1 Introduction
In 2010, Masjed-Jamei [1] obtained the following inequality:
[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 \(\sinh ^{-1}x=\ln (x+\sqrt{1+x^{2}})\).
$$ ( \arctan x ) ^{2}\leq \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}},\quad \vert x \vert < 1. $$
(1.1)
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.
Anzeige
Theorem 1.1
The inequality
holds for all
\(x\in (-\infty ,\infty )\), and the power number 2 is the best in (1.2).
$$ ( \arctan x ) ^{2}\leq \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}} $$
(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
holds for all
\(x\in (-r,r)\), where
λ
and
μ
are the best constants in (1.3).
$$ \lambda ( \arctan x ) ^{2}\leq \frac{x\ln (x+\sqrt{1+x ^{2}})}{\sqrt{1+x^{2}}}\leq \mu ( \arctan x ) ^{2} $$
(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
holds for all
\(x\in (-1,1)\), and the power number 2 is the best in (1.6).
$$ ( \operatorname{arctanh}x ) ^{2}\leq \frac{x\arcsin x}{\sqrt{1-x ^{2}}} $$
(1.6)
Anzeige
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
holds for all
\(x\in (-r,r)\), where
α
and
β
are the best constants in (1.7).
$$ \alpha ( \operatorname{arctanh}x ) ^{2}\leq \frac{x\arcsin x}{\sqrt{1-x ^{2}}}\leq \beta ( \operatorname{arctanh}x ) ^{2} $$
(1.7)
Theorem 1.6
Let
n, N
be two integers, \(n,N\geq 3\), and
Then the inequality
holds for all
\(x\in (-1,1)\).
$$ v_{n}=\frac{1}{n} \biggl( \frac{n!2^{n-1}}{ ( 2n-1 ) !!}- \biggl( 1+\frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) \biggr) . $$
(1.8)
$$ \frac{x\arcsin x}{\sqrt{1-x^{2}}}- ( \operatorname{arctanh}x ) ^{2}\geq \sum_{n=3}^{N}v_{n}x^{2n} $$
(1.9)
2 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
for \(t\neq 0\) since the equality in (1.2) holds for \(x=0\). Let
Then
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
the proof of Theorem 1.1 is complete.
$$ \ln (\tan t+\sec t)=\ln \frac{1+\sin t}{\cos t}>\frac{t^{2}}{\sin t} $$
(2.1)
$$ F_{1}(t)=\ln \frac{1+\sin t}{\cos t}-\frac{t^{2}}{\sin t}=\ln (1+ \sin t)- \ln \cos t-\frac{t^{2}}{\sin t}. $$
$$\begin{aligned} F_{1}^{\prime }(t) =& \frac{\cos t}{\sin t+1}+ \frac{1}{\cos t}\sin t+ \frac{1}{\sin ^{2}t} \bigl( t^{2}\cos t-2t\sin t \bigr) \\ =& \frac{1}{\cos t} + \frac{1}{\sin ^{2}t} \bigl( t^{2}\cos t-2t\sin t \bigr) = \frac{ ( -\sin t+t\cos t ) ^{2}}{\cos t\sin ^{2}t}, \end{aligned}$$
$$ \lim_{x\rightarrow 0}\frac{\ln \frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}}}{\ln \tan ^{-1}x}=2, $$
3 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
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
where
Then
and
Since
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
the proof of Theorem 1.2 is complete.
