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<title>About bounds for the elliptic integral<br />of the first kind: About bounds for the elliptic integral<br />of the first kind</title>
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<h1>About bounds for the elliptic integral<br />of the first kind</h1>
<p class="authors">
<span class="author">Pál A. Kupán\(^\ast \) Róbert Szász\(^\ast \)</span>
</p>
<p class="date">February 6, 2012</p>
</div>
<p>\(^\ast \)Department of Mathematics, Sapientia University, Sighişoarei st., no. 1C, Tg. Mureş, Romania, e-mail: <span class="tt">kupanp@ms.sapientia.ro, szasz_robert2001@yahoo.com</span>. </p>

<div class="abstract"><p> We deduce an inequality using elementary methods which makes it possible to prove a conjecture regarding the upper bound of the elliptic integral of the first kind, furthermore we also improve the lower bound. </p>
<p><b class="bf">MSC.</b> 33C05 </p>
<p><b class="bf">Keywords.</b> Hypergeometric function; Elliptic integral; Inequality; Bounds. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>Legendre’s complete elliptic integral of the first kind is defined for \(r\in (0,1)\) by </p>
<div class="displaymath" id="a0000000003">
  \[ \mathcal{K}(r)=\int _{0}^{\pi /2}\tfrac {1}{\sqrt{1-r^2\sin {t}}}{\rm d}t. \]
</div>
<p> This integral is a special case of Gauss’s hypergeometric function </p>
<div class="displaymath" id="a0000000004">
  \[ _2F_1(a,b;c;x)=\sum _{n=0}^\infty \tfrac {(a,n)(b,n)}{(c,n)}\tfrac {x^n}{n!}, \  \  x\in (-1,1),  \]
</div>
<p> where \((a,n)=\prod _{k=0}^{n-1}(a+k).\) We have </p>
<div class="equation" id="2">
<p>
  <div class="equation_content">
    \begin{equation} \label{2}\mathcal{K}(r)=\tfrac {\pi }{2} \  {_2F_1}(\tfrac {1}{2},\tfrac {1}{2};1;r^2)=\tfrac {\pi }{2}\Big[1+\sum _{n=1}^\infty \Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)} {2\cdot 4\cdot ...\cdot (2n)}\Big)^2r^{2n}\Big].\end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> In <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span> the authors posed the problem to determine the best values \(\alpha ^*\) and \(\beta ^*\) such that </p>
<div class="displaymath" id="1">
  \begin{align}  \label{1} \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{3/4+\alpha ^*r}{\lt}K(r) {\lt} \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{3/4+\beta ^*r}, \  \  r\in (0,1). \end{align}
</div>
<p> This problem is equivalent to the following: determine the best values \(\alpha ^*\) and \(\beta ^*\) such that </p>
<div class="displaymath" id="a0000000005">
  \[ \alpha ^*{\lt}\Big[G(r)-\tfrac {3}{4}\Big]/r{\lt}\beta ^*, \  \  r\in (0,1), \   \textrm{where} \  \  G(r)=\tfrac {\log (2\mathcal{K}(r)/\pi )}{\log ([{\rm arth}(r)]/r)}. \]
</div>
<p> The first part of this problem had been solved by the authors in <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span> showing that \(\alpha ^*=0.\) Concerning the second part they conjectured that the mapping \(G:(0,1)\rightarrow {\mathbb {R}}\) is strictly increasing and convex. Since \(\lim _{r\nearrow 1}G(r)=1,\) this conjecture would imply \(\beta ^*=1/4.\) The result \(\beta ^*=1/4\) has been proved recently in <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>. The basic tool used in their proof is Theorem 1.25 from <span class="cite">
	[
	<a href="#2" >2</a>
	]
