also [7].Various approximations of the arctangent function can be found in [4–6, 8–21]; see also [22, 23]. One of the inequalities that attracted attention of many authors is Shafer’s inequality [9]:(1)3x1+21+x2<arctan⁡x,which holds for x>0; see also [10–12].Recently, in [8], Mortici and Srivastava proved the following results, cited here as Statements 1, 2, and 3, related to the above inequality. These results are the starting point of our research.Statement 1 (Theorem 1, [8]).
For everyx>0, the following two-sided inequality holds:(2)3x1+21+x2+ax<arctan⁡x<3x1+21+x2+bx,where a(x)=1/180x5-13/1512x7 and b(x)=1/180x5.Statement 2 (Theorem 2, [8]).
For everyx>0, it is asserted that(3)3x+cx1+21+x2<arctanx<3x+dx1+21+x2,where c(x)=1/60x5-17/840x7 and d(x)=1/60x5.Statement 3 (Theorem 4, [8]).
For everyx>0, it is asserted that(4)-112x3<arctanx-2x1+1+x2<-112x3+340x5.The main results of this paper are refined estimates and generalizations of the inequalities given in Statements1, 2, and 3. Although inequalities (2), (3), and (4) hold for x>0, considering them in a neighborhood of zero is of primary importance, as noted in [8].


## 2. Main Results

First, let us recall some well-known power series expansions that will be used in our proofs.Forx≤1,(5)arctan⁡x=∑m=0∞-1mAmx2m+1,where(6)Am=12m+1.Forx≤1,(7)1+x2=1+∑m=0∞-1mKmx2m+2,where(8)Km=2m!m!m+1!22m+1.The following power series expansion holds:(9)3x1+21+x2=∑m=0∞Bmx2m+1, where |x|≤3/2, with B(0)=1 and B(1)=-1/3, and for m≥2,(10)Bm=-1m4m-13m1-8∑i=2m2i-2!i-1!i!22i-134i.Power series coefficients are calculated by applying Cauchy’s product to the power series expansions arising from the following transformation of the corresponding function:(11)3x1+21+x2=-x1-21+x21+4/3x2.It is easy to prove that sequenceBmm∈N0 for m≥1 satisfies the recurrence equation:(12)Bm+1+43Bm=-1m2m-1!22m-1m-1!m+1!=2-1m2m-1!!2m+2!!.


