% AFT, epub/pagination, 7/10/00 % DXR PRETEX 3/1/00 %%%APT-TXCX(janet)04/26/00 %%%%%%%%%%%%%%%%%%%%%%%%%%%% ADG FINALS 7.5.00 %%%%%%%%%%% \documentclass[final,oneeqnum]{siamltex} \usepackage{amsfonts} %\oddsidemargin=-.35in %\evensidemargin=-.35in %\topmargin=.25in %\textwidth=7.2in %\textheight=8.3in %\headsep=.3in %\advance\textheight by 1.3in %\advance\topmargin by -.9in %\def\baselinestretch{1.5} %\advance\textwidth by 2.2in %\pagestyle{headings} %\setcounter{secnumdepth}{-2} \title{A Monotonicity Property Involving \protect\boldmath$\,_{3}F_{2}$ and Comparisons of the Classical Approximations of Elliptical Arc Length\thanks{Received by the editors January 15, 1999; accepted for publication (in revised form) December 8, 1999; published electronically July 11, 2000. \URL sima/32-2/35050.html}} \author{Roger W.\ Barnard\thanks{Department of Mathematics, Texas Tech University, Lubbock, TX 79409 (barnard@\break math.ttu.edu, pearce@math.ttu.edu)} \and Kent Pearce\footnotemark[2] \and Kendall C.\ Richards\thanks {Department of Mathematics, Southwestern University, Georgetown, TX 78626 (richards@\break southwestern.edu) }} \begin{document} \maketitle \vspace{-1.35in} \slugger{sima} {2000}{32}{2}{403--419} \vspace{1.05in} \setcounter{page}{403} \begin{abstract} Conditions are determined under which $ \,_{3}F_{2}\left(-n,a,b;a+b+2,\varepsilon -n+1;1\right)$ is a monotone function of $n$ satisfying $a b\cdot\,_{3}F_{2}\left(-n,a,b;a+b+2,\varepsilon -n+1;1\right) \geq a b\cdot\,_{2}F_{1}\left(a,b;a+b+2;1\right).$ Motivated by a conjecture of Vuorinen [{\it Proceedings of Special Functions and Differential Equations}, K. S. Rao, R. Jagannathan, G. Vanden Berghe, J. Van der Jeugt, eds., Allied Publishers, New Delhi, 1998], the corollary that $\,_{3}F_{2}(-n,-\frac{1}{2},-\frac{1}{2};1,\varepsilon -n+1;1) \geq \frac{4}{\pi},$ for $1>\epsilon \geq\frac{1}{4}$ and $n\geq 2,$ is used to determine surprising hierarchical relationships among the 13 known historical approximations of the arc length of an ellipse.\ This complete list of inequalities compares the Maclaurin series coefficients of $\,_{2}F_{1}\,$ with the coefficients of each of the known approximations, for which maximum errors can then be established.\ These approximations range over four centuries from Kepler's in 1609 to Almkvist's in 1985 and include two from Ramanujan. \end{abstract} \begin{keywords} hypergeometric, approximations, elliptical arc length \end{keywords} \begin{AMS} 33C, 41A \end{AMS} \begin{PII} S003614109935050X \end{PII} \pagestyle{myheadings} \thispagestyle{plain} \markboth{ROGER W.\ BARNARD, KENT PEARCE, AND KENDALL C.\ RICHARDS} {A MONOTONICITY PROPERTY INVOLVING $\,_{3}F_{2}$} \section{Introduction} \noindent Let $L(x,y)$ be the arc length of an ellipse with \label{`semiaxes'} semiaxes of length $x$ and $y$ (with $x \geq y > 0$) and let $\displaystyle \lambda \equiv \frac{x-y}{x+y}.$ In 1742, Maclaurin \cite{MC} determined that \begin{eqnarray} L(x,y) = \pi (x+y)\,\cdot\,_{2}F_{1}\left(-\frac{1}{2},-\frac{1}{2};1;\lambda^2\right), \label{AX} \end{eqnarray} where ${\,}_{2}F_{1}$ is the hypergeometric function defined by \[ {\,}_{2}F_{1}\left(a,b;c;z\right)\equiv 1+\sum_{n=1}^{ \infty }\frac{(a)_{n} (b)_{n}z^{n}}{ (c)_{n}n!} \] with the Appell (or Pochhammer) symbol $(a)_{n}\equiv a(a+1)\cdots(a+n-1)$ for $n\geq 1$ and $(a)_{0}\equiv 1,$ $a\neq 0$.\ (For more background information, see \cite{AB}, \cite{R}, \cite{BPR}, and the recent survey article \cite{BARN} by the first author.) In \cite{AB}, Almkvist and Berndt compiled and presented the list of the approximations in Table \ref{Approximations} for \[ G(\lambda) \equiv \,_{2}F_{1}\left(-\frac{1}{2},-\frac{1}{2};1;\lambda^2\right)=\frac{L(x,y)}{ \pi(x+y)}. \] %%%%%%%Table 1.1 \noindent These approximations and their historical and recent connections to the approximations of $\pi$ can be found in the Borweins' book \cite{BB}.\ Another excellent source for historical and current studies of these topics is the book \cite{AVV} by Anderson, Vamanamurthy, and Vuorinen. Recently, several inequalities between various mean values and the hypergeometric function were proved in \cite{BB}, \cite{VV}, and the dependence of the hypergeometric function $\,_{2}F_{1}(a,b;c;z)$ on its parameters was studied in \cite{ABRVV}, \cite{AQV}.\ These results led to a conjecture of Vuorinen (see \cite{MV}) concerning Muir's approximation $A_{5}$. Vuorinen conjectured (see \cite{MV}) that \begin{table} %%%{\scriptsize %%%%%% TOO BIG --ADG {\tiny \caption{\rm{Approximations of $G(\lambda)\equiv\,_{2}F_{1}(-\frac{1}{2},-\frac{1}{2}; 1;\lambda^2)$ (see [2]).