</span>. It seems very difficult to prove the conjecture regarding the monotonicity of \(G.\) In the following we will show that a different elementary approach leads to a result, which improves the upper bound conjectured in <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>.<br />In order to prove our results, we need certain lemmas, which will be exposed in the next section. </p>
<h1 id="a0000000006">2 Preliminaries</h1>
<p><div class="lema_thmwrapper " id="a0000000007">
  <div class="lema_thmheading">
    <span class="lema_thmcaption">
    Lemma
    </span>
    <span class="lema_thmlabel">1</span>
  </div>
  <div class="lema_thmcontent">
  <p>If \(a_n=\Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)}{2\cdot 4\cdot ...\cdot (2n)}\Big)^2,\) \(b_n=\tfrac {4}{\pi (4n+1)},\) \(c_n=\tfrac {1}{\pi (n+\tfrac {4}{\pi }-1)},\) \(x_n=\tfrac {a_n}{b_n}, \  \  n\in {\mathbb {N}^*},\) \(y_n=\tfrac {a_n}{c_n}, \  \  n\in {\mathbb {N}^*},\) then the sequence \(\big(x_{n}\big)_{n\in \mathbb {N}^*}\) is strictly increasing, the sequence \(\big(y_{n}\big)_{n\in \mathbb {N}^*}\) is strictly decreasing for  \(n\geq 2\) and \(\lim _{n\rightarrow \infty }x_n=\lim _{n\rightarrow \infty }y_n=1.\) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000008">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Since </p>
<div class="displaymath" id="a0000000009">
  \[ \tfrac {x_{n+1}}{x_n}=\tfrac {(4n+5)(2n+1)^2}{(4n+1)(2n+2)^2}=\tfrac {16n^3+36n^2+24n+5}{16n^3+36n^2+24n+4}{\gt}1 \]
</div>
<p> it follows that \(x_{n+1}{\gt}x_n, \  n\in {\mathbb {N^*}}.\) On the other hand, we have \(a_n=\Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)}{2\cdot 4\cdot ...\cdot (2n)}\Big)^2{\lt}\Big(\tfrac {2\cdot 4\cdot 6\cdot ...\cdot (2n)}{3\cdot 5\cdot ...\cdot (2n+1)}\Big)^2.\) This implies \(a_n{\lt}\tfrac {1}{2n+1}\) and finally we get \(x_n{\lt}\tfrac {\pi (4n+1)}{4(2n+1)}{\lt}\tfrac {\pi }{2},  n\in {\mathbb {N^*}}.\) Consequently, \((x_n)_{n\in {\mathbb {N}^*}}\) is convergent. Wallis product formula implies that \(\lim _{n\rightarrow \infty }x_n=1.\) Thus we have </p>
<div class="displaymath" id="a0000000010">
  \[ 1{\gt}x_{n+1}{\gt}x_n\geq {x_1}=\tfrac {5\pi }{16}=0.981\dots . \]
</div>
<p> An analogous calculation implies the assertion regarding \(\big(y_{n}\big)_{n\in \mathbb {N}^*}.\) <div class="proof_wrapper" id="a0000000011">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="lema_thmwrapper " id="a0000000012">
  <div class="lema_thmheading">
    <span class="lema_thmcaption">
    Lemma
    </span>
    <span class="lema_thmlabel">2</span>
  </div>
  <div class="lema_thmcontent">
  <p>For all real numbers \(r\in (0,1),\) we have </p>
<div class="displaymath" id="6">
  \begin{eqnarray} \label{6} \mathcal{K}(r){\lt}\tfrac {\pi }{2}\bigg\{ 1+\tfrac {1}{4}r^2+\tfrac {4}{\pi }\sum _{n=2}^\infty \tfrac {r^{2n}}{4n+1}\bigg\} . \end{eqnarray}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000013">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>We use the notations and the results of Lemma 1 </p>
<div class="displaymath" id="a0000000014">
  \[ \tfrac {2}{\pi }\mathcal{K}(r)=1+\sum _{n=1}^\infty {a}_nr^{2n}=1+\sum _{n=1}^\infty \Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)} {2\cdot 4\cdot ...\cdot (2n)}\Big)^2r^{2n}, \]
</div>
<p> and let </p>
<div class="displaymath" id="a0000000015">
  \[ h(r)=1+a_1r^2+\sum _{n=2}^\infty {b}_n{r^{2n}}=1+\tfrac {1}{4}r^2+\tfrac {4}{\pi }\sum _{n=2}^\infty \tfrac {r^{2n}}{4n+1}. \]
</div>
