Inequalities between the inverse hyperbolic tangent and the inverse sine and the analogue for corresponding functions

Abstract

In this paper, we obtain some new inequalities which reveal the further relationship between the inverse tangent function arctanx and the inverse hyperbolic sine function sinh⁻¹x. At the same time, we give the analogue for inverse hyperbolic tangent and inverse sine.

Introduction

In 2010, Masjed-Jamei [1] obtained the following inequality:

$$ ( \arctan x ) ^{2}\leq \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}},\quad \vert x \vert < 1. $$

(1.1)

[1] also reminded us that the above inequality is established in a larger interval $(-\infty ,\infty )$ because it was detected by Maple software. Inequality (1.1) gives the upper bound for the square of the inverse tangent function arctanx by the inverse hyperbolic sine function $\sinh ^{-1}x=\ln (x+\sqrt{1+x^{2}})$.

In this paper, we first affirm Masjed-Jamei’s quest, conclude that the scope of the inequality is indeed the large interval $(-\infty , \infty )$, and give a simple proof of this result. Second, we get the strengthening of the inequality that we have just given. Then, we obtain some natural generalizations of this inequality. At the same time, we show the analogue for inverse hyperbolic tangent function $\operatorname{arctanh}x= ( 1/2 ) \ln ((1+x)/(1-x))$ and inverse sine function arcsinx. Finally, we propose a conjecture on this topic.

Theorem 1.1

The inequality

$$ ( \arctan x ) ^{2}\leq \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}} $$

(1.2)

holds for all $x\in (-\infty ,\infty )$, and the power number 2 is the best in (1.2).

Theorem 1.2

Let $0< r<\infty$, $\lambda = 1$, and $\mu =r\ln ( r+\sqrt{r^{2}+1} ) / ( \sqrt{r^{2}+1} ( \arctan r ) ^{2} )$. Then the double inequality

$$ \lambda ( \arctan x ) ^{2}\leq \frac{x\ln (x+\sqrt{1+x ^{2}})}{\sqrt{1+x^{2}}}\leq \mu ( \arctan x ) ^{2} $$

(1.3)

holds for all $x\in (-r,r)$, where λ and μ are the best constants in (1.3).

Theorem 1.3

Let $-\infty <$ $x<\infty$. Then we have

$$\begin{aligned}& -\frac{1}{45}x^{6} \leq ( \arctan x ) ^{2}-\frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}}\leq -\frac{1}{45}x^{6}+ \frac{4}{105}x^{8}, \end{aligned}$$

(1.4)

$$\begin{aligned}& -\frac{1}{45}x^{6}+\frac{4}{105}x^{8}- \frac{11}{225}x^{10} \leq ( \arctan x ) ^{2}- \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}} \\& \hphantom{-\frac{1}{45}x^{6}+\frac{4}{105}x^{8}-\frac{11}{225}x^{10}}\leq -\frac{1}{45}x^{6}+ \frac{4}{105}x^{8}-\frac{11}{225}x^{10}+ \frac{586}{10 {,}395}x^{12}. \end{aligned}$$

(1.5)

Theorem 1.4

The inequality

$$ ( \operatorname{arctanh}x ) ^{2}\leq \frac{x\arcsin x}{\sqrt{1-x ^{2}}} $$

(1.6)

holds for all $x\in (-1,1)$, and the power number 2 is the best in (1.6).

Theorem 1.5

Let $0< r<1$, $\alpha = 1$, and $\beta =r ( \arcsin r ) / ( \sqrt{1-r^{2}} ( \operatorname{arctanh}r ) ^{2} )$. Then the double inequality

$$ \alpha ( \operatorname{arctanh}x ) ^{2}\leq \frac{x\arcsin x}{\sqrt{1-x ^{2}}}\leq \beta ( \operatorname{arctanh}x ) ^{2} $$

(1.7)

holds for all $x\in (-r,r)$, where α and β are the best constants in (1.7).

Theorem 1.6

Let n, N be two integers, $n,N\geq 3$, and

$$ v_{n}=\frac{1}{n} \biggl( \frac{n!2^{n-1}}{ ( 2n-1 ) !!}- \biggl( 1+\frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) \biggr) . $$

(1.8)

Then the inequality

$$ \frac{x\arcsin x}{\sqrt{1-x^{2}}}- ( \operatorname{arctanh}x ) ^{2}\geq \sum_{n=3}^{N}v_{n}x^{2n} $$

(1.9)

holds for all $x\in (-1,1)$.