$$\begin{aligned} G_{1}(t) =&\frac{t^{2}}{\frac{ ( \tan t ) \ln (\tan t+ \sec t)}{\sec t}}=\frac{t^{2}}{ ( \sin t ) ( \ln ( 1+\sin t ) -\ln \cos t ) } \\ =&\frac{\frac{t^{2}}{\sin t}}{\ln ( 1+\sin t ) -\ln \cos t}:=\frac{f_{1}(t)}{g_{1}(t)}, \end{aligned}$$
$$ f_{1}(t)=\frac{t^{2}}{\sin t},\qquad g_{1}(t)=\ln ( 1+\sin t ) - \ln \cos t. $$
$$ f_{1}^{\prime }(t)=\frac{2t\sin t-t^{2}\cos t}{\sin ^{2}t}, \qquad g_{1}^{\prime }(t)= \frac{1}{\cos t}, $$
$$ \frac{f_{1}^{\prime }(t)}{g_{1}^{\prime }(t)}=2t\cot t-t^{2}\cot ^{2}t. $$
$$\begin{aligned} \biggl( \frac{f_{1}^{\prime }(t)}{g_{1}^{\prime }(t)} \biggr) ^{ \prime } =& 2t^{2} \cot ^{3}t+2t^{2}\cot t-4t\cot ^{2}t-2t+2\cot t \\ =& 2 ( t\cot t-1 ) \bigl( t-\cot t+t\cot ^{2}t \bigr) \\ =& 2 ( t\cot t-1 ) \biggl( \frac{t}{\sin ^{2}t}-\frac{\cos t}{ \sin t} \biggr) \\ =&2 ( t\cot t-1 ) \frac{t-\sin t\cos t}{\sin ^{2}t}< 0, \end{aligned}$$
$$\begin{aligned}& \frac{1}{\lambda } :=\lim_{t\rightarrow 0^{+}}G_{1}(t)= 1, \\& \frac{1}{\mu } :=\lim_{t\rightarrow \arctan r}G_{1}(t)= \frac{\sqrt{r^{2}+1} ( \arctan r ) ^{2}}{r\ln ( r+\sqrt{r ^{2}+1} ) }, \end{aligned}$$
4 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
Then we get \(h_{i}(0^{+})=0\), \(i=1,2,3,4\), and
where
and \(g_{i}(0)=0\), \(i=1,2,3,4\). We compute to get
with \(f_{i}(0)=0\), \(i=1,2,3,4\). Then
Through differential deduction, we complete the proof of Theorem 1.3.
$$\begin{aligned}& h_{1}(t) =\frac{ t^{2}- ( \sin t ) \ln \frac{1+\sin t}{\cos t}+\frac{1}{45} \tan ^{6}t}{\sin t}=\frac{ t^{2}}{\sin t}-\ln \frac{1+\sin t}{\cos t}+\frac{1}{45}\frac{\tan ^{6}t}{ \sin t}, \\& h_{2}(t) =\frac{t^{2}}{\sin t}-\ln \frac{1+\sin t}{\cos t}+ \frac{1}{45}\frac{\tan ^{6}t}{\sin t}-\frac{4}{105}\frac{\tan ^{8}t}{ \sin t}, \\& h_{3}(t) =\frac{ t^{2}}{\sin t}-\ln \frac{1+\sin t}{\cos t}+ \frac{1}{45}\frac{\tan ^{6}t}{ \sin t}-\frac{4}{105}\frac{\tan ^{8}t}{\sin t}+ \frac{11}{225}\frac{ \tan ^{10}t}{\sin t}, \\& h_{4}(t) =\frac{ t^{2}}{\sin t}-\ln \frac{1+\sin t}{\cos t}+ \frac{1}{45}\frac{\tan ^{6}t}{ \sin t}-\frac{4}{105}\frac{\tan ^{8}t}{\sin t}+ \frac{11}{225}\frac{ \tan ^{10}t}{\sin t}-\frac{586}{10\text{,}395}\frac{ \tan ^{12}t}{\sin t}. \end{aligned}$$
$$\begin{aligned}& h_{1}^{\prime }(t) =\frac{\cos t}{45\sin ^{2}t}g_{1}(t), \\& h_{2}^{\prime }(t) =-\frac{\cos t}{315\sin ^{2}t}g_{2}(t), \\& h_{3}^{\prime }(t)=\frac{\cos t}{1575\sin ^{2}t} g_{3}(t), \\& h_{4}^{\prime }(t) =-\frac{\cos t}{51\text{,}975\sin ^{2}t} g_{4}(t), \end{aligned}$$