<p> We introduce the notations \(\tfrac {2}{\pi }\mathcal{K}(r)=1+u(r)\) and \(h(r)=1+v(r).\) Lemma 1 implies \(a_{n}{\lt}{b}_{n},  n\in {\mathbb {N}^*, \  n\geq 2},\) and consequently \(u(r){\lt}v(r)\) for all \(r\in (0,1).\) Thus, inequality (<a href="#6">3</a>) holds. <div class="proof_wrapper" id="a0000000016">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="lema_thmwrapper " id="a0000000017">
  <div class="lema_thmheading">
    <span class="lema_thmcaption">
    Lemma
    </span>
    <span class="lema_thmlabel">3</span>
  </div>
  <div class="lema_thmcontent">
  <p>For all real numbers \(r\in (0,1),\) we have </p>
<div class="displaymath" id="68">
  \begin{eqnarray} \label{68} \tfrac {\pi }{2}\bigg\{ 1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}\bigg\} {\lt}\mathcal{K}(r). \end{eqnarray}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000018">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>We use the notations and the results of Lemma1 in our proof again. We recall that </p>
<div class="displaymath" id="a0000000019">
  \[ \tfrac {2}{\pi }\mathcal{K}(r)=1+\sum _{n=1}^\infty {a}_nr^{2n}=1+\sum _{n=1}^\infty \Big(\tfrac {1\cdot 3\cdot ...\cdot (2n-1)} {2\cdot 4\cdot ...\cdot (2n)}\Big)^2r^{2n}, \]
</div>
<p> and let </p>
<div class="displaymath" id="a0000000020">
  \[ k(r)=1+\tfrac {1}{\pi }\sum _{n=2}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}. \]
</div>
<p> Lemma 1 implies \(c_n{\lt}a_{n},  n\in {\mathbb {N}^*, \  n\geq 2},\) and consequently \(k(r){\lt}\tfrac {2}{\pi }\mathcal{K}(r)\) for all \(r\in (0,1).\) Thus, inequality (<a href="#68">4</a>) holds. <div class="proof_wrapper" id="a0000000021">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="lema_thmwrapper " id="a0000000022">
  <div class="lema_thmheading">
    <span class="lema_thmcaption">
    Lemma
    </span>
    <span class="lema_thmlabel">4</span>
  </div>
  <div class="lema_thmcontent">
  <p>(Bernoulli’s inequality) If \(\alpha \geq 1\) and \(a{\gt}-1\), then </p>
<div class="equation" id="mng">
<p>
  <div class="equation_content">
    \begin{equation} \label{mng} (1+a)^\alpha \geq 1+a\alpha . \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p> If \(b\in [0,1]\) and \(\alpha \in (1,2),\) then </p>
<div class="equation" id="qw2">
<p>
  <div class="equation_content">
    \begin{equation} \label{qw2}(1+b)^\alpha \geq 1+\alpha {b}+\tfrac {\alpha (\alpha -1)}{2}b^2+\tfrac {\alpha (\alpha -1)(\alpha -2)}{6}b^3.\end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000023">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We prove the second inequality. Let \(g:[0,1]\rightarrow {\mathbb {R}}\) be the function defined by \(g(b)=(1+b)^\alpha -1-\alpha {b}-\tfrac {\alpha (\alpha -1)}{2}b^2-\tfrac {\alpha (\alpha -1)(\alpha -2)}{6}b^3.\) We have \(g'(b)=\alpha (1+b)^{\alpha -1}-\alpha -{\alpha (\alpha -1)}b-\tfrac {\alpha (\alpha -1)(\alpha -2)}{2}b^2,\) \(g''(b)=\alpha (\alpha -1)[(1+b)^{\alpha -2}-1-(\alpha -2)b]\) and \(g'''(b)=\alpha (\alpha -1)(\alpha -2)[(1+b)^{\alpha -3}-1].\) Thus \(g''(0)=0\) implies that \(g''(b){\gt}0, \  b\in (0,1).\) An analogous argumentation shows that \(g'\) and \(g\) are strictly increasing on \((0,1)\) and so \(g(0)=0\) implies inequality (<a href="#qw2">6</a>). <div class="proof_wrapper" id="a0000000024">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="lema_thmwrapper " id="a0000000025">
  <div class="lema_thmheading">
    <span class="lema_thmcaption">
    Lemma
    </span>
    <span class="lema_thmlabel">5</span>
  </div>