Simple proof of Theorem 1.1

Let $\arctan x=t$, $x\in (-\infty ,\infty )$. Then $x=\tan t$, $t\in (- \pi /2,\pi /2)$, and (1.2) is equivalent to

$$ \ln (\tan t+\sec t)=\ln \frac{1+\sin t}{\cos t}>\frac{t^{2}}{\sin t} $$

(2.1)

for $t\neq 0$ since the equality in (1.2) holds for $x=0$. Let

$$ F_{1}(t)=\ln \frac{1+\sin t}{\cos t}-\frac{t^{2}}{\sin t}=\ln (1+ \sin t)- \ln \cos t-\frac{t^{2}}{\sin t}. $$

Then

$$\begin{aligned} F_{1}^{\prime }(t) =& \frac{\cos t}{\sin t+1}+ \frac{1}{\cos t}\sin t+ \frac{1}{\sin ^{2}t} \bigl( t^{2}\cos t-2t\sin t \bigr) \\ =& \frac{1}{\cos t} + \frac{1}{\sin ^{2}t} \bigl( t^{2}\cos t-2t\sin t \bigr) = \frac{ ( -\sin t+t\cos t ) ^{2}}{\cos t\sin ^{2}t}, \end{aligned}$$

which means that $F_{1}^{\prime }(t)>0$ for all $t\in (0,\pi /2)$ and $F_{1}^{\prime }(t)<0$ for all $t\in (-\pi /2,0)$. So $F_{1}(t)>F_{1}(0^{+})=0$ for all $t\in (-\pi /2,0)\cup (0,\pi /2)$. In view of

$$ \lim_{x\rightarrow 0}\frac{\ln \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}}}{\ln \tan ^{-1}x}=2, $$

the proof of Theorem 1.1 is complete.

Proof of Theorem 1.2

In order to prove Theorem 1.2, we use a key method as follows, which is called the monotone form of l’Hospital’s rule.

Lemma 3.1

([2, 3])

For $-\infty < a< b<\infty$, let $f,g:[a,b]\rightarrow \mathbb{R}$ be continuous functions that are differentiable on $( a,b )$, with $f ( a ) =g ( a ) =0$ or $f ( b ) =g ( b ) =0$. If $f^{\prime }/g^{\prime }$ with $g^{\prime }(x)\neq 0$ for each x in $(a,b)$ is increasing (decreasing) on $(a,b)$, then so is $f/g$.

Now, we are in the state of proving Theorem 1.2. After making the same transformation with the second section, we obtain that $t\in (-\arctan r,\arctan r)\subset (-\pi /2,\pi /2)$. Considering that the two functions involved in (1.3) are even functions, we can discuss problems in the range $(0,\arctan r)$. Let

$$\begin{aligned} G_{1}(t) =&\frac{t^{2}}{\frac{ ( \tan t ) \ln (\tan t+ \sec t)}{\sec t}}=\frac{t^{2}}{ ( \sin t ) ( \ln ( 1+\sin t ) -\ln \cos t ) } \\ =&\frac{\frac{t^{2}}{\sin t}}{\ln ( 1+\sin t ) -\ln \cos t}:=\frac{f_{1}(t)}{g_{1}(t)}, \end{aligned}$$

where

$$ f_{1}(t)=\frac{t^{2}}{\sin t},\qquad g_{1}(t)=\ln ( 1+\sin t ) - \ln \cos t. $$

Then

$$ f_{1}^{\prime }(t)=\frac{2t\sin t-t^{2}\cos t}{\sin ^{2}t}, \qquad g_{1}^{\prime }(t)= \frac{1}{\cos t}, $$

and

$$ \frac{f_{1}^{\prime }(t)}{g_{1}^{\prime }(t)}=2t\cot t-t^{2}\cot ^{2}t. $$

Since

$$\begin{aligned} \biggl( \frac{f_{1}^{\prime }(t)}{g_{1}^{\prime }(t)} \biggr) ^{ \prime } =& 2t^{2} \cot ^{3}t+2t^{2}\cot t-4t\cot ^{2}t-2t+2\cot t \\ =& 2 ( t\cot t-1 ) \bigl( t-\cot t+t\cot ^{2}t \bigr) \\ =& 2 ( t\cot t-1 ) \biggl( \frac{t}{\sin ^{2}t}-\frac{\cos t}{ \sin t} \biggr) \\ =&2 ( t\cot t-1 ) \frac{t-\sin t\cos t}{\sin ^{2}t}< 0, \end{aligned}$$