### 2.1. Refinements of the Inequalities in Statement1

Before we proceed to Theorem9, which represents an improvement and generalization of Statement 1, we need the following lemmas.Lemma 4.
Letβ(0)=1, β(1)=-1/3, and β(m)=3-1m/22m+1∑k=0∞2k+2m-1!/k+m-1!k+m+1!3/16k, for m≥2. The sequence β(m)m∈N0 for m≥1 satisfies the recurrence relation (12).Proof.
In the proof of this lemma we use the Wilf- Zeilberger method [24–26]. (The same approach we used in [27].)
The assertion is obviously true form=1.
Letm≥2 and (13)gk,m=2k+2m-1!k+m-1!k+m+1!andϕm=3-1m22m+1.Then we have (14)βm=ϕm∑k=0∞gk,m316k.Further we have(15)βm+1+43βm=∑k=0∞ϕm+1gk,m+1+43ϕmgk,m316k=-1m22m-1·∑k=0∞2k+2m+132k+2m-1!8k+8m+16k+m-1!k+m+1!316k.Consider now the sequence S(m)m∈N,m≥2, where(16)Sm=∑k=0m-1Fm,kandFm,k=2m+2k+132m+2k-1!8m+8k+16m+k-1!m+k+1!316k.Consider the function (an algorithm for determining function G(m,k) for a given function F(m,k) is described in [24]. Note that the pair of discrete functions Fm,k,Gm,k is the so-called Wilf- Zeilberger pair)(17)Gm,k=-8k-8m-162k+2m-1!8k+8m+16k+m-1!k+m+1!316k,where m∈N0 and k∈N0. It is not hard to verify that functions F(m,k) and G(m,k) satisfy the following relation:(18)Fm,k=Gm,k+1-Gm,k.If we sum both sides of (18) over all k∈N0, we get the following relation: (19)Sm=-Gm,0. Finally, as (20)Gm,0=-2m-1!m-1!m+1! we have(21)Sm=2m-1!m-1!m+1!.Therefore from (15) and (21) we conclude that (22)βm+1+43βm=-1m22m-1·Sm=-1m22m-1·2m-1!m-1!m+1!.Corollary 5.
Given that the sequencesB(m)m∈N0 and β(m)m∈N0 satisfy the same recurrence relation and as they agree for m=0 and m=1, we conclude that(23)Bm=βm,form∈N0.Let us introduce the notation:(24)Cm=Am-Bm, where C(0)=C(1)=0, and for m≥2 the following holds:(25)Cm=12m+1-4m-13m1-8∑i=2m2i-2!i-1!i!22i-134i.Thus, we have the power series expansion:(26)fx=arctanx-3x2+1+x2=∑m=0∞-1mCmx2m+1,for m∈N0 and |x|≤3/2.Let us introduce the notation:(27)β+m=322m+1∑k=0∞2k+2m-1!k+m-1!k+m+1!22k34k,m∈N0β1m=∑k=0+∞∏n=0k-1m+1/2+n∏n=0k-1m+2+n34k,β1km=∏n=0k-1m+1/2+n∏n=0k-1m+2+n,m∈N0.Lemma 6.
Form∈N0 the following holds:(28)β+m=322m-1!!2m+2!!β1m.Proof.
(29) 3 2 2 m + 1 2 m + 2 k - 1 ! m + k - 1 ! m + k + 1 ! 2 2 k = 3 2 1 2 m + 2 ! ! 2 m + 2 k - 1 ! ! m + k + 1 ! m + 1 ! 2 k = 3 2 2 m - 1 ! ! 2 m + 2 ! ! 2 m + 1 2 m + 3 … 2 m + 2 k - 1 m + 1 ! m + 2 m + 3 … m + k + 1 m + 1 ! 2 k = 3 2 2 m - 1 ! ! 2 m + 2 ! ! ∏ n = 0 k - 1 m + 1 / 2 + n ∏ n = 0 k - 1 m + 2 + n .Lemma 7.