  <div class="lema_thmcontent">
  <p>Let \(w:(0,1)\rightarrow {\mathbb {R}}\) be the function defined by \(\tfrac {{\rm arth}(r)}{r}=1+\sum ^\infty _{n=1}\tfrac {1}{2n+1}r^{2n}=1+w(r).\) If \(w(r)+w(r)v(r)=\sum ^\infty _{n=1}\delta _nr^{2n},\) then \(\delta _n\leq \tfrac {1}{3}, \  n\in \mathbb {N}, \  n\geq 1.\) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000026">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Indeed \(\delta _1=\tfrac {1}{3}, \  \delta _2=\tfrac {17}{60},\) and if \(n\geq 3,\) then </p>
<div class="displaymath" id="a0000000027">
  \begin{align*}  \delta _n& =\tfrac {1}{2n+1}+\tfrac {1}{8n-4}+\tfrac {4}{\pi }\sum ^{n-2}_{k=1}\tfrac {1}{(2k+1)(4(n-k)+1)}\\ & =\tfrac {1}{2n+1}+\tfrac {1}{8n-4}+\tfrac {4}{\pi (4n+3)}\sum ^{n-2}_{k=1}\Big(\tfrac {1}{2k+1}+\tfrac {2}{4(n-k)+1}\Big)\\ & =\tfrac {1}{2n+1}+\tfrac {1}{8n-4}+\tfrac {4}{\pi (4n+3)}\Big(\sum ^{n-2}_{k=1}\tfrac {1}{2k+1}+\sum ^{n-1}_{k=2}\tfrac {2}{4k+1}\Big) \end{align*}
</div>
<div class="displaymath" id="a0000000028">
  \begin{align*} & {\lt}\tfrac {1}{2n+1}+\tfrac {1}{8n-4}+\tfrac {4}{\pi (4n+3)}\tfrac {3(n-2)}{5}\\ & \leq \tfrac {1}{5}+\tfrac {1}{20}+\tfrac {4}{25\pi }{\lt}\tfrac {1}{3}. \end{align*}
</div>
<p> <div class="proof_wrapper" id="a0000000029">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="lema_thmwrapper " id="a0000000030">
  <div class="lema_thmheading">
    <span class="lema_thmcaption">
    Lemma
    </span>
    <span class="lema_thmlabel">6</span>
  </div>
  <div class="lema_thmcontent">
  <p><span class="cite">
	[
	<a href="#4" >4</a>
	]
</span> Let \((a_n)_{n\geq 1}\) and \((b_n)_{n\geq 1}\) be two sequences of real numbers, and let the power series </p>
<div class="displaymath" id="a0000000031">
  \[ u(x)=\sum _{n=1}^\infty {a}_nx^n \  \  \textrm{and} \   v(x)=\sum _{n=1}^\infty {b}_nx^n \]
</div>
<p> be convergent for \(|x|{\lt}1.\) If \(b_n{\gt}0, \  \  n=1,2,3,...,\) and if the sequence \(\Big(\tfrac {a_n}{b_n}\Big)_{n\geq 1}\) is strictly increasing (resp. decreasing), then the function<br />\(\tfrac {u}{v}:(0,1)\rightarrow {\mathbb {R}}\) is strictly increasing (resp. decreasing). </p>

  </div>
</div> </p>
<h1 id="a0000000032">3 The main result</h1>
<p>Recall that </p>
<div class="displaymath" id="a0000000033">
  \begin{eqnarray} v(r)=\tfrac {1}{4}r^2+\tfrac {4}{\pi }\sum _{n=2}^\infty \tfrac {1}{4n+1}r^{2n},\nonumber \end{eqnarray}
</div>
<p> and </p>
<div class="displaymath" id="a0000000034">
  \[ w(r)=\tfrac {{\rm arth}(r)}{r}-1=\sum ^\infty _{n=1}\tfrac {1}{2n+1}r^{2n}. \]
</div>
<p><div class="theorem_thmwrapper " id="a0000000035">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p>If \(r\in (0,1),\) then the following inequality holds: </p>
<div class="equation" id="123">
<p>
  <div class="equation_content">
    \begin{equation} \label{123}1+v(r)<\big(1+w(r)\big)^{\tfrac {3}{4}+\tfrac {1}{4}r^2}.\end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000036">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We begin with the remark that (<a href="#123">7</a>) is equivalent to </p>
<div class="equation" id="32">
<p>
  <div class="equation_content">
    \begin{equation} \label{32}\bigg(1+\tfrac {w(r)-v(r)}{1+v(r)}\bigg)^{\tfrac {4}{1-r^2}}>1+w(r), \   \  r\in (0,1).\end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p> The inequality (<a href="#mng">5</a>) from Lemma 4 implies that </p>
<div class="displaymath" id="a0000000037">