}} \label{Approximations} \begin{tabular}{||l|l|c||} \hline \hspace*{.4in }Discoverer(s) &\hspace*{.6in} Approximation & {$\delta_{p} =$ first nonzero term} \\ \hspace*{.1in} and year of discovery& \hspace*{.9in} $A_{p}(\lambda)$ & in the Maclaurin series for \\ \hspace*{.0in} & & $\Delta_{p}(\lambda) \equiv A_{p}(\lambda)-G(\lambda)$ \\\hline\hline &&\\[-.05in] Kepler, 1609 & $\displaystyle A_{1}(\lambda) \equiv (1-\lambda^2)^{1/2}$ & $\displaystyle \delta_{1}=-\frac{3}{4}\lambda^2$\\[-.05in] &&\\\hline &&\\[-.05in] Euler, 1773 &$\displaystyle A_{2}(\lambda) \equiv(1+\lambda^2)^{1/2}$& $\displaystyle \delta_{2}=\frac{1}{4}\lambda^2$\\[-.05in] &&\\\hline &&\\[-.05in] Sipos, 1792&$ \displaystyle A_{3}(\lambda) \equiv\frac{2}{1+\sqrt{1-\lambda^2}}$ &$\displaystyle \delta_{3}=\frac{7}{64}\lambda^4$\\[-.1in] Ekwall, 1973&&\\\hline &&\\[-.05in] Peano, 1889 & $\displaystyle A_{4}(\lambda) \equiv\frac{3}{2}-\frac{1}{2}(1-\lambda^2)^{1/2}$ &$\displaystyle \delta_{4}=\frac{3}{64}\lambda^4$\\[-.05in] &&\\\hline &&\\[-.05in] Muir, 1883& $\displaystyle A_{5}(\lambda) \equiv \left(\frac{(1+\lambda)^{3/2} +(1-\lambda)^{3/2}}{2}\right)^{2/3} $&$\displaystyle\delta_{5}= -\frac{1}{64}\lambda^{4}$\\[-.05in] &&\\\hline &&\\[-.05in] Lindner, 1904-1920 & $\displaystyle A_{6}(\lambda) \equiv \left(1+\frac{\lambda^2}{8}\right)^2 $&$\displaystyle\delta_{6}= -\frac{1}{2^8}\lambda^{6}$\\[-.05in] Nyvoll, 1978&&\\\hline &&\\[-.05in] Selmer, 1975 & $\displaystyle A_{7}(\lambda) \equiv 1+\frac{\lambda^2/4}{1-\lambda^2/16} $&$\displaystyle\delta_{7}= -\frac{3}{2^{10}}\lambda^{6}$\\[-.05in] &&\\\hline &&\\[-.05in] Ramanujan, 1914 & $\displaystyle A_{8}(\lambda) \equiv 3-\sqrt{4-\lambda^2} $&$\displaystyle \delta_{8}=-\frac{1}{2^9}\lambda^{6}$\\[-.05in] Fergestad, 1951 &&\\\hline &&\\[-.05in] Almkvist, 1978 &$\displaystyle A_{9}(\lambda) \equiv 2\frac{\left(1+\sqrt{1-\lambda^2}\right)^2 +\lambda^2\sqrt{1-\lambda^2}}{\left(1+\sqrt{1-\lambda^2}\right)\left(1+\sqrt[4]{1-\lambda^2} \right)^2}$&$\displaystyle \delta_{9}=\frac{15}{2^{14}}\lambda^{8}$\\[-.05in] &&\\\hline &&\\[-.05in] Bronshtein $\mbox{and}$ Semendyayev, 1964 & $\displaystyle A_{10}(\lambda) \equiv\frac{64-3\lambda^4}{64-16\lambda^2} $&$\displaystyle\delta_{10}= -\frac{9}{2^{14}}\lambda^{8}$\\[-.05in] Selmer, 1975&&\\ \hline &&\\[-.05in] Selmer, 1975 & $\displaystyle A_{11}(\lambda) \equiv \frac{3}{2}+\frac{\lambda^2}{8}-\frac{1}{2}\left(1-\frac{\lambda^2}{2}\right)^{1 /2} $&$\displaystyle\delta_{11}= -\frac{5}{2^{14}}\lambda^{8}$\\[-.05in] &&\\\hline &&\\[-.05in] Jacobsen $\mbox{and}$ Waadeland, 1985 & $\displaystyle A_{12}(\lambda) \equiv \frac{256-48\lambda^2 -21\lambda^4}{256-112\lambda^2 +3\lambda^4} $&$\displaystyle\delta_{12}= -\frac{33}{2^{18}}\lambda^{10}$\\[-.05in] &&\\\hline &&\\[-.05in] Ramanujan, 1914 & $\displaystyle A_{13}(\lambda) \equiv 1+\frac{3\lambda^2 }{10+\sqrt{4-3\lambda^2}} $&$\displaystyle \delta_{13}=-\frac{3}{2^{17}}\lambda^{10}$\\[-.05in] &&\\\hline \end{tabular} \\[0.05in] %\label{Approximations} %\caption{\rm{(see [2]) Approximations of } %$G(\lambda)\equiv\,_{2}F_{1}\left(-\frac{1}{2},-\frac{1}{2};1;\lambda^2\right)$. } \end{table} \begin{eqnarray} A_{5}(\lambda) \leq G(\lambda)\quad \mbox{for all}\,\lambda \,\in \,[0,1] \label{eq1}. \end{eqnarray} That is, Vuorinen conjectured that $A_{5}$ is a {\em lower bound} for $G$. This conjecture was recently proved by the authors in \cite{BPR} which has become the genesis of the present article.\ Moreover, the results here attest to the adage that a single conjecture may have many ramifications. Also, note that $A_5$ is one of the mean values studied in \cite{VV}.\ More approximations for hypergeometric functions in terms of such mean values are actively being sought. For example, let $\nu \,\in \,{\bf R}\backslash\{0\}$ and define \[ M_{\nu}(\lambda) \equiv \left[\frac{(1+\lambda)^\nu +(1-\lambda)^\nu}{2}\right ]^{1/\nu}. \] H.\ Alzer \cite{ALZER} originally made the following conjecture. {\sc Conjecture}.\ {\em The inequalities \begin{eqnarray} M_{\alpha}(\lambda) \leq G(\lambda) \leq M_{\beta}(\lambda) \ \ \mbox{hold for all $\lambda \,\in\, (0,1)$} \label{MALZ} \end{eqnarray} if and only if} \[ \alpha \leq 3/2\quad \mbox{{\em and}}\quad \beta \geq (\ln 2)/ \left(\ln \frac{\pi}{2}\right) \approx 1.53. \] As noted by Alzer \cite{ALZER}, it follows from our results (see the set of inequalities in expression (\ref{eqzz})) that (\ref{MALZ}) holds with $\alpha = 3/2$ and $\beta =2$. Moreover, for a fixed $\lambda,$ $M_{\nu}(\lambda)$ is an increasing function of $\nu$.\ Thus it follows that (\ref{MALZ}) holds for all $\alpha \leq 3/2$ and $\beta \geq 2$.\ It can be shown that $\alpha = 3/2$ is sharp.% while the sharpness of %$\beta = (\ln 2)/(\ln \frac{\pi}{2})$ was recently verified by Alzer and Qiu \cite{ALZER}. \section{Main results} \noindent In an earlier paper (see \cite{BPR}), the authors were able to verify inequality (\ref{eq1}) by working with the original version of Vuorinen's conjecture in terms of the eccentricity (see (\ref{eqzx}) and (\ref{eqVz})).\ In this direction, a generating function argument (motivated by \cite{AG}) was used to obtain the following general result (which will also be applied in this paper to obtain Theorem \ref{thm2}). \begin{theorem}[see \cite{BPR}] \label{thm1}%\noindent{\bf Theorem 1.} Suppose $a,b>0$.\ Then for any $\varepsilon $ satisfying $1>\varepsilon \geq\frac{ab}{a+b+1},$ it follows that \[ \,_{3}F_{2}\left(-n,a,b;a+b+1,\varepsilon -n +1;1\right)\geq 0, \] for all integers $n \geq 1,$ where $\,_{3}F_{2}$ is the generalized hypergeometric function.\end{theorem} In light of the conjecture in (\ref{eq1}), the following question naturally arises: \begin{quote} {\em Which of the remaining approximations given in Table {\rm\ref{Approximations}} are upper bounds or lower bounds for $G$}? \end{quote} \noindent An attempt to compare an approximation $A_{p}$ with $G$ motivates an analysis of the term $\delta_{p}$ (the first nonzero term in the Maclaurin series representation for the error function $\Delta_{p}(\lambda)\equiv A_{p}(\lambda)-G(\lambda)$). What information does $\delta_{p}$ provide? Certainly the leading term can be viewed as a measure of accuracy of the given approximation, and the error function $\Delta_{p}(\lambda)$ will have the same sign as $\delta_{p}$ for {\em sufficiently small} $\lambda$.