$$\begin{aligned}& g_{1}(t) =-45t^{2}+ 90t\tan t-45\tan ^{2}t+5 \tan ^{6}t+6\tan ^{8}t, \\& g_{2}(t) =315t^{2}- 630t\tan t+315\tan ^{2}t -35\tan ^{6}t+42\tan ^{8}t+96 \tan ^{10}t, \\& g_{3}(t) =-1575t^{2}+3150t\tan t-1575\tan ^{2}t+ 175\tan ^{6}t-210\tan ^{8}t \\& \hphantom{g_{3}(t) =}{}+ 213\tan ^{10}t+770\tan ^{12}t, \\& g_{4}(t) =51\text{,}975t^{2}-103\text{,}950 t \tan t+51\text{,}975 \tan ^{2}t -5775\tan ^{6}t+ 6930\tan ^{8}t \\& \hphantom{g_{4}(t) =}{}-7029\tan ^{10}t+6820\tan ^{12}t+35\text{,}160\tan ^{14}t, \end{aligned}$$
$$\begin{aligned}& f_{1}(t) :=\frac{g_{1}^{\prime }(t)}{6\tan ^{2}t}=8\tan ^{7}t+13\tan ^{5}t+5 \tan ^{3}t-15\tan t+15t , \\& f_{2}(t) :=\frac{g_{2}^{\prime }(t)}{6\tan ^{2}t}=160\tan ^{9}t+216 \tan ^{7}t+21\tan ^{5}t-35\tan ^{3}t+105\tan t-105t , \\& f_{3}(t) :=\frac{g_{3}^{\prime }(t)}{30\tan ^{2}t} \\& \hphantom{f_{3}(t)\,}=308\tan ^{11}t+379 \tan ^{9}t+15\tan ^{7}t-21\tan ^{5}t+35\tan ^{3}t-105\tan t+105t, \\& f_{4}(t) :=\frac{g_{3}^{\prime }(t)}{30\tan ^{2}t}=16\text{,}408\tan ^{13}t+19 \text{,}136\tan ^{11}t+385\tan ^{9}t-495\tan ^{7}t+693\tan ^{5}t \\& \hphantom{f_{4}(t) :=}{}-1155\tan ^{3}t+3465\tan t-3465t \end{aligned}$$
$$\begin{aligned}& f_{1}^{\prime }(t) = \bigl( \tan ^{4}t \bigr) \bigl( 56\tan ^{4}t+121 \tan ^{2}t+80 \bigr) >0, \\& f_{2}^{\prime }(t) = \bigl( \tan ^{6}t \bigr) \bigl( 1440\tan ^{4}t+2952\tan ^{2}t+1617 \bigr) >0, \\& f_{3}^{\prime }(t) = \bigl( \tan ^{8}t \bigr) \bigl( 3388\tan ^{4}t+6799 \tan ^{2}t+3516 \bigr) >0, \\& f_{4}^{\prime }(t) = \bigl( \tan ^{10}t \bigr) \bigl( 213\text{,}304\tan ^{4}t+423 \text{,}800\tan ^{2}t+213 \text{,}961 \bigr) >0. \end{aligned}$$
5 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
and (1.6) is equivalent to
$$\begin{aligned} \operatorname{arctanh} ( \sin t ) =&\frac{1}{2}\ln \frac{1+\sin t}{1- \sin t}= \frac{1}{2}\ln \frac{ ( 1+\sin t ) ^{2}}{ ( 1- \sin t ) ( 1+\sin t ) } \\ =&\frac{1}{2}\ln \biggl( \frac{1+\sin t}{\cos t} \biggr) ^{2}=\ln \frac{1+ \sin t}{\cos t}, \end{aligned}$$
$$ \biggl( \ln \frac{1+\sin t}{\cos t} \biggr) ^{2}< \frac{t\sin t}{ \cos t},\quad 0< t< \frac{\pi }{2}. $$
(5.1)
Let
Then
or
$$ F_{2}(t)=\frac{t\sin t}{\cos t}- \biggl( \ln \frac{1+\sin t}{\cos t} \biggr) ^{2}. $$
$$ F_{2}^{\prime }(t)= \frac{t+\cos t\sin t}{\cos ^{2}t} - \frac{2}{\cos t}\ln \frac{\sin t+1}{\cos t}, $$
$$ ( \cos t ) F_{2}^{\prime }(t)=\frac{t+\cos t\sin t}{ \cos t}-2\ln \frac{\sin t+1}{\cos t}. $$
We can compute to obtain
which implies that
for all \(t\in (0,\pi /2)\). Then
for all \(t\in (0,\pi /2)\).