  \[ \bigg(1+\tfrac {w(r)-v(r)}{1+v(r)}\bigg)^{\tfrac {4}{1-r^2}}{\gt}1+\tfrac {4}{1-r^2}\tfrac {w(r)-v(r)}{1+v(r)}, \   \  r\in (0,1). \]
</div>
<p> Thus, in order to prove (<a href="#32">8</a>), we have to show that </p>
<div class="equation" id="54">
<p>
  <div class="equation_content">
    \begin{equation} \label{54}\tfrac {4(w(r)-v(r))}{(1-r^2)(1+v(r))}>w(r), \  r\in (0,1).\end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p> We have \(w(r)-v(r){\gt}\tfrac {1}{12}r^2, \  r\in (0,1).\) Thus </p>
<div class="displaymath" id="a0000000038">
  \[ \tfrac {r^2}{3(1-r^2)(1+v(r))}{\gt}w(r), \  r\in (0,1) \]
</div>
<p> implies (<a href="#54">9</a>). This inequality is equivalent to </p>
<div class="displaymath" id="a0000000039">
  \[ \tfrac {r^2}{3(1-r^2)}{\gt}w(r)+w(r)v(r), \  r\in (0,1). \]
</div>
<p> According to Lemma 5, we have </p>
<div class="displaymath" id="a0000000040">
  \[ \tfrac {r^2}{3(1-r^2)}=\sum _{n=1}^\infty \tfrac {1}{3}r^{2n}{\gt}\sum ^\infty _{n=1}c_nr^{2n}=w(r)+w(r)v(r),  \]
</div>
<p> and the proof is completed. <div class="proof_wrapper" id="a0000000041">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Theorem 1 and Lemma 2 imply the following result. </p>
<p><div class="corollary_thmwrapper " id="a0000000042">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">8</span>
  </div>
  <div class="corollary_thmcontent">
  <p>If \( \  r\in (0,1),\) then </p>
<div class="displaymath" id="a0000000043">
  \[ \mathcal{K}(r){\lt}\tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\tfrac {1}{4}r^2}. \]
</div>

  </div>
</div> <div class="theorem_thmwrapper " id="a0000000044">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">9</span>
  </div>
  <div class="theorem_thmcontent">
  <p>If \(r\in (0,1),\)   then </p>
<div class="equation" id="8nm9">
<p>
  <div class="equation_content">
    \begin{equation} \label{8nm9}1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}>\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\tfrac {r^4}{200}}.\end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000045">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We introduce the notations \(\mu _1=\tfrac {4}{\pi }-1\) and \(z(r)=\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}=\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\mu _1}.\) Using this notation, (<a href="#8nm9">10</a>) will be equivalent to </p>
<div class="equation" id="1qwn2">
<p>
  <div class="equation_content">
    \begin{equation} \label{1qwn2}\big(1+z(r)\big)^{\tfrac {200}{r^4+150}>1+w(r)}. \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p> We shall prove this inequality in three steps. First assume that \(r\in [0,\tfrac {1}{5}].\)<br />In this case we use the second inequality of Lemma 4 putting \(\alpha =\tfrac {200}{r^4+150}\) and \(b=z(r),\) and we obtain </p>
<div class="equation" id="qwn2">
<p>
  <div class="equation_content">
    \begin{equation} \label{qwn2}(1+z(r))^\alpha \geq 1+\alpha {z(r)}+\tfrac {\alpha (\alpha -1)}{2}(z(r))^2+\tfrac {\alpha (\alpha -1)(\alpha -2)}{6}(z(r))^3.\end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<p> On the other hand we have  \(\tfrac {\alpha (\alpha -1)(2-\alpha )}{6}{\lt}\tfrac {1}{20},\) \(\tfrac {\alpha (\alpha -1)}{2}{\gt}0.22,\) \((z(r))^3{\lt}\tfrac {r^6}{50},  r\in (0,\tfrac {1}{5}).\)   Thus, inequality (<a href="#qwn2">12</a>) implies </p>
<div class="displaymath" id="a0000000046">
  \[ (1+z(r))^\alpha \geq 1+\alpha {z(r)}+0.22(z(r))^2-\tfrac {r^6}{1000}, \  \  r\in [0,\tfrac {1}{5}], \]