\ For example, $\delta_{1} <0$ and it follows directly that $A_{1}$ is a lower bound for $G,$ as Kepler intended (see \cite[p. 599]{AB}).\ In this case, the sign of $\delta_{1}$ is indicative of the sign of $\Delta_{1}(\lambda)$ {\em for all} $\lambda \, \in \,[0,1]$.\ Almkvist and Berndt proved (see \cite[p. 603]{AB}) that Ramanujan's first estimate $A_{8}$ is a lower bound for $G$ by proving the significantly stronger result that the nonzero Maclaurin series coefficients of $\Delta_{8}$ all have the same (negative) sign.\ A numerical investigation suggests that a similar trait might be shared by other approximations given in Table \ref{Approximations}.\ In this article, it will be shown that all of the approximations given in Table \ref{Approximations} satisfy the following property: \begin{quote} {\em The sign of the error function $\Delta_{p}(\lambda)$ coincides with the sign of the leading term $\delta_{p}$ \underline{for all $\lambda\, \in \,[0,1]$}.} \end{quote} Moreover, for all but two of the approximations, it will be established that the nonzero Maclaurin series coefficients of $\Delta_{p}$ all have the same sign as $\delta_{p}$.\ (Only Euler's approximation and Muir's approximation fail to satisfy this condition.) As a consequence of the forthcoming results, each function $|\Delta_{p}|$ is a strictly increasing function of $\lambda,$ for $p=1,\ldots,13$.\ Therefore, $0=|\Delta_{p}(0)| < |\Delta_{p}(\lambda)|< |\Delta_{p}(1)|$ for all $\lambda \, \in \,(0,1)$. For example, the maximum error for Ramanujan's second estimate is $|\Delta_{13}(1)| = |\frac{14}{11} - \frac{4}{\pi}| \approx 0.000512 $ and satisfies $|\Delta_{13}(1)|< |\Delta_{p}(1)|$ for $p=1,\ldots,12$.\ In this direction, we will prove the following three propositions. %\pagebreak \begin{proposition} \label{prop1}%\noindent{\bf Proposition 1.} \ Let $G(\lambda) \equiv \sum_{n=0}^{\infty}\alpha_{n}\lambda^{2n} $ and $A_{p}(\lambda) \equiv \sum_{n=0}^{\infty}\beta^{(p)}_{n}\lambda^{2n}$ where $\alpha_{n} \equiv (\frac{(-1/2)_{n}}{n!})^2$ and each $A_{p}$ is defined as in Table {\rm\ref{Approximations}}.\ Then \begin{eqnarray} \beta^{(12)}_{n}\leq \alpha_{n} \leq \beta^{(9)}_{n}\quad \mbox{ for all integers $n \geq 0$.} \nonumber \end{eqnarray} Therefore, the error functions $|\Delta_{9}|$ and $|\Delta_{12}|$ are strictly increasing and \end{proposition} \[ A_{12}(\lambda) \leq G(\lambda)\leq A_{9}(\lambda) \quad \mbox{for all $\lambda$} \,\in \,[0,1]. \] \begin{proposition} \label{prop2}%\noindent{\bf Proposition 2.} \ Let $G(\lambda) \equiv \sum_{n=0}^{\infty}\alpha_{n}\lambda^{2n} $ and $A_{p}(\lambda) \equiv \sum_{n=0}^{\infty}\beta^{(p)}_{n}\lambda^{2n}$ where $\alpha_{n} \equiv (\frac{(-1/2)_{n}}{n!})^2$ and each $A_{p}$ is defined as in Table {\rm\ref{Approximations}}.\ Then \begin{eqnarray} \beta^{(1)}_{n}\leq \beta^{(6)}_{n}\leq \beta^{(7)}_{n}\leq \beta^{(8)}_{n}\leq \beta^{(10)}_{n} \leq \beta^{(11)}_{n}\leq \beta^{(13)}_{n}\leq \alpha_{n} \leq\beta^{(4)}_{n}\leq\beta^{(3)}_{n} \nonumber \end{eqnarray} \noindent for all integers $n \geq 0$. Therefore, the corresponding error functions $|\Delta_{p}|$ are strictly increasing and \[ A_{1}(\lambda) \leq A_{6}(\lambda) \leq A_{7}(\lambda) \leq A_{8}(\lambda) \leq A_{10}(\lambda) \leq A_{11}(\lambda) \leq A_{13}(\lambda) \leq G(\lambda)\leq A_{4}(\lambda) \leq A_{3}(\lambda) \] \noindent {for all $\lambda$} \, $\in$ \,$[0,1]$. \end{proposition} The next proposition addresses the two remaining estimates: Euler's approximation $A_{2}$ and Muir's approximation $A_{5}$. The claim will be made that \begin{eqnarray} A_{5}(\lambda) \equiv \left(\frac{(1+\lambda)^{3/2} + (1-\lambda)^{3/2}}{2}\right)^{2/3}\leq G(\lambda)\leq (1+\lambda^2)^{1/2}\equiv A_{2}(\lambda) \label{eqzz} \end{eqnarray} \noindent {for all}\ $\lambda$ \, $\in$ \,[0,1].\ As we have noted, the nonzero Maclaurin series coefficients of $\Delta_{2}$ and $\Delta_{5}$ (as functions of $\lambda$) do not have constant sign.\ In order to verify the inequalities in (\ref{eqzz}), we make use of the known fact due to Landen and Ivory (e.g., see \cite[p. 598]{AB}) that \begin{eqnarray} G(\lambda)\equiv\,_{2}F_{1}\left(-\frac{1}{2},-\frac{1}{2};1;\lambda^2\right) =\frac{2x}{x+y} \cdot\,_{2}F_{1}\left(\frac{1}{2},-\frac{1}{2};1;\xi^2\right), \label{eqzx} \end{eqnarray} where $\lambda \equiv (x-y)/(x+y)$ and $\xi\equiv (1/x){\textstyle\sqrt{x^2-y^2}}$ is the \label{`eccentricity'} eccentricity of the original ellipse (see (\ref{AX})). Without loss of generality, assume that $1=x\geq y\geq 0$.