Form∈N0 the following holds:(30)8m+22m+13=∑k=0∞m+1/2m+2k34k<β1m<∑k=0∞34k=4.Proof.
The statement immediately follows from the inequalities:(31)m+1/2m+2k<β1km<1.Lemma 8.
Form∈N0 the following holds:(32)β1m+1>β1mm+2m+1/21-321m+2.Proof.
(33) β 1 k m + 1 = β 1 k m m + 2 m + 1 / 2 + k 1 / 2 + m m + 2 + k = β 1 k m m + 2 m + 1 / 2 1 - 3 2 1 m + 2 + k > β 1 k m m + 2 m + 1 / 2 1 - 3 2 1 m + 2 .Theorem 9.
For the real analytic function(34)fx=arctanx-3x2+1+x2the following inequalities hold for k∈N and x∈(0,3/2]:(35)∑m=02k+1-1mCmx2m+1<fx<∑m=02k-1mCmx2m+1.where C(0)=C(1)=0, and for m≥2 the following holds: (36)Cm=12m+1-4m-13m1-8∑i=2m2i-2!i-1!i!22i-134i.Proof.
We will prove that the sequence{C(m)}m∈N0 is positive and monotonically decreasing and tends to zero as m tends to infinity. We will use Lemmas 7 and 8. (37)Cm=12m+1-322m-1!!2m+2!!β1m>12m+1-4322m-1!!2m+2!!=12m+11-62m+1!!2m+2!!.It is easy to verify that 6(2m+1)!!/(2m+2)!!<1 for m≥11; therefore C(m)>0. Let us note that 0<C(m)<1/2m+1, so we can conclude that limm→+∞⁡C(m)=0.
Let us now prove that{C(m)}m∈N0 is a monotonically decreasing sequence. (38)Cm+1-Cm=-22m+12m+3-322m+1!!2m+4!!β1m+1+322m-1!!2m+2!!β1m=-22m+12m+3+322m-1!!2m+2!!β1m-2m+12m+4β1m+1<-22m+12m+3+322m-1!!2m+2!!β1m-2m+12m+4·m+2m+1/21-321m+2β1m=-22m+12m+3+β1m942m-1!!m+22m+2!!<-22m+12m+3+4942m-1!!m+22m+2!!=-22m+12m+31-92m+3!!2m+4!!.It is easy to prove that C(m+1)-C(m)<0 for m≥8, i.e., the sequence is monotonically decreasing. Since Cmm∈N0 is positive for m≥2 and monotonically decreasing (for m≥8) and tends to zero, the same holds true for the sequence C(m)x2m+1m∈N0 for a fixed x∈(0,3/2] (noting that it is decreasing for m≥3 ), so we can apply Leibniz’s theorem for alternating series [28], thus proving the claim of Theorem 9: (39)∑m=02k+1-1mCmx2m+1<fx<∑m=02k-1mCmx2m+1,k∈N.Examples. For k=1 and x∈(0,3/2] we get Statement 1.Fork=2 and x∈(0,3/2],(40)x5180-13x71512+53x95184-3791x11342144<arctanx-3x1+21+x2<x5180-13x71512+53x95184.For k=3 and x∈(0,3/2],(41)x5180-13x71512+53x95184-3791x11342144+55801x134852224-130591x1511197440<arctanx-3x1+21+x2<x5180-13x71512+53x95184-3791x11342144+55801x134852224,etc.