we have that the function $f_{1}^{\prime }(t)/g_{1}^{\prime }(t)$ is decreasing on $(0,\arctan r)$. Then $G_{1}(t)=f_{1}(t)/g_{1}(t)$ is decreasing on $(0,\arctan r)$ too by Lemma 3.1. In view of

$$\begin{aligned}& \frac{1}{\lambda } :=\lim_{t\rightarrow 0^{+}}G_{1}(t)= 1, \\& \frac{1}{\mu } :=\lim_{t\rightarrow \arctan r}G_{1}(t)= \frac{\sqrt{r^{2}+1} ( \arctan r ) ^{2}}{r\ln ( r+\sqrt{r ^{2}+1} ) }, \end{aligned}$$

the proof of Theorem 1.2 is complete.

Remark 3.1

Letting $r\rightarrow \infty$ in Theorem 1.2, we can obtain Theorem 1.1.

Proof of Theorem 1.3

Because the functions involved in this section are all even functions, we only assume $x>0$. After doing the same transformation with the second section, we will only discuss problems in the situation $t\in (0,\pi /2)$. Let

$$\begin{aligned}& h_{1}(t) =\frac{ t^{2}- ( \sin t ) \ln \frac{1+\sin t}{\cos t}+\frac{1}{45} \tan ^{6}t}{\sin t}=\frac{ t^{2}}{\sin t}-\ln \frac{1+\sin t}{\cos t}+\frac{1}{45}\frac{\tan ^{6}t}{ \sin t}, \\& h_{2}(t) =\frac{t^{2}}{\sin t}-\ln \frac{1+\sin t}{\cos t}+ \frac{1}{45}\frac{\tan ^{6}t}{\sin t}-\frac{4}{105}\frac{\tan ^{8}t}{ \sin t}, \\& h_{3}(t) =\frac{ t^{2}}{\sin t}-\ln \frac{1+\sin t}{\cos t}+ \frac{1}{45}\frac{\tan ^{6}t}{ \sin t}-\frac{4}{105}\frac{\tan ^{8}t}{\sin t}+ \frac{11}{225}\frac{ \tan ^{10}t}{\sin t}, \\& h_{4}(t) =\frac{ t^{2}}{\sin t}-\ln \frac{1+\sin t}{\cos t}+ \frac{1}{45}\frac{\tan ^{6}t}{ \sin t}-\frac{4}{105}\frac{\tan ^{8}t}{\sin t}+ \frac{11}{225}\frac{ \tan ^{10}t}{\sin t}-\frac{586}{10\text{,}395}\frac{ \tan ^{12}t}{\sin t}. \end{aligned}$$

Then we get $h_{i}(0^{+})=0$, $i=1,2,3,4$, and

$$\begin{aligned}& h_{1}^{\prime }(t) =\frac{\cos t}{45\sin ^{2}t}g_{1}(t), \\& h_{2}^{\prime }(t) =-\frac{\cos t}{315\sin ^{2}t}g_{2}(t), \\& h_{3}^{\prime }(t)=\frac{\cos t}{1575\sin ^{2}t} g_{3}(t), \\& h_{4}^{\prime }(t) =-\frac{\cos t}{51\text{,}975\sin ^{2}t} g_{4}(t), \end{aligned}$$

where

$$\begin{aligned}& g_{1}(t) =-45t^{2}+ 90t\tan t-45\tan ^{2}t+5 \tan ^{6}t+6\tan ^{8}t, \\& g_{2}(t) =315t^{2}- 630t\tan t+315\tan ^{2}t -35\tan ^{6}t+42\tan ^{8}t+96 \tan ^{10}t, \\& g_{3}(t) =-1575t^{2}+3150t\tan t-1575\tan ^{2}t+ 175\tan ^{6}t-210\tan ^{8}t \\& \hphantom{g_{3}(t) =}{}+ 213\tan ^{10}t+770\tan ^{12}t, \\& g_{4}(t) =51\text{,}975t^{2}-103\text{,}950 t \tan t+51\text{,}975 \tan ^{2}t -5775\tan ^{6}t+ 6930\tan ^{8}t \\& \hphantom{g_{4}(t) =}{}-7029\tan ^{10}t+6820\tan ^{12}t+35\text{,}160\tan ^{14}t, \end{aligned}$$