\ A change of variable from $\lambda$ to $\xi$ can be accomplished in (\ref{eqzz}) by using (\ref{eqzx}) and the substitutions $\lambda =(1-y)/(1+y)$ and $y=\textstyle\sqrt{1-\xi^2}$. Multiplying through by $(1+y)/2$ and simplifying, we see that the inequalities in (\ref{eqzz}) are equivalent to \begin{eqnarray} \left( \frac{1+(1-\xi^2)^{3/4}}{2}\right) ^{2/3} \leq \,_{2}F_{1}\left(\frac{1}{2},-\frac{1}{2};1;\xi^{2}\right)\leq (1-\xi^2/2)^{1/2} \label{eqVz} \end{eqnarray} \noindent {for all}\ $\xi$ \, $\in$ \,[0,1]. (The first inequality in (\ref{eqVz}) is the original version of Vuorinen's conjecture \cite{MV}.) It is interesting to note that one can show that the functions in (\ref{eqVz}) can be shown to satisfy the stated inequalities by establishing that the coefficients of their respective Maclaurin series, {\em expanded in powers of $\xi$}, satisfy the corresponding inequality relationships. \noindent In view of the preceding discussion, we now state the following proposition. \vfil \begin{proposition}[see \cite{BPR}] \label{prop3}%\noindent{\bf Proposition 3.} (see \cite{BPR}) \ Let $G$ and $A_{p}$ be as defined in Table {\rm\ref{Approximations}} and let \begin{eqnarray} 1+\sum_{n=1}^{\infty}b_{n}\xi^{2n} &\equiv& \left( \frac{1+(1-\xi^2)^{3/4}}{2}\right) ^{2/3} \quad \mbox{and} \label{eqRR}\\ 1+\sum_{n=1}^{\infty}c_{n}\xi^{2n} &\equiv& (1-\xi^2/2)^{1/2}.\ \label{eqRR3} \end{eqnarray} It follows that \begin{eqnarray} b_{n}\leq \frac{(1/2)_{n}(-1/2)_{n}}{n!\cdot n!} \leq c_{n} \quad \mbox{for all integers $n\geq 1$.} \nonumber \end{eqnarray} Therefore, $(\ref{eqVz})$ holds and is equivalent to $A_{5}(\lambda) \leq G(\lambda)\leq A_{2}(\lambda)\ \mbox{for all $\lambda$} \,\in \,[0,1].\ $\end{proposition} %\vspace{.1in} {\it Remark.} If we apply the identity in (\ref{eqzx}) with $\lambda = (1-\textstyle\sqrt{1-\xi^2})/(1+\textstyle\sqrt{1-\xi^2}),$ the definition of $A_2,$ and simplify, we obtain $\Delta_{2}(\lambda) = 2[(1-\xi^2/2)^{1/2} -\,_{2}F_{1}(\frac{1}{2},-\frac{1}{2};1;\xi^{2})]/(1\break +\textstyle\sqrt{1-\xi^2})$. Proposition \ref{prop3} implies that $(1-\xi^2/2)^{1/2} -\,_{2}F_{1}(\frac{1}{2},-\frac{1}{2};1;\xi^{2})$ is a strictly increasing function of $\displaystyle \xi$.\ Therefore $\Delta_{2}(\lambda)$ is a strictly increasing function of $\displaystyle \xi$. Since $\xi = \frac{2\sqrt{\lambda}}{1+\lambda}$ is a strictly increasing function of $\lambda$ on [0,1], it follows that $|\Delta_{2}|$ is a strictly increasing function of $\lambda$. A similar argument can be applied to $|\Delta_{5}|$. Although some of the inequalities in the above propositions are straightforward, several proved to be surprisingly challenging to verify.\ In particular, the effort involving Almkvist's approximation $A_{9}$ precipitated the discovery of some deeper results involving the generalized hypergeometric function $\,_{3}F_{2}\,,$ which are also of independent interest.\ In this direction, our main general results are as follows. %\vspace{.1in} \begin{theorem} \label{thm2}%\noindent{\bf Theorem 2.} Let $1>a\geq b>-1$ and $1>\varepsilon\geq\,\frac{(a+1)(b+2)}{a+b+4}$.\ Then $\displaystyle T_{n}\equiv\,_{3}F_{2}\left(-n,a,b;a+b+2,\varepsilon -n+1;1\right)$ satisfies \end{theorem} \[ a b (T_{n}- T_{n+1})\geq 0 \quad \mbox{for all integers $n \geq 2$.} \] \begin{corollary} \label{cor1}%\noindent{\bf Corollary 1.} \ Let $1>a\geq b>-1$ and $1>\varepsilon\geq\,\frac{(a+1)(b+2)}{a+b+4}$.\ Then $\displaystyle T_{n}\equiv \\ \,_{3}F_{2}\left(-n,a,b;a+b+2,\varepsilon -n+1;1\right)$ satisfies \end{corollary} \[ a b T_{n} \geq a b T_{n+1} \geq a b\cdot\,_{2}F_{1}\left(a,b;a+b+2;1\right) \quad \mbox{for all integers $n \geq 2$.} \] \begin{corollary} \label{cor2}%\noindent{\bf Corollary 2.} \ Let $1>\varepsilon\geq\frac{1}{4}$. Then $ T_{n}\equiv\,_{3}F_{2}\left(-n,-\frac{1}{2},-\frac{1}{2};1,\varepsilon -n+1;1\right)$ satisfies \end{corollary} \[ T_{n} \geq T_{n+1}\geq \frac{4}{\pi} \quad \mbox{for all integers $n \geq 2$.} \] \section{Verification of coefficient inequalities} \mbox{} {\it Proof of Proposition {\rm\ref{prop1}}}.\ {\em Part {\rm I:} Almkvist's Approximation $A_{9}$.} \ Let $s\equiv (1-\lambda^2)^{1/2}$ and $\beta_{n}\equiv \beta^{(9)}_{n}$.\ It follows that \begin{eqnarray} A_{9}(\lambda) = 2\left[\frac{(1+s) +(1-s)s}{(1+\sqrt{s})^2}\right] = \sum_{n=0}^{\infty}\beta_{n}\lambda^{2n}, \nonumber \end{eqnarray} which implies that \begin{eqnarray} 2(1+2s-s^2) = (1 + 2\sqrt{s} + s)\sum_{n=0}^{\infty}\beta_{n}\lambda^{2n}. \label{eq8} \end{eqnarray} By replacing $s$ by $(1-\lambda^2)^{1/2}$ and applying $ (1-\lambda^2)^{q} = \sum_{n=0}^{\infty}\frac{(-q)_{n}}{n!}\lambda^{2n}, $ we may change (\ref{eq8}) to the form \begin{eqnarray} 2\lambda^2 &+& 4\sum_{n=0}^{\infty}\frac{(-1/2)_{n}}{n!}\lambda^{2n} \nonumber \\ &&=\sum_{n=0}^{\infty}\beta_{n}\lambda^{2n}+ 2\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1/4)_{n-k}}{(n-k)!}\beta_{k}\lambda^{2 n}+ \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1/2)_{n-k}}{(n-k)!}\beta_{k}\lambda ^{2n}.\ \nonumber \end{eqnarray} Equating the coefficients of $\lambda^{2n},$ we obtain $\beta_{0}=1,$ $\beta_{1}=1/4,$ and \begin{eqnarray} 4\frac{(-1/2)_{n}}{n!}= \beta_{n} + 2\sum_{k=0}^{n}\frac{(-1/4)_{n-k}}{(n-k)!}\beta_{k} +\sum_{k=0}^{n}\frac{(-1/2)_{n-k}}{(n-k)!}\beta_{k}\quad \mbox{for $n\geq 2$}.\ \nonumber \end{eqnarray} Solving for $\beta_{n},$ we have the recursive relationship \begin{eqnarray} \beta_{n} = \frac{(-1/2)_{n}}{n!} - \frac{1}{2}\sum_{k=0}^{n-1}\frac{(-1/4)_{n-k}}{(n-k)!}\beta_{k}- \frac{1}{4}\sum_{k=0}^{n-1}\frac{(-1/2)_{n-k}}{(n-k)!}\beta_{k} \quad \mbox{for $n\geq 2$}.\ \label{eq9} \end{eqnarray} \noindent We will use (\ref{eq9}) and induction to show that \begin{eqnarray} \beta_{n} \geq \alpha_{n} \quad \mbox{for all $n\geq 0$}.