$$ \bigl( ( \cos t ) F_{2}^{\prime }(t) \bigr) ^{\prime }= \frac{ ( \sin t ) ( 2t-\sin 2t ) }{2\cos ^{2}t}>0,\quad 0< t< \frac{ \pi }{2}, $$
$$ ( \cos t ) F_{2}^{\prime }(t)>\lim_{t\rightarrow 0^{+}} ( \cos t ) F_{2}^{\prime }(t)=0 $$
$$ F_{2}^{\prime }(t)>0\quad \Longrightarrow\quad F_{2}(t)>F_{2} \bigl(0^{+}\bigr)=0 $$
In view of
the proof of Theorem 1.4 is complete.
$$ \lim_{x\rightarrow 0}\frac{\ln \frac{x\arcsin x}{\sqrt{1-x^{2}}}}{ \ln \operatorname{arctanh}x}=2, $$
6 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
where
Then
and
Since
we have
So
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 \((0,\arcsin r)\).
$$ G_{2}(t)=\frac{\frac{t ( \sin t ) }{\cos t}}{ ( \ln \frac{1+\sin t}{\cos t} ) ^{2}}:=\frac{f_{2}(t)}{g_{2}(t)}, $$
$$ f_{2}(t)=\frac{t ( \sin t ) }{\cos t},\qquad g_{2}(t)= \biggl( \ln \frac{1+\sin t}{\cos t} \biggr) ^{2}. $$
$$ f_{2}^{\prime }(t)=\frac{t+\cos t\sin t}{\cos ^{2}t},\qquad g_{2}^{\prime }(t)= \frac{2}{\cos t}\ln \frac{\sin t+1}{\cos t} , $$
$$ \frac{f_{2}^{\prime }(t)}{g_{2}^{\prime }(t)}=\frac{ \frac{t+\cos t\sin t}{\cos ^{2}t} }{\frac{2}{\cos t}\ln \frac{\sin t+1}{\cos t}}=\frac{\frac{t+\cos t \sin t}{\cos t}}{2\ln \frac{\sin t+1}{\cos t}}:= \frac{f_{3}(t)}{g_{3}(t)}. $$
$$\begin{aligned}& f_{3}^{\prime }(t) = \biggl( \frac{t+\cos t\sin t}{\cos t} \biggr) ^{\prime }= \frac{1}{\cos ^{2}t} \bigl( \cos ^{3}t+\cos t+t\sin t \bigr) , \\& g_{3}^{\prime }(t) = \biggl( 2\ln \frac{\sin t+1}{\cos t} \biggr) ^{\prime }=\frac{2}{\cos t}, \end{aligned}$$
$$ \frac{f_{3}^{\prime }(t)}{g_{3}^{\prime }(t)}= \frac{1}{2\cos t} \bigl( \cos ^{3}t+ \cos t+t\sin t \bigr) . $$
$$ \biggl( \frac{f_{3}^{\prime }(t)}{g_{3}^{\prime }(t)} \biggr) ^{ \prime }= \frac{1}{8\cos ^{2}t} ( 4t-\sin 4t ) >0, $$
In view of
the proof of Theorem 1.5 is complete.
$$\begin{aligned}& \alpha :=\lim_{t\rightarrow 0^{+}}G_{2}(t)= 1, \\& \beta :=\lim_{t\rightarrow \arcsin r}G_{2}(t)=\frac{r\arcsin r}{ ( \operatorname{arctanh}r ) ^{2}\sqrt{1-r^{2}}}, \end{aligned}$$
7 Proof of Theorem 1.6
In order to prove Theorem 1.6, we need the following lemma.