### 2.2. Refinements of the Inequalities in Statement2

We propose the following improvement and generalization of Statement2.Theorem 10.
For everyx∈(0,1] and k∈N, it is asserted that(42)3x+∑m=22k+1-1mEmx2m+11+21+x2<arctanx<3x+∑m=22k-1mEmx2m+11+21+x2,where (43)Em=32m+1-∑i=0m-12m-2i-2!22m-2i-22i+1m-i-1!m-i!.Examples. For x∈(0,1] and k=1 we get inequality (3) from Statement 2.Forx∈(0,1] and k≥2 inequality (42) refines inequality (3) from Statement 2 and we have the following new results:(i)
Takingk=2 in (42) gives (44)3x+1/60x5-17/840x7+139/6720x9-8947/443520x111+21+x2<arctanx<3x+1/60x5-17/840x7+139/6720x91+21+x2.(ii)
Takingk=3 in (42) gives (45)3x+1/60x5-17/840x7+139/6720x9-8947/443520x11+89279/4612608x13-851677/46126080x151+21+x2<arctanx<3x+1/60x5-17/840x7+139/6720x9-8947/443520x11+89279/4612608x131+21+x2,etc.Proof of Theorem10.
Based on Cauchy’s product of power series (7) and (5), the real analytical function,(46)fx=1+21+x2·arctanx-3x,for x∈(0,1] has the following power series:(47)fx=∑m=2∞-1mEmx2m+1,where(48)Em=32m+1-∑i=0m-12m-2i-2!2i+1m-i-1!m-i!22m-2i-2.
We aim to show that sequenceE(m)m∈N,m≥2 decreases monotonically and that limm→+∞⁡E(m)=0. It is easy to verify that sequence E(m)m∈N,m≥2 satisfies the following recurrence relation:(49)-2mEm+2m+3Em+1=12m+1-m+12m!m+1!24m.Consider the sequence e(m)m∈N,m≥2 where (50)em=gm·Sm and (51)Sm=∑j=1m-12j+2!2j!!2j+1-2j+1!22j+32j+1j+1!22j!!2 and (52)gm=m!m-1!22m+12m+1!.
It is easy to verify that sequencee(m)m∈N,m≥2 satisfies the recurrence relation (49). Given that sequences E(m)m∈N,m≥2 and e(m)m∈N,m≥2 agree for m=2 and m=3, we conclude that(53)Em=em,form∈N,m≥2.
We prove that sequencee(m)m∈N,m≥2 is a monotonically decreasing sequence and limm→+∞⁡e(m)=0.
By the principle of mathematical induction, it follows that(54)2j+1!<2j!!2j+1 is true for all j∈N. Therefore S(m)>0 for m≥2, i.e.,(55)em>0,form≥2.To prove that emm∈N,m≥2 is a monotonically decreasing sequence, let us use the following notation: (56)Sm=∑j=1m-1hj where (57)hj=2j+2!2j!!2j+1-2j+1!22j+32j+1j+1!22j!!2.
Consider the following equivalences form≥2:(58)em+1em<1⇔gm+1Sm+1gmSm<1⇔gm+1gm·Sm+hmSm<1⇔2m2m+3·Sm+hmSm<1⇔2mhm<3Sm.
Consider the last inequality. It is easy to verify that it is true form=2. Observing that(59)3Sm+1=3Sm+hm=3Sm+3hmand using the induction hypothesis 3Sm>2mhm for some positive integer m≥2, we conclude that (60)3Sm+1>2m+3hm>2m+1hm. Therefore, by the principle of mathematical induction, the inequality(61)2mhm<3Sm is true for m≥2, i.e.,(62)em+1em<1,form≥2.
Let us further consider the positive addend ofS(m), i.e.,(63)S+m=∑j=1m-12j+2!2j‼2j+122j+32j+1j+1!22j‼2=∑j=1m-12j!42j!!2.By the principle of mathematical induction, it follows that(64)S+m=m2m!22m!!2-14.
Finally, given that form≥2(65)0<em≤gm·S+m=12m+1 we have limm→+∞⁡e(m)=0.
Finally, based on (53) we conclude that Emm∈N,m≥2 is a positive monotonically decreasing sequence and that it tends to zero. The same holds true for the sequence E(m)x2m+1m∈N,m≥2 for a fixed x∈(0,1] so we can apply Leibniz’s theorem for alternating series [28], thus proving the claim of Theorem 10.


### 2.3. Refinements of the Inequalities in Statement3

We propose the following improvement and generalization of Statement3.Theorem 11.
For everyx∈(0,1] and k∈N, it is asserted that(66)∑m=12k-1-1mCmx2m+1<arctanx-2x1+1+x2<∑m=12k-1mCmx2m+1,where (67)Cm=12m+1-2m-1!!m+1!2m.Examples. For x∈(0,1] and k=1 we get inequality (4) from Statement 3.Forx∈(0,1] and k≥2 inequality (66) from Theorem 11 refines inequality (4) from Statement 3 and we have the following new results:(i)
Takingk=2 in (66) gives (68)-112x3+340x5-29448x7<arctanx-2x1+1-x2<-112x3+340x5-29448x7+651152x9.(ii)
Takingk=3 in (66) gives (69)-112x3+340x5-29448x7+651152x9-2815632x11<arctanx-2x1+1-x2<-112x3+340x5-29448x7+651152x9-2815632x11+59513312x13,etc.Proof of Theorem11.
Forx∈(0,1] the following power series expansion holds: (70)arctanx-2x1+1+x2=arctanx+21-x2+1x=∑m=1∞-1mCmx2m+1 where(71)Cm=12m+1-2m-1!!m+1!2m.
We prove that the sequenceC(m)m∈N is positive and monotonically decreasing and tends to zero as m tends to infinity.
It is easy to verify that(2m+2)!!>2(2m+1)!! is true for m∈N. Thus, the following equivalences hold true for every m∈N: (72)Cm>0⇔12m+1>2m-1!!m+1!2m⇔2m+2!!>22m+1!!,and we conclude that C(m)>0 for every m∈N.
Let us now prove thatC(m)m∈N is a monotonically decreasing sequence. We have (73)Cm-Cm+1>0⇔22m+12m+3-32m-1!!2m+2!!m+2>0⇔2m+4!!>32m+3!!.
As it is easy to show (by the principle of mathematical induction) that the last inequality holds true form∈N, we may conclude that C(m)m∈N is a monotonically decreasing sequence.
Finally, as0<C(m)<1/2m+1, we conclude that limm→+∞⁡C(m)=0.
SinceCmm∈N is a positive monotonically decreasing sequence, and it tends to zero, the same holds true for the sequence C(m)x2m+1m∈N for a fixed x∈(0,1]. So we can apply Leibniz’s theorem for alternating series [28] and thus prove the claim of Theorem 11.