\ \label{eq10} \end{eqnarray} First note that $\beta_{n} = \alpha_{n}$ for $n=0,1,2$. Now let $n\geq 2$ and suppose that $\beta_{k} \geq \alpha_{k}$ for all $k=0,\ldots, n-1$.\ Since the coefficients of $\beta_{k}$ in (\ref{eq9}) are all positive, it follows that \begin{eqnarray} \beta_{n} \geq \frac{(-1/2)_{n}}{n!} - \frac{1}{2}\sum_{k=0}^{n-1}\frac{(-1/4)_{n-k}}{(n-k)!}\alpha_{k}- \frac{1}{4}\sum_{k=0}^{n-1}\frac{(-1/2)_{n-k}}{(n-k)!}\alpha_{k}.\ \nonumber \end{eqnarray} Thus (\ref{eq10}) will be established if we can verify that \begin{eqnarray} \quad \frac{(-1/2)_{n}}{n!} - \frac{1}{2}\sum_{k=0}^{n-1}\frac{(-1/4)_{n-k}}{(n-k)!}\alpha_{k}- \frac{1}{4}\sum_{k=0}^{n-1} \frac{(-1/2)_{n-k}}{(n-k)!}\alpha_{k}\geq \alpha_{n} \quad \mbox{for $n\geq 2$}.\ \label{eq12} \end{eqnarray} Next we use the identities $(c)_{n-k} = \frac{(-1)^{k}(c)_{n}}{(1-c-n)_{k}}$ and $(1)_{n} = n!$ and add the corresponding $n$th term of each summation to both sides.\ Then (\ref{eq12}) becomes \begin{eqnarray} \quad \frac{(-1/2)_{n}}{n!} - \frac{(-1/4)_{n}}{2\cdot n!}\sum_{k=0}^{n}\frac{(-n)_{k}}{(5/4-n)_{k}}\alpha_{k}- \frac{(-1/2)_{n}}{4\cdot n!}\sum_{k=0}^{n}\frac{(-n)_{k}}{(3/2-n)_{k}}\alpha_{k}\geq \frac{\alpha_{n}}{4}. \label{eq13} \end{eqnarray} Now we apply $\alpha_{k} \equiv \big(\frac{(-1/2)_{k}}{k!}\big)^2$ and the definition of $\,_{3}F_{2},$ then divide both sides of (\ref{eq13}) by $\frac{-(-1/2)_{n}}{4\cdot n!},$ and simplify.\ Then inequality (\ref{eq13}) becomes \begin{eqnarray} P(n)\cdot\,_{3}&F&_{2}\left(-n,-\frac{1}{2},-\frac{1}{2};1,\frac{5}{4} -n;1\right) \nonumber \\&+& \,_{3}F_{2}\left(-n,-\frac{1}{2},-\frac{1}{2};1,\frac{3}{2} -n;1\right) \geq Q(n), \label{eq16} \end{eqnarray} where $P(n) \equiv 2\frac{(-1/4)_{n}}{(-1/2)_{n}}$ and $Q(n) \equiv 4-\frac{(-1/2)_{n}}{n!}$. For $n\geq 2,$ these can be shown to satisfy \begin{eqnarray} P(n) &\leq& P(n+1) \quad \mbox{and} \label{eq16.1}\\ Q(n) &\geq& Q(n+1). \label{eq16.2} \end{eqnarray} We first note that inequality (\ref{eq16}) can be confirmed directly for $n=2,\ldots,6$.\ An application of Corollary \ref{cor2} (to be proved in the following section), with the respective values of $\varepsilon = 1/4$ and $\varepsilon = 1/2,$ yields \begin{eqnarray} \,_{3}F_{2}\left(-n,-\frac{1}{2},-\frac{1}{2};1,\frac{5}{4} -n;1\right) &\geq& \frac{4}{\pi} \quad \mbox{and} \label{eq16.3}\\ \,_{3}F_{2}\left(-n,-\frac{1}{2},-\frac{1}{2};1,\frac{3}{2} -n;1\right) &\geq& \frac{4}{\pi} \label{eq16.4} \end{eqnarray} for all $n\geq 2$.\ From inequalities (\ref{eq16.1})--(\ref{eq16.4}) with $n\geq 6,$ it follows that \begin{eqnarray} P(n)\cdot\,_{3}F_{2}\left(-n,-\frac{1}{2},-\frac{1}{2};1,\frac{5}{4} -n;1\right) &+& \,_{3}F_{2}\left(-n,-\frac{1}{2},-\frac{1}{2};1,\frac{3}{2} -n;1\right) \nonumber \\ &\geq& P(6)\frac{4}{\pi} + \frac{4}{\pi} \geq Q(6) \geq Q(n).\ \nonumber \end{eqnarray} Therefore, inequality (\ref{eq16}) holds for all $n\geq 2$ and hence $\beta^{(9)}_{n}\equiv \beta_{n}\geq \alpha_{n}$ for all $n\geq 0$.\ That is, Almkvist's approximation satisfies the property that all of the nonzero Maclaurin series coefficients of $\Delta_{9}$ are positive.\ This concludes the proof of Part I of Proposition \ref{prop1}. {\it Proof of Proposition} \ref{prop1}. {\em Part {\rm II:} Jacobsen $\mbox{and}$ Waadeland's Approximation $A_{12}$.} \ Now we seek to show that the approximation $A_{12}$ satisfies the property that all of the nonzero Maclaurin series coefficients of $\Delta_{12}$ are negative.\ Let $a=3,$ $b=-112,$ $c=256,$ and $D=\sqrt{b^2-4ac}$.\ It follows that \[ \frac{1}{au^2 + bu + c}=\frac{2a}{D}\left[\frac{1}{2au + b-D} -\frac{1}{2au + b+D}\right] = \sum_{n=0}^{\infty}d_{n}u^n \quad \mbox{for} \quad |u| < \left|\frac{D+b}{2a}\right|, \] where \begin{eqnarray} d_{n}\equiv \frac{2a}{D}\left[\frac{(-1)^n(2a)^n}{(b-D)^{n+1}}-\frac{(-1)^n(2a)^n} {(b+D)^{n+1}}\right]= \frac{1}{D}\left(\frac{2a}{D-b}\right)^{n+1}\left[\left(\frac{b-D}{b+D}\right)^ {n+1}-1\right].\ \nonumber \end{eqnarray} It follows that $d_{n} >0$ for all $n\geq 0$ and \begin{eqnarray} A_{12}(\lambda) &\equiv& \frac{256-48\lambda^2 -21\lambda^4}{256-112\lambda^2 +3\lambda^4}\nonumber\\ &=&-7 +\frac{2048 - 832\lambda^2}{256-112\lambda^2 +3\lambda^4} \nonumber \\ &=& -7 +(2048 - 832\lambda^2) \sum_{n=0}^{\infty}d_{n}\lambda^{2n}.\ \nonumber \end{eqnarray} Now let $\beta_{n} \equiv \beta^{(12)}_{n}$.\ Then the nonzero Maclaurin series coefficients for $A_{12}$ are given by $\beta_{0} = 1$ and \begin{eqnarray} \beta_{n} = 2048d_{n} - 832d_{n-1} \quad \mbox{for all $n\geq 1$.} \nonumber \end{eqnarray} Since $(x^{n+1}-1)/(x-1) > x$ for $x\equiv (b-D)/(b+D)> 1,$ it follows easily that $(2048d_{n})/(832d_{n-1})>1$ for all $n\geq 1$.\ Thus \begin{equation} \beta_{n} >0 \quad \mbox{for all $n\geq 0$.} \label{eq21} \end{equation} Direct calculation reveals that $\beta_{n} = \alpha_{n}$ for $n=0,\ldots,4$. Also note that \begin{eqnarray} (256 - 112\lambda^2 + 3\lambda^4) \sum_{n=0}^{\infty}\beta_{n}\lambda^{2n} = 256 - 48\lambda^2 -21\lambda^4.\ \nonumber \end{eqnarray} Hence \begin{eqnarray} \sum_{n=3}^{\infty}(256\beta_{n} - 112\beta_{n-1} + 3\beta_{n-2})\lambda^{2n} = 0.\label{eq22} \end{eqnarray} Thus the coefficients of $\lambda^{2n}$ in (\ref{eq22}) are zero for all $n\geq 3$.