</div>
<p> and consequently, in order to prove (<a href="#1qwn2">11</a>) we have to show that </p>
<div class="displaymath" id="a0000000047">
  \[ 1+\alpha {z(r)}+0.22(z(r))^2-\tfrac {r^6}{1000}\geq 1+w(r), \  \  r\in [0,\tfrac {1}{5}]. \]
</div>
<p> This inequality is equivalent to </p>
<div class="equation" id="dfg21">
<p>
  <div class="equation_content">
    \begin{equation} \label{dfg21}0.22\Big(\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\mu _1}\Big)^2>\tfrac {r^6}{1000}+\sum _{n=1}^\infty \Big(\tfrac {1}{2n+1}-\tfrac {\alpha }{\pi (n+\mu _1)}\Big)r^{2n}, \  \  r\in [0,\tfrac {1}{5}].\end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p> Let us denote the coefficient of \(r^{2n}\) in \(0.22\Big(\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\mu _1}\Big)^2\) by \(d_n, \  n\geq 2.\)<br />In order to prove inequality (<a href="#dfg21">13</a>), we will show that </p>
<div class="equation" id="d1fg21">
<p>
  <div class="equation_content">
    \begin{equation} \label{d1fg21}d_2r^4\geq \tfrac {r^6}{1000}+\Big(\tfrac {1}{3}-\tfrac {\alpha }{\pi (1+\mu _1)}\Big)r^{2}+\Big(\tfrac {1}{5}-\tfrac {\alpha }{\pi (2+\mu _1)}\Big)r^{4}, \  \  r\in [0,\tfrac {1}{5}],\end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p> and </p>
<div class="equation" id="df2g21">
<p>
  <div class="equation_content">
    \begin{equation} \label{df2g21}d_n\geq \tfrac {1}{2n+1}-\tfrac {\alpha }{\pi (n+\mu _1)}, \  \  n\geq 3.\end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p> The inequality (<a href="#d1fg21">14</a>) holds, because </p>
<div class="displaymath" id="a0000000048">
  \begin{align*}  d_2r^4& =0.22\tfrac {1}{16}r^4\geq \Big[\tfrac {1}{25000}+\tfrac {1}{450\cdot 25}+\Big(\tfrac {1}{5}-\tfrac {200}{(4+\pi )(\tfrac {1}{625}+150)}\Big)\Big]r^4\nonumber \\ & \geq \tfrac {r^6}{1000}+\tfrac {1}{450+3r^4}r^6+\Big(\tfrac {1}{5}-\tfrac {200}{(4+\pi )(r^4+150)}\Big)r^4\nonumber \\ & =\tfrac {r^6}{1000}+(\tfrac {1}{3}-\tfrac {\alpha }{\pi (1+\mu _1)}\Big)r^{2}+\Big(\tfrac {1}{5}-\tfrac {\alpha }{\pi (2+\mu _1)}\Big)r^{4}, \  \  r\in [0,\tfrac {1}{5}].\nonumber \end{align*}
</div>
<p> It is sufficient to prove (<a href="#df2g21">15</a>) for \(r=\tfrac {1}{5}.\) We have </p>
<div class="displaymath" id="a0000000049">
  \begin{eqnarray} d_n=\tfrac {0.22}{\pi ^2}\sum _{k=1}^{n-1}\tfrac {1}{(n-k+\mu _1)(k+\mu _1)}=\tfrac {0.44}{\pi ^2}\tfrac {1}{n+2\mu _1}\sum _{k=1}^{n-1}\tfrac {1}{k+\mu _1}. \nonumber \end{eqnarray}
</div>
<p> If \(r=\tfrac {1}{5},\) inequality (<a href="#df2g21">15</a>) is equivalent to </p>
<div class="equation" id="qqq">
<p>
  <div class="equation_content">
    \begin{equation} \label{qqq}t_n=\tfrac {2\cdot 62500(n+\tfrac {1}{2})}{46876\pi (n+\mu _1)}+\tfrac {0.88}{\pi ^2}\tfrac {n+\tfrac {1}{2}}{n+2\mu _1}\sum _{k=1}^{n-1}\tfrac {1}{k+\mu _1}>1, \  n\in {\mathbb {N}^*}, \  n\geq 3.\end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p> We prove now that the sequence \((t_n)_{n\geq 3}\) is strictly increasing. </p>
<div class="displaymath" id="a0000000050">