## 2.1. Refinements of the Inequalities in Statement1

Before we proceed to Theorem9, which represents an improvement and generalization of Statement 1, we need the following lemmas.Lemma 4.
Letβ(0)=1, β(1)=-1/3, and β(m)=3-1m/22m+1∑k=0∞2k+2m-1!/k+m-1!k+m+1!3/16k, for m≥2. The sequence β(m)m∈N0 for m≥1 satisfies the recurrence relation (12).Proof.
In the proof of this lemma we use the Wilf- Zeilberger method [24–26]. (The same approach we used in [27].)
The assertion is obviously true form=1.
Letm≥2 and (13)gk,m=2k+2m-1!k+m-1!k+m+1!andϕm=3-1m22m+1.Then we have (14)βm=ϕm∑k=0∞gk,m316k.Further we have(15)βm+1+43βm=∑k=0∞ϕm+1gk,m+1+43ϕmgk,m316k=-1m22m-1·∑k=0∞2k+2m+132k+2m-1!8k+8m+16k+m-1!k+m+1!316k.Consider now the sequence S(m)m∈N,m≥2, where(16)Sm=∑k=0m-1Fm,kandFm,k=2m+2k+132m+2k-1!8m+8k+16m+k-1!m+k+1!316k.Consider the function (an algorithm for determining function G(m,k) for a given function F(m,k) is described in [24]. Note that the pair of discrete functions Fm,k,Gm,k is the so-called Wilf- Zeilberger pair)(17)Gm,k=-8k-8m-162k+2m-1!8k+8m+16k+m-1!k+m+1!316k,where m∈N0 and k∈N0. It is not hard to verify that functions F(m,k) and G(m,k) satisfy the following relation:(18)Fm,k=Gm,k+1-Gm,k.If we sum both sides of (18) over all k∈N0, we get the following relation: (19)Sm=-Gm,0. Finally, as (20)Gm,0=-2m-1!m-1!m+1! we have(21)Sm=2m-1!m-1!m+1!.Therefore from (15) and (21) we conclude that (22)βm+1+43βm=-1m22m-1·Sm=-1m22m-1·2m-1!m-1!m+1!.Corollary 5.
Given that the sequencesB(m)m∈N0 and β(m)m∈N0 satisfy the same recurrence relation and as they agree for m=0 and m=1, we conclude that(23)Bm=βm,form∈N0.Let us introduce the notation:(24)Cm=Am-Bm, where C(0)=C(1)=0, and for m≥2 the following holds:(25)Cm=12m+1-4m-13m1-8∑i=2m2i-2!i-1!i!22i-134i.Thus, we have the power series expansion:(26)fx=arctanx-3x2+1+x2=∑m=0∞-1mCmx2m+1,for m∈N0 and |x|≤3/2.Let us introduce the notation:(27)β+m=322m+1∑k=0∞2k+2m-1!k+m-1!k+m+1!22k34k,m∈N0β1m=∑k=0+∞∏n=0k-1m+1/2+n∏n=0k-1m+2+n34k,β1km=∏n=0k-1m+1/2+n∏n=0k-1m+2+n,m∈N0.Lemma 6.
Form∈N0 the following holds:(28)β+m=322m-1!!2m+2!!β1m.Proof.
(29) 3 2 2 m + 1 2 m + 2 k - 1 ! m + k - 1 ! m + k + 1 ! 2 2 k = 3 2 1 2 m + 2 ! ! 2 m + 2 k - 1 ! ! m + k + 1 ! m + 1 ! 2 k = 3 2 2 m - 1 ! ! 2 m + 2 ! ! 2 m + 1 2 m + 3 … 2 m + 2 k - 1 m + 1 ! m + 2 m + 3 … m + k + 1 m + 1 ! 2 k = 3 2 2 m - 1 ! ! 2 m + 2 ! ! ∏ n = 0 k - 1 m + 1 / 2 + n ∏ n = 0 k - 1 m + 2 + n .Lemma 7.