\ Solving for $\beta_{n}$ and using (\ref{eq21}), we have \begin{eqnarray} \beta_{n} = (112\beta_{n-1} - 3\beta_{n-2})/256 < \frac{112}{256}\beta_{n-1} \quad \mbox{for all $n\geq 3$.} \nonumber \end{eqnarray} Now suppose that $\beta_{n} \leq \alpha_{n}$ for some integer $n\geq4,$ where ${\textstyle \alpha_{n} \equiv (\frac{(-1/2)_{n}}{n!})^2}$.\ Then \begin{eqnarray} \beta_{n+1} < \frac{112}{256}\beta_{n}\leq \frac{112}{256}\alpha_{n}= \frac{112}{256}\frac{\alpha_{n}}{\alpha_{n+1}}\alpha_{n+1}= \frac{112}{256}\left(\frac{n+1}{n-\frac{1}{2}}\right)^{2}\alpha_{n+1} \leq\alpha_{n+1}.\ \nonumber \end{eqnarray} Thus $\beta^{(12)}_{n}\equiv \beta_{n}\leq \alpha_{n}$ for all integers $n\geq 0$. This concludes the proof of Part II of Proposition \ref{prop1}.\, \qquad\endproof%\rule{2.4mm}{2.4mm} Before proving Proposition \ref{prop2}, we first observe that the nine approximations involved have the following respective Maclaurin series representations (recursive relationships satisfied by $\beta^{(13)}_{n}$ and $\beta^{(3)}_{n}$ are developed in the appendix): \begin{eqnarray} A_{1}(\lambda) &\equiv&(1-\lambda^2)^{1/2}=1 + \sum_{n=1}^{\infty}\frac{(-1/2)_{n}}{n!}\lambda^{2n}, \label{eqA1}\\ \vspace*{-.2in}\nonumber\\ A_{6}(\lambda) &\equiv&\left(1+\frac{\lambda^2}{8}\right)^2 = 1 + \frac{\lambda^2}{4}+ \frac{\lambda^4}{64},\label{eqA6}\\ \vspace*{-.2in}\nonumber\\ A_{7}(\lambda) &\equiv&1+\frac{\lambda^2/4}{1-\lambda^2/16} = 1 + \frac{\lambda^2}{4} + \sum_{n=2}^{\infty}\frac{1}{2^{4n-2}}\lambda^{2n},\label{eqA7}\\ \vspace*{-.2in}\nonumber\\ A_{8}(\lambda) &\equiv&3-\sqrt{4-\lambda^2}=1 + \frac{\lambda^2}{4} - \sum_{n=2}^{\infty}\frac{(-1/2)_{n}}{n!2^{2n-1}}\lambda^{2n},\label{eqA8}\\ \vspace*{-.2in}\nonumber\\ A_{10}(\lambda) &\equiv&\frac{64-3\lambda^4}{64-16\lambda^2}=1 + \frac{\lambda^2}{4} + \sum_{n=2}^{\infty}\frac{1}{2^{2n+2}}\lambda^{2n},\label{eqA10}\\ \vspace*{-.2in}\nonumber\\ A_{11}(\lambda) &\equiv&\frac{3}{2}+\frac{\lambda^2}{8}-\frac{1}{2}\left(1-\frac{\lambda^2}{2}\right)^{1/2}=1 + \frac{\lambda^2}{4} - \sum_{n=2}^{\infty}\frac{(-1/2)_{n}}{n!2^{n+1}}\lambda^{2n},\label{eqA11}\\ \vspace*{-.2in}\nonumber\\ A_{13}(\lambda) &\equiv&1+\frac{3\lambda^2 }{10+\sqrt{4-3\lambda^2}}=1 + \frac{\lambda^2}{4} + \sum_{n=2}^{\infty}\beta^{(13)}_{n}\lambda^{2n},\label{eqA13}\\ \vspace*{-.2in}\nonumber\\ A_{4}(\lambda) &\equiv&\frac{3}{2}-\frac{1}{2}(1-\lambda^2)^{1/2}=1 + \frac{\lambda^2}{4} - \frac{1}{2}\sum_{n=2}^{\infty}\frac{(-1/2)_{n}}{n!}\lambda^{2n},\label{eqA4}\\ \vspace*{-.2in}\nonumber\\ A_{3}(\lambda) &\equiv&\frac{2}{1+\sqrt{1-\lambda^2}}=1 + \frac{\lambda^2}{4} + \sum_{n=2}^{\infty}\beta^{(3)}_{n}\lambda^{2n}.\label{eqA3} \end{eqnarray} {\it Proof of Proposition {\rm\ref{prop2}}}.\ We seek to establish the following inequalities regarding the specified Maclaurin series coefficients: \begin{eqnarray} \beta^{(1)}_{n}\leq \beta^{(6)}_{n}\leq \beta^{(7)}_{n}\leq \beta^{(8)}_{n}\leq \beta^{(10)}_{n} \leq \beta^{(11)}_{n}\leq \beta^{(13)}_{n}\leq \alpha_{n}\leq\beta^{(4)}_{n}\leq\beta^{(3)}_{n} \label{eqBB1} \end{eqnarray} for all $n\geq 0$. Referring to (\ref{eqA1})--(\ref{eqA3}), we note that the inequalities in (\ref{eqBB1}) are trivial for $n=0$ and $n=1$.\ Thus we must verify (\ref{eqBB1}) for all $n\geq 2$.\ The first two inequalities are immediate while the next three inequalities follow directly by induction.\ We now proceed to prove the remaining inequalities in (\ref{eqBB1}). \noindent$\bullet$ Claim I. {\em $\displaystyle \beta^{(11)}_{n} \leq \beta^{(13)}_{n}\leq \alpha_{n}$ for all $n\geq 2$.} %\vspace{.15in} \noindent Let $\beta_{n}\equiv \beta^{(13)}_{n}$ and $\gamma_{n} \equiv \beta^{(11)}_{n},$ where $\beta^{(11)}_{n} \equiv \frac{-(-1/2)_{n}}{n!2^{n+1}}$ for $n\geq 2$ (see (\ref{eqA11})) and recall that $\alpha_{n}\equiv (\frac{(-1/2)_{n}}{n!})^{2}$. The nonzero Maclaurin series coefficients of Ramanujan's second estimate $A_{13}$ can be shown to satisfy (see the appendix) $\beta_{0}=1,$ $\beta_{1}=1/4,$ $\beta_{2}=1/64,$ and \begin{eqnarray} \beta_{n} = \phi_{n-1} - 2^{-5}\beta_{n-1} \quad \mbox{for all} \ n\geq 3, \quad \mbox{where} \quad \phi_{n} \equiv - \frac{(-1/2)_{n}(3/4)^n}{16\cdot n!}.\label{eqR} \end{eqnarray} Applying (\ref{eqR}) twice, we have \begin{eqnarray} \beta_{n} &=&\phi_{n-1} - 2^{-5}\phi_{n-2} + 2^{-10}\beta_{n-2} \quad \mbox{for all} \ n\geq 4.\ \label{eqR2} \end{eqnarray} Direct calculation reveals that Claim I holds for $n=2,3, 4$.\ That is, $\gamma_{n} \leq \beta_{n} \leq \alpha_{n}$ for $n=2,3, 4$.\ Now let $n\geq 5$ and suppose that \begin{eqnarray} \gamma_{k} \leq \beta_{k}\leq \alpha_{k} \quad \mbox{for all $k=2,\ldots,n-1.$} \label{eqih} \end{eqnarray} Then (\ref{eqR2}) and (\ref{eqih}) together imply that \begin{eqnarray} \phi_{n-1} &-& 2^{-5}\phi_{n-2} + 2^{-10}\gamma_{n-2} \nonumber \\ &\leq& \overbrace{\phi_{n-1} - 2^{-5}\phi_{n-2} + 2^{-10}\beta_{n-2}}^{\beta_{n}} \leq \phi_{n-1} - 2^{-5}\phi_{n-2} + 2^{-10}\alpha_{n-2}.\ \label{eqw3} \end{eqnarray} It can be shown (see the appendix) that \begin{eqnarray} \gamma_{n}&\leq& \phi_{n-1} - 2^{-5}\phi_{n-2} + 2^{-10}\gamma_{n-2} \quad \mbox{and} \label{eqw3.1}\\ \alpha_{n}&\geq& \phi_{n-1} - 2^{-5}\phi_{n-2} + 2^{-10}\alpha_{n-2} \label{eqw3.2} \end{eqnarray} for all $n\geq 5$.