and $g_{i}(0)=0$, $i=1,2,3,4$. We compute to get

$$\begin{aligned}& f_{1}(t) :=\frac{g_{1}^{\prime }(t)}{6\tan ^{2}t}=8\tan ^{7}t+13\tan ^{5}t+5 \tan ^{3}t-15\tan t+15t , \\& f_{2}(t) :=\frac{g_{2}^{\prime }(t)}{6\tan ^{2}t}=160\tan ^{9}t+216 \tan ^{7}t+21\tan ^{5}t-35\tan ^{3}t+105\tan t-105t , \\& f_{3}(t) :=\frac{g_{3}^{\prime }(t)}{30\tan ^{2}t} \\& \hphantom{f_{3}(t)\,}=308\tan ^{11}t+379 \tan ^{9}t+15\tan ^{7}t-21\tan ^{5}t+35\tan ^{3}t-105\tan t+105t, \\& f_{4}(t) :=\frac{g_{3}^{\prime }(t)}{30\tan ^{2}t}=16\text{,}408\tan ^{13}t+19 \text{,}136\tan ^{11}t+385\tan ^{9}t-495\tan ^{7}t+693\tan ^{5}t \\& \hphantom{f_{4}(t) :=}{}-1155\tan ^{3}t+3465\tan t-3465t \end{aligned}$$

with $f_{i}(0)=0$, $i=1,2,3,4$. Then

$$\begin{aligned}& f_{1}^{\prime }(t) = \bigl( \tan ^{4}t \bigr) \bigl( 56\tan ^{4}t+121 \tan ^{2}t+80 \bigr) >0, \\& f_{2}^{\prime }(t) = \bigl( \tan ^{6}t \bigr) \bigl( 1440\tan ^{4}t+2952\tan ^{2}t+1617 \bigr) >0, \\& f_{3}^{\prime }(t) = \bigl( \tan ^{8}t \bigr) \bigl( 3388\tan ^{4}t+6799 \tan ^{2}t+3516 \bigr) >0, \\& f_{4}^{\prime }(t) = \bigl( \tan ^{10}t \bigr) \bigl( 213\text{,}304\tan ^{4}t+423 \text{,}800\tan ^{2}t+213 \text{,}961 \bigr) >0. \end{aligned}$$

Through differential deduction, we complete the proof of Theorem 1.3.

Proof of Theorem 1.4

Since the two functions showed in (1.6) are even functions, we can discuss problems in the range $(0,1)$. Let $\arcsin x=t$, $x\in (0,1)$. Then $x=\sin t$, $t\in (0,\pi /2)$. We find that

$$\begin{aligned} \operatorname{arctanh} ( \sin t ) =&\frac{1}{2}\ln \frac{1+\sin t}{1- \sin t}= \frac{1}{2}\ln \frac{ ( 1+\sin t ) ^{2}}{ ( 1- \sin t ) ( 1+\sin t ) } \\ =&\frac{1}{2}\ln \biggl( \frac{1+\sin t}{\cos t} \biggr) ^{2}=\ln \frac{1+ \sin t}{\cos t}, \end{aligned}$$

and (1.6) is equivalent to

$$ \biggl( \ln \frac{1+\sin t}{\cos t} \biggr) ^{2}< \frac{t\sin t}{ \cos t},\quad 0< t< \frac{\pi }{2}. $$

(5.1)

Let

$$ F_{2}(t)=\frac{t\sin t}{\cos t}- \biggl( \ln \frac{1+\sin t}{\cos t} \biggr) ^{2}. $$

Then

$$ F_{2}^{\prime }(t)= \frac{t+\cos t\sin t}{\cos ^{2}t} - \frac{2}{\cos t}\ln \frac{\sin t+1}{\cos t}, $$

or

$$ ( \cos t ) F_{2}^{\prime }(t)=\frac{t+\cos t\sin t}{ \cos t}-2\ln \frac{\sin t+1}{\cos t}. $$

We can compute to obtain

$$ \bigl( ( \cos t ) F_{2}^{\prime }(t) \bigr) ^{\prime }= \frac{ ( \sin t ) ( 2t-\sin 2t ) }{2\cos ^{2}t}>0,\quad 0< t< \frac{ \pi }{2}, $$

which implies that

$$ ( \cos t ) F_{2}^{\prime }(t)>\lim_{t\rightarrow 0^{+}} ( \cos t ) F_{2}^{\prime }(t)=0 $$

for all $t\in (0,\pi /2)$. Then

$$ F_{2}^{\prime }(t)>0\quad \Longrightarrow\quad F_{2}(t)>F_{2} \bigl(0^{+}\bigr)=0 $$

for all $t\in (0,\pi /2)$.