Form∈N0 the following holds:(30)8m+22m+13=∑k=0∞m+1/2m+2k34k<β1m<∑k=0∞34k=4.Proof.
The statement immediately follows from the inequalities:(31)m+1/2m+2k<β1km<1.Lemma 8.
Form∈N0 the following holds:(32)β1m+1>β1mm+2m+1/21-321m+2.Proof.
(33) β 1 k m + 1 = β 1 k m m + 2 m + 1 / 2 + k 1 / 2 + m m + 2 + k = β 1 k m m + 2 m + 1 / 2 1 - 3 2 1 m + 2 + k > β 1 k m m + 2 m + 1 / 2 1 - 3 2 1 m + 2 .Theorem 9.
For the real analytic function(34)fx=arctanx-3x2+1+x2the following inequalities hold for k∈N and x∈(0,3/2]:(35)∑m=02k+1-1mCmx2m+1<fx<∑m=02k-1mCmx2m+1.where C(0)=C(1)=0, and for m≥2 the following holds: (36)Cm=12m+1-4m-13m1-8∑i=2m2i-2!i-1!i!22i-134i.Proof.
We will prove that the sequence{C(m)}m∈N0 is positive and monotonically decreasing and tends to zero as m tends to infinity. We will use Lemmas 7 and 8. (37)Cm=12m+1-322m-1!!2m+2!!β1m>12m+1-4322m-1!!2m+2!!=12m+11-62m+1!!2m+2!!.It is easy to verify that 6(2m+1)!!/(2m+2)!!<1 for m≥11; therefore C(m)>0. Let us note that 0<C(m)<1/2m+1, so we can conclude that limm→+∞⁡C(m)=0.
Let us now prove that{C(m)}m∈N0 is a monotonically decreasing sequence. (38)Cm+1-Cm=-22m+12m+3-322m+1!!2m+4!!β1m+1+322m-1!!2m+2!!β1m=-22m+12m+3+322m-1!!2m+2!!β1m-2m+12m+4β1m+1<-22m+12m+3+322m-1!!2m+2!!β1m-2m+12m+4·m+2m+1/21-321m+2β1m=-22m+12m+3+β1m942m-1!!m+22m+2!!<-22m+12m+3+4942m-1!!m+22m+2!!=-22m+12m+31-92m+3!!2m+4!!.It is easy to prove that C(m+1)-C(m)<0 for m≥8, i.e., the sequence is monotonically decreasing. Since Cmm∈N0 is positive for m≥2 and monotonically decreasing (for m≥8) and tends to zero, the same holds true for the sequence C(m)x2m+1m∈N0 for a fixed x∈(0,3/2] (noting that it is decreasing for m≥3 ), so we can apply Leibniz’s theorem for alternating series [28], thus proving the claim of Theorem 9: (39)∑m=02k+1-1mCmx2m+1<fx<∑m=02k-1mCmx2m+1,k∈N.Examples. For k=1 and x∈(0,3/2] we get Statement 1.Fork=2 and x∈(0,3/2],(40)x5180-13x71512+53x95184-3791x11342144<arctanx-3x1+21+x2<x5180-13x71512+53x95184.For k=3 and x∈(0,3/2],(41)x5180-13x71512+53x95184-3791x11342144+55801x134852224-130591x1511197440<arctanx-3x1+21+x2<x5180-13x71512+53x95184-3791x11342144+55801x134852224,etc.