\ Therefore, using inequalities (\ref{eqw3})--(\ref{eqw3.2}) and induction, we have $\gamma_{n} \leq \beta_{n} \leq \alpha_{n}$ for all $n\geq 2$. This completes the proof of Claim I. \noindent$\bullet$ Claim II. {\em $\displaystyle \alpha_{n}\leq \beta^{(4)}_{n} \leq \beta^{(3)}_{n}$ for all $n\geq 2$}. \vspace{.15in} \noindent If we now apply (\ref{eqA4}), the first inequality in Claim II becomes \begin{eqnarray} \alpha_{n}\equiv \left(\frac{(-1/2)_{n}}{n!}\right)^{2}\leq \frac{-(-1/2)_{n}}{2\cdot n!}\equiv \beta^{(4)}_{n}\quad \mbox{for all $n\geq 2$.} \nonumber \end{eqnarray} This is equivalent to \begin{eqnarray} \frac{-2(-1/2)_{n}}{n!}\leq 1\quad \mbox{for all $n\geq 2$} \nonumber \end{eqnarray} which follows by induction.\ The second inequality in Claim II involves the Maclaurin series coefficients of Sipos and Ekwall's approximation $A_{3}$ which can be shown to satisfy the following recursive relationship (see the appendix): $\beta^{(3)}_{0}=1,$ $\beta^{(3)}_{1}=1/4,$ $\beta^{(3)}_{2}=1/8,$ and \begin{eqnarray} \beta^{(3)}_{n} = -\frac{1}{2}\sum_{k=0}^{n-1}\frac{(-1/2)_{n-k}}{(n-k)!}\beta^{(3)}_{k} \quad \mbox{for all} \ n\geq 2. \label{eqSE} \end{eqnarray} Note that \begin{eqnarray} -\frac{1}{2}\sum_{k=0}^{n-1}\frac{(-1/2)_{n-k}}{(n-k)!}\beta^{(3)}_{k} = \frac{-(-1/2)_{n}}{2\cdot n!}-\frac{1}{2}\sum_{k=1}^{n-1}\frac{(-1/2)_{n-k}}{(n-k)!}\beta^{(3)}_{k} \label{eqDD} \end{eqnarray} \noindent{for all} \ $n \geq$ 2, and \begin{eqnarray} \frac{-(-1/2)_{n-k}}{2\cdot(n-k)!}\beta^{(3)}_{k} > 0 \quad \mbox{for} \ k=1,\ldots,n-1.\ \label{eqD2} \end{eqnarray} Therefore, (\ref{eqSE})--(\ref{eqD2}) together yield \begin{eqnarray} \beta^{(4)}_{n}\equiv\frac{-(-1/2)_{n}}{2\cdot n!}\leq \beta^{(3)}_{n} \quad \mbox{for all} \ n\geq 2.\ \nonumber \end{eqnarray} \noindent This concludes the proof of Claim II and Proposition \ref{prop2}.\, \qquad\endproof%\rule{2.4mm}{2.4mm} {\it Remarks on the Proof of Proposition {\rm\ref{prop3}}}.\ From (\ref{eqRR3}), we have that $\displaystyle c_{n} \equiv \frac{(1/2)^{n}(-1/2)_{n}}{n!}$ for all $n\geq 1$. By induction, it can be shown that \begin{eqnarray} \frac{(1/2)_{n}(-1/2)_{n}}{n!\cdot n!} \leq c_{n} \quad \mbox{for all $n\geq 1$.} \nonumber \end{eqnarray} In an earlier paper (see \cite{BPR}), the authors use the logarithmic derivative and Cauchy products to obtain the recursive relationship for $b_{n}$ (with $b_n$ as defined in (\ref{eqRR})) given by \begin{equation} b_{n+1}=\frac{1}{2(n+1)}\left[ \left( \frac{5}{4}n-\frac{1}{2}\right) b_{n}-\sum\limits_{k=0}^{n-2}(k+1)b_{k+1}\frac{\left( -\frac{1}{4}\right) _{n-k}}{(n-k)!}\right].\ \label{eq4} \end{equation} Theorem \ref{thm1}, together with (\ref{eq4}), was then used (see \cite{BPR}) to establish that \begin{eqnarray} b_{n} \leq \frac{(1/2)_{n}(-1/2)_{n}}{n!\cdot n!} \quad \mbox{for all } n\geq 1.\ \nonumber \, \qquad\endproof%$\rule{2.4mm}{2.4mm}$} \end{eqnarray} \section{Proofs of general results involving \protect\boldmath$\,_{3}F_{2}$} \noindent We will make use of the following classical identities which we include for the reader's convenience ($F\,\equiv{\,}_{3}F_{2}$). \noindent {\sc Identity 1} \{see \cite[p. 440, eq. (33)]{Pr}\}. \[ \hspace{-15pc} F\left(\rho,a,b;c,\sigma;1\right)-F\left(\rho+1,a,b;c,\sigma+1;1\right) \] \vspace{-2pc} \[ = \frac{-ab(\sigma-\rho)}{c\sigma(\sigma+1)} \cdot F\left(\rho+1,a+1,b+1;c+1,\sigma+2;1\right). \] \noindent {\sc Identity 2} \{see \cite[p. 59, eq. (3.1.1)]{GR}\}. \[ F(-n,a,b;c,d;1)=\frac{(d-b)_{n}}{(d)_{n}}\cdot F(-n,c-a,b;c,1+b-d-n;1). \] \noindent {\sc Identity 3} \{see \cite[p. 440, eq. (26)]{Pr}\}. \[ \sigma\cdot F\left(\rho,a,b;c,\sigma;1\right) = \rho\cdot F\left(\rho+1,a,b;c,\sigma+1;1\right) + (\sigma -\rho)\cdot F\left(\rho,a,b;c,\sigma+1;1\right). \] \noindent {\sc Identity 4} \{see \cite[p. 82, eq.\ (14)]{R}\}. \[ (a_{1}-a_{2})\cdot F(a_{1},a_{2},a_{3};b_{1},b_{2};z)= a_{1}\cdot F(a_{1}+1,a_{2},a_{3};b_{1},b_{2};z)-a_{2}\cdot F(a_{1},a_{2}+1,a_{3};b_{1},b_{2};z). \] \noindent {\sc Identity 5} \{see \cite[p. 440, eq. (30)]{Pr}\}. \[ F\left(\sigma,a,b;c,d;1\right)-F\left(\sigma+1,a,b;c,d;1\right) = \frac{-ab}{cd}\cdot F\left(\sigma+1,a+1,b+1;c+1,d+1;1\right). \] {\it Proof of Theorem {\rm\ref{thm2}}}.\ Define $T_{n}\equiv F\left(-n,a,b;a+b+2,\varepsilon -n +1;1\right),$ where $F\equiv \,_{3}F_{2}$.\ Let $1>a\geq b>-1$ and $1>\varepsilon\geq\,\frac{(a+1)(b+2)}{a+b+4}$.\ For $n\geq2,$ it follows that \begin{eqnarray} T_{n+1} - T_{n} &=& F\left(-n-1,a,b;a+b+2,\varepsilon -n;1\right)-F\left(-n,a,b;a+b+2,\varepsilon -n +1;1\right) \nonumber\\ \ &=& \frac{-a b(\varepsilon+1)}{(\varepsilon-n) (\varepsilon-n+1)(a+b+2)}F\left(-n,a+1,b+1;a+b+3,\varepsilon -n+2;1\right) \nonumber \\ &\ & \quad \{ \mbox{using Identity 1 with $\rho = -n-1,$ $\sigma=\varepsilon -n$}\} \nonumber\\ \ &=& \frac{-a b (\varepsilon+1)}{(n-\varepsilon)(n-\varepsilon-1)(a+b+2)} \frac{(\varepsilon-n-b+1)_{n}}{(\varepsilon-n+2)_{n}} \nonumber \\ &\ & \quad \times F\left(-n,b+2,b+1;a+b+3,b-\varepsilon ;1\right) \nonumber \\ &\ & \quad \{ \mbox{using Identity 2}\}\nonumber\\ \ &=& \frac{-a b (\varepsilon+1)(b-\varepsilon)_{n}}{(n-\varepsilon) (n-\varepsilon-1)(a+b+2)(-1-\varepsilon)_{n}(b-\varepsilon)} \nonumber \\ &\ & \quad \times [(b+1)F\left(-n,b+2,b+2;a+b+3,b+1-\varepsilon;1\right)\nonumber \\ \ &+&(-\varepsilon -1)F\left(-n,b+2,b+1;a+b+3,b+1-\varepsilon;1\right)], \label{eqT4} \end{eqnarray} where (\ref{eqT4}) follows from Identity 3 (with $\rho = b+1,$ $\sigma=b-\varepsilon$) and the identity $(1-\alpha-n)_{n} = (-1)^n(\alpha)_{n}$. \noindent Identity 4 (with $a_{1} = -n$ and $a_{2} = b+1$) implies that \begin{eqnarray} &F&\left(-n,b+2,b+2;a+b+3,b+1-\varepsilon;1\right) \nonumber \\ &\ & \quad \quad =\frac{1}{b+1}[(n+b+1)F\left(-n,b+2,b+1;a+b+3,b+1-\varepsilon;1 \right) \nonumber \\ &\ & \quad \quad + (-n)F\left(-n+1,b+2,b+1;a+b+3,b+1-\varepsilon;1\right)].