  \begin{align*} & t_{n+1}-t_n\nonumber \\ & {\gt}\tfrac {2\cdot 62500}{46876\pi }\Big(\tfrac {n+\tfrac {3}{2}}{n+\tfrac {4}{\pi }}-\tfrac {n+\tfrac {1}{2}} {n+\tfrac {4}{\pi }-1}\Big)+\tfrac {0.88}{\pi ^2}\tfrac {n+\tfrac {3}{2}}{n+\tfrac {8}{\pi }-1} \Big(\sum _{k=1}^{n}\tfrac {1}{k+\mu _1}-\sum _{k=1}^{n-1}\tfrac {1}{k+\mu _1}\Big)\nonumber \\ & =\tfrac {0.88}{\pi ^2} \tfrac {n+\tfrac {3}{2}}{(n+\tfrac {8}{\pi }-1)(n+\tfrac {4}{\pi }-1)}-\tfrac {2\cdot 62500}{46876} \tfrac {\tfrac {3}{2\pi }-\tfrac {4}{\pi ^2}}{(n+\tfrac {4}{\pi })(n+\tfrac {4}{\pi }-1)}\nonumber \\ & =\tfrac {1}{n+\tfrac {4}{\pi }-1}\Big(\tfrac {0.88}{\pi ^2}\tfrac {n+\tfrac {3}{2}}{n+\tfrac {8}{\pi }-1}-\tfrac {2\cdot 62500}{46876}\tfrac {\tfrac {3}{2\pi }-\tfrac {4}{\pi ^2}}{n+\tfrac {4}{\pi }}\Big){\gt}0, \  n\geq 3.\nonumber \end{align*}
</div>
<p> Consequently, inequality (<a href="#qqq">13</a>) holds, and the proof of inequality (<a href="#8nm9">10</a>) is done for \(r\in [0,\tfrac {1}{5}].\) </p>
<p>In the second step we will prove that inequality (<a href="#8nm9">10</a>) holds if \(r\in [\tfrac {1}{5},\tfrac {97}{100}].\) Let \(r_k=\tfrac {1}{5}+\tfrac {k}{200000}, \  k=\overline{0,154000}.\) The functions \(1+z(r)\) and \(\big(1+w(r)\big)^{\tfrac {3}{4}+\tfrac {r^4}{200}}\) are strictly increasing on \([\tfrac {1}{5},\tfrac {97}{100}].\) Thus, if the inequalities </p>
<div class="equation" id="951">
<p>
  <div class="equation_content">
    \begin{equation} \label{951}1+z(r_{k-1})\geq \big(1+w(r_k)\big)^{\tfrac {3}{4}+\tfrac {r_k^4}{200}}, \  \  k=\overline{1,154000}\end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p> hold, then the inequality-chains </p>
<div class="displaymath" id="a0000000051">
  \[ 1+z(r)\geq 1+z(r_{k-1})\geq \big(1+w(r_k)\big)^{\tfrac {3}{4}+\tfrac {r_k^4}{200}}\geq \big(1+w(r)\big)^{\tfrac {3}{4}+\tfrac {r^4}{200}}, \  \  r\in [r_{k-1},r_k], \]
</div>
<p> imply (<a href="#8nm9">10</a>) for \(r\in [\tfrac {1}{5},\tfrac {97}{100}].\) The inequalities (<a href="#951">10</a>) can be verified easily using a computer program. </p>
<p>The third case is \(r\in [\tfrac {97}{100},1).\) In this case we will prove the following inequality, which is stronger than (<a href="#8nm9">10</a>): </p>
<div class="displaymath" id="a0000000052">
  \[ 1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}{\gt}\bigg(\tfrac {arth(r)}{r}\bigg)^{\tfrac {151}{200}}, \  \  \  r\in [\tfrac {97}{100},1). \]
</div>
<p> We define the function \(m:[\tfrac {97}{100},1)\rightarrow \mathbb {R}\) by \(m(r)=1+z(r)-\big(1+w(r)\big)^{\tfrac {151}{200}}.\) We have </p>
<div class="displaymath" id="a0000000053">
  \[ m'(r)=w'(r)\Big(\tfrac {z’(r)}{w’(r)}-\tfrac {151}{200}\tfrac {1}{(1+w(r))^{\tfrac {49}{200}}}\Big). \]
</div>
<p> According to Lemma 6 the function \(\tfrac {z’}{w’}:(0,1)\rightarrow \mathbb {R}\) is strictly decreasing, and \(\lim _{r\nearrow 1}\tfrac {z’(r)}{w’(r)}=\tfrac {2}{\pi }.\) Thus </p>
<div class="displaymath" id="a0000000054">
  \[ \tfrac {z’(r)}{w’(r)}{\gt}\tfrac {2}{\pi }{\gt}\tfrac {151}{200}\tfrac {1}{(1+w(\tfrac {97}{100}))^{\tfrac {49}{200}}}\geq \tfrac {151}{200}\tfrac {1}{(1+w(r))^{\tfrac {49}{200}}}, \  r\in [\tfrac {97}{100},1), \]
</div>
<p> and it follows that the mapping \(m\) is strictly increasing. Consequently, the inequality \(m(\tfrac {97}{100}){\gt}0\) implies \(m(r){\gt}0, \  r\in [\tfrac {97}{100},1)\) and the proof is complete. <div class="proof_wrapper" id="a0000000055">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000056">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">10</span>
  </div>
  <div class="remark_thmcontent">
  <p>In order to prove the inequalities (<a href="#951">10</a>) we used the estimations </p>
<div class="displaymath" id="a0000000057">
  \[ 0{\lt}z(r)-\tfrac {1}{\pi }\sum _{n=1}^p\tfrac {r^{2n}}{n+\mu _1}{\lt}\tfrac {r^{2p+2}}{\pi (p+\mu _1+1)(1-r^2)}, \]