## 2.2. Refinements of the Inequalities in Statement2

We propose the following improvement and generalization of Statement2.Theorem 10.
For everyx∈(0,1] and k∈N, it is asserted that(42)3x+∑m=22k+1-1mEmx2m+11+21+x2<arctanx<3x+∑m=22k-1mEmx2m+11+21+x2,where (43)Em=32m+1-∑i=0m-12m-2i-2!22m-2i-22i+1m-i-1!m-i!.Examples. For x∈(0,1] and k=1 we get inequality (3) from Statement 2.Forx∈(0,1] and k≥2 inequality (42) refines inequality (3) from Statement 2 and we have the following new results:(i)
Takingk=2 in (42) gives (44)3x+1/60x5-17/840x7+139/6720x9-8947/443520x111+21+x2<arctanx<3x+1/60x5-17/840x7+139/6720x91+21+x2.(ii)
Takingk=3 in (42) gives (45)3x+1/60x5-17/840x7+139/6720x9-8947/443520x11+89279/4612608x13-851677/46126080x151+21+x2<arctanx<3x+1/60x5-17/840x7+139/6720x9-8947/443520x11+89279/4612608x131+21+x2,etc.Proof of Theorem10.
Based on Cauchy’s product of power series (7) and (5), the real analytical function,(46)fx=1+21+x2·arctanx-3x,for x∈(0,1] has the following power series:(47)fx=∑m=2∞-1mEmx2m+1,where(48)Em=32m+1-∑i=0m-12m-2i-2!2i+1m-i-1!m-i!22m-2i-2.
We aim to show that sequenceE(m)m∈N,m≥2 decreases monotonically and that limm→+∞⁡E(m)=0. It is easy to verify that sequence E(m)m∈N,m≥2 satisfies the following recurrence relation:(49)-2mEm+2m+3Em+1=12m+1-m+12m!m+1!24m.Consider the sequence e(m)m∈N,m≥2 where (50)em=gm·Sm and (51)Sm=∑j=1m-12j+2!2j!!2j+1-2j+1!22j+32j+1j+1!22j!!2 and (52)gm=m!m-1!22m+12m+1!.
It is easy to verify that sequencee(m)m∈N,m≥2 satisfies the recurrence relation (49). Given that sequences E(m)m∈N,m≥2 and e(m)m∈N,m≥2 agree for m=2 and m=3, we conclude that(53)Em=em,form∈N,m≥2.
We prove that sequencee(m)m∈N,m≥2 is a monotonically decreasing sequence and limm→+∞⁡e(m)=0.
By the principle of mathematical induction, it follows that(54)2j+1!<2j!!2j+1 is true for all j∈N. Therefore S(m)>0 for m≥2, i.e.,(55)em>0,form≥2.To prove that emm∈N,m≥2 is a monotonically decreasing sequence, let us use the following notation: (56)Sm=∑j=1m-1hj where (57)hj=2j+2!2j!!2j+1-2j+1!22j+32j+1j+1!22j!!2.
Consider the following equivalences form≥2:(58)em+1em<1⇔gm+1Sm+1gmSm<1⇔gm+1gm·Sm+hmSm<1⇔2m2m+3·Sm+hmSm<1⇔2mhm<3Sm.
Consider the last inequality. It is easy to verify that it is true form=2. Observing that(59)3Sm+1=3Sm+hm=3Sm+3hmand using the induction hypothesis 3Sm>2mhm for some positive integer m≥2, we conclude that (60)3Sm+1>2m+3hm>2m+1hm. Therefore, by the principle of mathematical induction, the inequality(61)2mhm<3Sm is true for m≥2, i.e.,(62)em+1em<1,form≥2.
Let us further consider the positive addend ofS(m), i.e.,(63)S+m=∑j=1m-12j+2!2j‼2j+122j+32j+1j+1!22j‼2=∑j=1m-12j!42j!!2.By the principle of mathematical induction, it follows that(64)S+m=m2m!22m!!2-14.
Finally, given that form≥2(65)0<em≤gm·S+m=12m+1 we have limm→+∞⁡e(m)=0.
Finally, based on (53) we conclude that Emm∈N,m≥2 is a positive monotonically decreasing sequence and that it tends to zero. The same holds true for the sequence E(m)x2m+1m∈N,m≥2 for a fixed x∈(0,1] so we can apply Leibniz’s theorem for alternating series [28], thus proving the claim of Theorem 10.