In view of

$$ \lim_{x\rightarrow 0}\frac{\ln \frac{x\arcsin x}{\sqrt{1-x^{2}}}}{ \ln \operatorname{arctanh}x}=2, $$

the proof of Theorem 1.4 is complete.

Proof of Theorem 1.5

After making the same transformation as in the section above, we obtain that $t\in (-\arcsin r,\arcsin r)\subset (-\pi /2,\pi /2)$. Considering that the two functions involved in (1.7) are even functions, we can discuss problems in the range $(0,\arcsin r)$. Let

$$ G_{2}(t)=\frac{\frac{t ( \sin t ) }{\cos t}}{ ( \ln \frac{1+\sin t}{\cos t} ) ^{2}}:=\frac{f_{2}(t)}{g_{2}(t)}, $$

where

$$ f_{2}(t)=\frac{t ( \sin t ) }{\cos t},\qquad g_{2}(t)= \biggl( \ln \frac{1+\sin t}{\cos t} \biggr) ^{2}. $$

Then

$$ f_{2}^{\prime }(t)=\frac{t+\cos t\sin t}{\cos ^{2}t},\qquad g_{2}^{\prime }(t)= \frac{2}{\cos t}\ln \frac{\sin t+1}{\cos t} , $$

and

$$ \frac{f_{2}^{\prime }(t)}{g_{2}^{\prime }(t)}=\frac{ \frac{t+\cos t\sin t}{\cos ^{2}t} }{\frac{2}{\cos t}\ln \frac{\sin t+1}{\cos t}}=\frac{\frac{t+\cos t \sin t}{\cos t}}{2\ln \frac{\sin t+1}{\cos t}}:= \frac{f_{3}(t)}{g_{3}(t)}. $$

Since

$$\begin{aligned}& f_{3}^{\prime }(t) = \biggl( \frac{t+\cos t\sin t}{\cos t} \biggr) ^{\prime }= \frac{1}{\cos ^{2}t} \bigl( \cos ^{3}t+\cos t+t\sin t \bigr) , \\& g_{3}^{\prime }(t) = \biggl( 2\ln \frac{\sin t+1}{\cos t} \biggr) ^{\prime }=\frac{2}{\cos t}, \end{aligned}$$

we have

$$ \frac{f_{3}^{\prime }(t)}{g_{3}^{\prime }(t)}= \frac{1}{2\cos t} \bigl( \cos ^{3}t+ \cos t+t\sin t \bigr) . $$

So

$$ \biggl( \frac{f_{3}^{\prime }(t)}{g_{3}^{\prime }(t)} \biggr) ^{ \prime }= \frac{1}{8\cos ^{2}t} ( 4t-\sin 4t ) >0, $$

which leads to the fact that the function $f_{3}^{\prime }(t)/g_{3} ^{\prime }(t)$ is increasing on $(0,\arcsin r)$. Then $f_{3}(t)/g_{3}(t)$ is increasing on $(0,\arcsin r)$ too by Lemma 3.1. Using Lemma 3.1 again, we come to the conclusion that $G_{2}(t)=f_{2}(t)/g_{2}(t)$ is increasing on $(0,\arcsin r)$.

In view of

$$\begin{aligned}& \alpha :=\lim_{t\rightarrow 0^{+}}G_{2}(t)= 1, \\& \beta :=\lim_{t\rightarrow \arcsin r}G_{2}(t)=\frac{r\arcsin r}{ ( \operatorname{arctanh}r ) ^{2}\sqrt{1-r^{2}}}, \end{aligned}$$

the proof of Theorem 1.5 is complete.

Remark 6.1

Letting $r\rightarrow 1$ in Theorem 1.5, we can obtain Theorem 1.4.

Proof of Theorem 1.6

In order to prove Theorem 1.6, we need the following lemma.

Lemma 7.1

([4,5,6,7,8])

Let $\vert x \vert <1$. Then

$$ \frac{2x\arcsin x}{\sqrt{1-x^{2}}}=\sum_{n=1}^{\infty } \frac{ ( 2x ) ^{2n}}{n\binom{2n}{n}}. $$

(7.1)

We are in the state of proving Theorem 1.6.