Lemma 7.1
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
due to
$$ \frac{x\arcsin x}{\sqrt{1-x^{2}}}=\sum_{n=1}^{\infty } \frac{n!2^{n-1}}{n ( 2n-1 ) !!}x^{2n} $$
(7.2)
$$ \binom{2n}{n}=\frac{2^{n} ( 2n-1 ) !!}{n!}. $$
Second, we have
Integrating two sides of (7.3) on \([0,x]\), we can obtain
$$\begin{aligned} \frac{d}{dx} ( \operatorname{arctanh}x ) ^{2} =& \frac{2}{1-x^{2}}\operatorname{arctanh}x=2 \Biggl( \sum _{n=0}^{\infty }x^{2n} \Biggr) \Biggl( \sum _{n=0}^{\infty }\frac{x ^{2n+1}}{2n+1} \Biggr) \\ =&2\sum_{n=1}^{\infty } \biggl( 1+ \frac{1}{3}+\cdots + \frac{1}{2n-1} \biggr) x^{2n-1}. \end{aligned}$$
(7.3)
$$ ( \operatorname{arctanh}x ) ^{2}=\sum _{n=1}^{\infty } \frac{1}{n} \biggl( 1+ \frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) x^{2n}. $$
(7.4)
From (7.2) and (7.4) we have
where
$$\begin{aligned} \frac{x\arcsin x}{\sqrt{1-x^{2}}}- ( \operatorname{arctanh}x ) ^{2} =& \sum_{n=1}^{\infty } \biggl( \frac{n!2^{n-1}}{n ( 2n-1 ) !!}x^{2n}- \frac{1}{n} \biggl( 1+\frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) \biggr) x^{2n} \\ =&\sum_{n=3}^{\infty }\frac{1}{n} \biggl( \frac{n!2^{n-1}}{ ( 2n-1 ) !!}x^{2n}- \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2n-1} \biggr) \biggr) x^{2n} \\ :=&\sum_{n=3}^{\infty }v_{n}x^{2n}, \end{aligned}$$
(7.5)
$$ v_{n}=\frac{1}{n} \biggl( \frac{n!2^{n-1}}{ ( 2n-1 ) !!}- \biggl( 1+ \frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) \biggr) ,\quad n\geq 3. $$
Below we shall prove that
for \(n\geq 3\).
$$ u_{n}:=nv_{n}=\frac{n!2^{n-1}}{ ( 2n-1 ) !!}- \biggl( 1+ \frac{1}{3}+\cdots +\frac{1}{2n-1} \biggr) >0 $$
(7.6)
In fact, when \(n=3\), inequality (7.6) holds. Now, we assume that (7.6) holds for \(n=m\), that is,
Since
in order to prove that (7.6) is also true for \(n=m+1\), it suffices to show that
which is true due to
or
$$ \frac{m!2^{m-1}}{ ( 2m-1 ) !!}>1+\frac{1}{3}+\cdots +\frac{1}{2m-1}. $$
$$ \frac{ ( m+1 ) !2^{m}}{ ( 2m+1 ) !!}=\frac{2 ( m+1 ) }{2m+1} \frac{m!2^{m-1}}{ ( 2m-1 ) !!}> \frac{2 ( m+1 ) }{2m+1} \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2m-1} \biggr) , $$
$$ \frac{2 ( m+1 ) }{2m+1} \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2m-1} \biggr) > \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2m-1} \biggr) + \frac{1}{2m+1} , $$
$$ \frac{1}{2m+1} \biggl( 1+\frac{1}{3}+\cdots + \frac{1}{2m-1} \biggr) > \frac{1}{2m+1}, $$
$$ 1+\frac{1}{3}+\cdots +\frac{1}{2m-1}>1. $$
So, \(v_{n}>0\) for \(n\geq 3\), and
holds for all \(x\in (-1,1)\), where N is any integer greater than or equal to 3.
$$ \frac{x\arcsin x}{\sqrt{1-x^{2}}}- ( \operatorname{arctanh}x ) ^{2}\geq \sum _{n=3}^{N}v_{n}x^{2n} $$
8 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
holds.
$$ \sum_{n=3}^{2m+1} ( -1 ) ^{n}v_{n}x^{2n} \leq ( \arctan x ) ^{2}-\frac{x\ln (x+\sqrt{1+x^{2}})}{\sqrt{1+x ^{2}}}\leq \sum _{n=3}^{2m+2} ( -1 ) ^{n}v_{n}x^{2n} $$
(8.1)
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.
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