</div>
<div class="displaymath" id="a0000000058">
  \[ 0{\lt}w(r)-\sum ^p_{n=1}\tfrac {1}{2n+1}r^{2n}{\lt}\tfrac {r^{2p+2}}{(2p+3)(1-r^2)}, \]
</div>
<p> and applied numerical methods using the Matlab program. </p>

  </div>
</div> </p>
<h1 id="a0000000059">4 Final comments</h1>
<p> Theorem 2 and Corollary 1 imply the inequalities </p>
<div class="displaymath" id="a0000000060">
  \[ \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\tfrac {r^4}{200}}{\lt}\tfrac {\pi }{2}\Big(1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}\Big) {\lt}\mathcal{K}(r){\lt}\tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\tfrac {1}{4}r^2}, \]
</div>
<p> for \(r\in (0,1).\) Since </p>
<div class="displaymath" id="a0000000061">
  \[ \tfrac {\pi }{2}\Big(1+\tfrac {1}{\pi }\sum _{n=1}^\infty \tfrac {r^{2n}}{n+\tfrac {4}{\pi }-1}\Big)=\tfrac {\pi }{2}+\tfrac {1}{2\mu _1}[_2F_1(1,\mu _1,\mu _1+1,r^2)-1] \]
</div>
<p> it follows that the first inequality implies </p>
<div class="displaymath" id="a0000000062">
  \[ \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{3/4}{\lt}\tfrac {\pi }{2}+\tfrac {1}{2\mu _1}[_2F_1(1,\mu _1,\mu _1+1,r^2)-1], \  r\in (0,1), \]
</div>
<p> which was conjectured in <span class="cite">
	[
	<a href="#3" >3</a>
	]
</span>. The second inequality has been established for the first time in <span class="cite">
	[
	<a href="#3" >3</a>
	]
</span>. The third inequality implies a conjecture from <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>. The authors of <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span> proved that the following inequalities hold </p>
<div class="displaymath" id="a0000000063">
  \[ \tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\alpha ^*r}{\lt}\mathcal{K}(r){\lt}\tfrac {\pi }{2}\bigg(\tfrac {{\rm arth}(r)}{r}\bigg)^{\tfrac {3}{4}+\beta ^*r}, \  r\in (0,1), \]
</div>
<p> with the best possible constants \(\alpha ^*=0\) and \(\beta ^*=1/4.\) Our results are improvements of these inequalities. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="1">1</a></dt>
  <dd><p><i class="sc">H. Alzer</i> and <i class="sc">S.-L. Qiu</i>, <em>Monotonicity theorems and inequalities for the complete elliptic integrals</em>, Journal of Comp. Appl. Math., <b class="bfseries">172</b> (2004), pp. 289–312. </p>
</dd>
  <dt><a name="2">2</a></dt>
  <dd><p><i class="sc">G.D. Anderson, M.K. Vamanamurthy</i> and <i class="sc">M. Vourinen</i>, <em>Inequalities for quasiconformal mappings in spaces</em>, Pacific J. Math., <b class="bfseries">160</b> (1993), pp. 1–18. </p>
</dd>
  <dt><a name="3">3</a></dt>
  <dd><p><i class="sc">Sz. András</i> and <i class="sc">Á. Baricz</i>, <em>Bounds for complete elliptic integrals of the first kind</em>, Expo. Math., <b class="bfseries">28</b> (2010) no. 4, pp. 357–364. </p>
</dd>
  <dt><a name="4">4</a></dt>
  <dd><p><i class="sc">Y.-M. Chu, M.-K. Wang</i> and <i class="sc">Y.-F. Qiu</i>, <em>On Alzer and Qiu’s conjecture for complete elliptic integral and inverse hyperbolic tangent function</em>, Abstract and Applied Analysis, article ID. 697547, 2011, pp. 1–7. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><i class="sc">S. Ponnusamy</i> and <i class="sc">M. Vuorinen</i>, <em>Asymptotic expansions and inequalities for hypergeometric functions</em>, Mathematika, <b class="bfseries">44</b> (1997) no. 2, pp. 278–301. </p>
</dd>
</dl>


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