First, by Lemma 7.1 we get

$$ \frac{x\arcsin x}{\sqrt{1-x^{2}}}=\sum_{n=1}^{\infty } \frac{n!2^{n-1}}{n ( 2n-1 ) !!}x^{2n} $$

(7.2)

due to

$$ \binom{2n}{n}=\frac{2^{n} ( 2n-1 ) !!}{n!}. $$

Second, we have

$$\begin{aligned} \frac{d}{dx} ( \operatorname{arctanh}x ) ^{2} =& \frac{2}{1-x^{2}}\operatorname{arctanh}x=2 \Biggl( \sum _{n=0}^{\infty }x^{2n} \Biggr) \Biggl( \sum _{n=0}^{\infty }\frac{x ^{2n+1}}{2n+1} \Biggr) \\ =&2\sum_{n=1}^{\infty } \biggl( 1+ \frac{1}{3}+\cdots + \frac{1}{2n-1} \biggr) x^{2n-1}. \end{aligned}$$

(7.3)

Integrating two sides of (7.3) on $[0,x]$, we can obtain

$$ ( \operatorname{arctanh}x ) ^{2}=\sum _{n=1}^{\infty } \frac{1}{n} \biggl( 1+ \frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) x^{2n}. $$

(7.4)

From (7.2) and (7.4) we have

$$\begin{aligned} \frac{x\arcsin x}{\sqrt{1-x^{2}}}- ( \operatorname{arctanh}x ) ^{2} =& \sum_{n=1}^{\infty } \biggl( \frac{n!2^{n-1}}{n ( 2n-1 ) !!}x^{2n}- \frac{1}{n} \biggl( 1+\frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) \biggr) x^{2n} \\ =&\sum_{n=3}^{\infty }\frac{1}{n} \biggl( \frac{n!2^{n-1}}{ ( 2n-1 ) !!}x^{2n}- \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2n-1} \biggr) \biggr) x^{2n} \\ :=&\sum_{n=3}^{\infty }v_{n}x^{2n}, \end{aligned}$$

(7.5)

where

$$ v_{n}=\frac{1}{n} \biggl( \frac{n!2^{n-1}}{ ( 2n-1 ) !!}- \biggl( 1+ \frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) \biggr) ,\quad n\geq 3. $$

Below we shall prove that

$$ u_{n}:=nv_{n}=\frac{n!2^{n-1}}{ ( 2n-1 ) !!}- \biggl( 1+ \frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) >0 $$

(7.6)

for $n\geq 3$.

In fact, when $n=3$, inequality (7.6) holds. Now, we assume that (7.6) holds for $n=m$, that is,

$$ \frac{m!2^{m-1}}{ ( 2m-1 ) !!}>1+\frac{1}{3}+\cdots +\frac{1}{2m-1}. $$

Since

$$ \frac{ ( m+1 ) !2^{m}}{ ( 2m+1 ) !!}=\frac{2 ( m+1 ) }{2m+1} \frac{m!2^{m-1}}{ ( 2m-1 ) !!}> \frac{2 ( m+1 ) }{2m+1} \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2m-1} \biggr) , $$

in order to prove that (7.6) is also true for $n=m+1$, it suffices to show that

$$ \frac{2 ( m+1 ) }{2m+1} \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2m-1} \biggr) > \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2m-1} \biggr) + \frac{1}{2m+1} , $$

which is true due to

$$ \frac{1}{2m+1} \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2m-1} \biggr) > \frac{1}{2m+1}, $$

or

$$ 1+\frac{1}{3}+\cdots +\frac{1}{2m-1}>1. $$

So, $v_{n}>0$ for $n\geq 3$, and

$$ \frac{x\arcsin x}{\sqrt{1-x^{2}}}- ( \operatorname{arctanh}x ) ^{2}\geq \sum _{n=3}^{N}v_{n}x^{2n} $$

holds for all $x\in (-1,1)$, where N is any integer greater than or equal to 3.

Remark 7.1

Theorem 1.6 is obviously a natural extension of Theorem 1.4.

Conjecture

Inspired by [9], in the last section, we pose the following conjecture in the form of (1.4) and (1.5).

Conjecture 8.1

Let $x\in \mathbf{R}$, $m\geq 1$, and $v_{n}$ as defined by (1.8). Then the double inequality

$$ \sum_{n=3}^{2m+1} ( -1 ) ^{n}v_{n}x^{2n} \leq ( \arctan x ) ^{2}-\frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}}\leq \sum _{n=3}^{2m+2} ( -1 ) ^{n}v_{n}x^{2n} $$

(8.1)

holds.

Remark 8.1

There are several factors that lead to the fact that this double inequality cannot be proved by Leibniz’s theorem for alternating series. The first is that the interval we are discussing now is infinite, and the second is that the sequence $\{v_{n}\}_{n\geq 3}$ does not have the characteristic of monotone decreasing.