## 2.3. Refinements of the Inequalities in Statement3

We propose the following improvement and generalization of Statement3.Theorem 11.
For everyx∈(0,1] and k∈N, it is asserted that(66)∑m=12k-1-1mCmx2m+1<arctanx-2x1+1+x2<∑m=12k-1mCmx2m+1,where (67)Cm=12m+1-2m-1!!m+1!2m.Examples. For x∈(0,1] and k=1 we get inequality (4) from Statement 3.Forx∈(0,1] and k≥2 inequality (66) from Theorem 11 refines inequality (4) from Statement 3 and we have the following new results:(i)
Takingk=2 in (66) gives (68)-112x3+340x5-29448x7<arctanx-2x1+1-x2<-112x3+340x5-29448x7+651152x9.(ii)
Takingk=3 in (66) gives (69)-112x3+340x5-29448x7+651152x9-2815632x11<arctanx-2x1+1-x2<-112x3+340x5-29448x7+651152x9-2815632x11+59513312x13,etc.Proof of Theorem11.
Forx∈(0,1] the following power series expansion holds: (70)arctanx-2x1+1+x2=arctanx+21-x2+1x=∑m=1∞-1mCmx2m+1 where(71)Cm=12m+1-2m-1!!m+1!2m.
We prove that the sequenceC(m)m∈N is positive and monotonically decreasing and tends to zero as m tends to infinity.
It is easy to verify that(2m+2)!!>2(2m+1)!! is true for m∈N. Thus, the following equivalences hold true for every m∈N: (72)Cm>0⇔12m+1>2m-1!!m+1!2m⇔2m+2!!>22m+1!!,and we conclude that C(m)>0 for every m∈N.
Let us now prove thatC(m)m∈N is a monotonically decreasing sequence. We have (73)Cm-Cm+1>0⇔22m+12m+3-32m-1!!2m+2!!m+2>0⇔2m+4!!>32m+3!!.
As it is easy to show (by the principle of mathematical induction) that the last inequality holds true form∈N, we may conclude that C(m)m∈N is a monotonically decreasing sequence.
Finally, as0<C(m)<1/2m+1, we conclude that limm→+∞⁡C(m)=0.
SinceCmm∈N is a positive monotonically decreasing sequence, and it tends to zero, the same holds true for the sequence C(m)x2m+1m∈N for a fixed x∈(0,1]. So we can apply Leibniz’s theorem for alternating series [28] and thus prove the claim of Theorem 11.


## 3. Conclusion

In Theorems9, 10, and 11 of this paper we proved some new inequalities related to Shafer’s inequality for the arctangent function. These inequalities represent sharpening and generalization of the inequalities given in [8] (Theorems1,2,and4).Let us mention that it is possible to prove inequality (35), for any fixed k∈N and x∈(0,3/2], by substituting x=tant for t∈(0,π/3] using the algorithms and methods (see also [29, 30]) developed in [31, 32]. Also, inequalities (42) and (66) for any fixed k∈N and x∈0,1 can be proved by substituting x=tant for t∈0,π/4 using the algorithms and methods (see also [29, 30]) developed in [31, 32].



---

*Source: 4178629-2018-10-10.xml*