\ \label{eqT5} \end{eqnarray} Now let $G_{n} = F\left(-n,b+2,b+1;a+b+3,b+1-\varepsilon;1\right)$ and use (\ref{eqT4}) and (\ref{eqT5}).\ Then we have that \begin{eqnarray} T_{n+1} - T_{n} &=& \frac{-a b (\varepsilon+1)(b-\varepsilon)_{n}}{(n-\varepsilon)(n-\varepsilon- 1)(a+b+2)(-1-\varepsilon)_{n}(b-\varepsilon)}\nonumber \\ &\ & \quad \times [(n+b+1)G_{n} - nG_{n-1} +(-\varepsilon-1)G_{n}]\nonumber \\ \ &=& \frac{-a b (\varepsilon+1)(b-\varepsilon)_{n}}{(n-\varepsilon)(n-\varepsilon-1)(a+b+2)(-1 -\varepsilon)_{n}(b-\varepsilon)}\nonumber \\ &\ & \quad \times [n(G_{n} - G_{n-1}) +(b-\varepsilon)G_{n}].\label{eqT6} \end{eqnarray} \noindent Applications of Identity 5 (with $\sigma=-n$) followed by Identity 2 yield \begin{eqnarray} G_{n} - G_{n-1} &=& F\left(-n,b+2,b+1;a+b+3,b+1-\varepsilon;1\right) \nonumber \\ &-& F\left(-n+1,b+2,b+1;a+b+3,b+1-\varepsilon;1\right) \nonumber \\ &=& \frac{-(b+2)(b+1)}{(a+b+3)(b+1-\varepsilon)}F\left(-n+1,b+3,b+2;a+b+4,b+2-\varepsilon;1\right)\nonumber \\ &=& \frac{-(b+2)(b+1)}{(a+b+3)(b+1-\varepsilon)}\cdot\frac{(-\varepsilon)_{n-1}} {(b+2-\varepsilon)_{n-1}}\nonumber \\ &\ & \quad \times F\left(-n+1,a+1,b+2;a+b+4,\varepsilon-n+2;1\right).\ \label{eqT9} \end{eqnarray} Identity 2 also implies that \begin{eqnarray} G_{n} = \frac{(-\varepsilon)_{n}}{(b+1- \varepsilon)_{n}}F\left(-n,a+1,b+1;a+b+3,\varepsilon-n+1;1\right). \label{eqT10} \end{eqnarray} \noindent Combining (\ref{eqT6})--(\ref{eqT10}), we have \begin{eqnarray} T_{n+1} - T_{n} = \frac{-a b (\varepsilon+1)(b-\varepsilon)_{n}}{(n-\varepsilon)(n-\varepsilon-1)(a+b+2)(-1 -\varepsilon)_{n}(b-\varepsilon)}\hspace*{1.5in} \nonumber\\ \times\left[\frac{-n(b+2)(b+1)(-\varepsilon)_{n-1}}{(a+b+3)(b+1-\varepsilon)(b+2 -\varepsilon)_{n-1}} F\left(-n+1,a+1,b+2;a+b+4,\varepsilon-n+2 ;1\right) \right.\nonumber \\ \left. +(b-\varepsilon)\frac{(-\varepsilon)_{n}}{(b+1-\varepsilon)_{n}}F\left(-n,a+1,b+ 1;a+b+3,\varepsilon-n+1;1\right)\right]. \label{eqT11} \end{eqnarray} \noindent Now make use of $\frac{(b-\varepsilon)_{n}}{(b+1-\varepsilon)_{n}}= \frac{(b-\varepsilon)}{(n+ b-\varepsilon)},$ \ $\frac{(-\varepsilon)_{n}}{(-1-\varepsilon)_{n}}= \frac{(-1-\varepsilon+n)}{(-1-\varepsilon)},$ \ $\frac{(-\varepsilon)_{n-1}}{(-1-\varepsilon)_{n}}= \frac{1}{(-1-\varepsilon)},$ \ and multiply both sides by $-ab$.\ Then (\ref{eqT11}) becomes \begin{eqnarray} \hspace*{-1.1in}a b (T_{n} - T_{n+1}) = \frac{(a b)^2(\varepsilon+1)}{(n-\varepsilon)(n-\varepsilon-1)(a+b+2)(b-\varepsilon)} \hspace*{1.5in} \nonumber\\ \times\left[\frac{-n(b+2)(b+1)(b-\varepsilon)}{(a+b+3)(-1-\varepsilon)(n+ b-\varepsilon)} F\left(-n+1,a+1,b+2;a+b+4,\varepsilon-n+2 ;1\right) \right.\nonumber\\ \left.\ +(b-\varepsilon)\frac{(n-1-\varepsilon)(b-\varepsilon)}{(-1-\varepsilon)(n+b- \varepsilon)}F\left(-n,a+1,b+1;a+b+3,\varepsilon-n+1;1\right)\right] \nonumber \\ = \frac{(ab)^2}{(n-\varepsilon)(n-\varepsilon-1)(a+b+2)(n+ b-\varepsilon)}\hspace*{2.0in} \nonumber\\ \times\left[\frac{n(b+2)(b+1)}{(a+b+3)} F\left(-(n-1),a+1,b+2;a+b+4,\varepsilon-(n-1)+1 ;1\right) \right.\nonumber\\ \left.\ \frac{}{}+(\varepsilon-b)(n-\varepsilon-1)F\left(-n,a+1,b+1;a+b+3,\varepsilon-n+ 1;1\right)\right], \label{eqT13} \end{eqnarray} where $n+ b-\varepsilon > n-\varepsilon-1>n-2 \geq 0,$ $n-\varepsilon >0,$ and $\varepsilon-b> \varepsilon-\frac{(a+1)(b+2)}{a+b+4} \geq 0$. Since $1> \varepsilon \geq \frac{(a+1)(b+2)}{a+b+4}>\frac{(a+1)(b+1)}{a+b+3},$ Theorem \ref{thm1} implies that \begin{eqnarray} F\left(-(n-1),a+1,b+2;a+b+4,\varepsilon-(n-1)+1 ;1\right) &\geq& 0 \quad \mbox{and} \nonumber \\ F\left(-n,a+1,b+1;a+b+3,\varepsilon-n+1;1\right) &\geq& 0.\ \nonumber \end{eqnarray} Therefore, (\ref{eqT13}) is the product and sum of nonnegative quantities and thus \begin{eqnarray} a b (T_{n} - T_{n+1}) \geq 0 \quad \mbox{for all integers } n\geq 2.\ \nonumber \, \qquad\endproof %\rule{2.4mm}{2.4mm}} \nonumber \end{eqnarray} \noindent In order to prove Corollary \ref{cor1}, we will make use of the following two lemmas. \begin{lemma}\label{lem1}%\noindent{\bf Lemma 1.} \ Let $n$ be a positive integer and $0 < \varepsilon <1$.\ Then \[ \frac{(-n)_{k}}{(\varepsilon-n+1)_{k}} \geq 1 \quad \mbox{for all $k=0,\dots, n-1$.} \] \end{lemma} {\it Proof of Lemma {\rm\ref{lem1}}}.\ Note that the desired inequality holds at $k=0$.\ Now let $n \geq 2$ and suppose that \[ \frac{(-n)_{k}}{(\varepsilon-n+1)_{k}} \geq 1 \quad \mbox{for some $k$ with $0\le k\leq n-2$.}\] Then \[ \frac{(-n)_{k+1}}{(\varepsilon-n+1)_{k+1}}=\frac{(-n)_{k}(-n+k)}{(\varepsilon- n+1)_{k}(\varepsilon-n+1+k)}\geq \frac{(-n)_{k}}{(\varepsilon-n+1)_{k}} \geq 1. \qquad\endproof\]%\mbox{\, \rule{2.4mm}{2.4mm}}\] %\vspace{.1in} \begin{lemma} \label{lem2}%\noindent{\bf Lemma 2.} \ Define $\psi_{n}(a,b,c,\varepsilon)\equiv \frac{(a)_{n}(b)_{n}(-n)_{n}}{n!(c)_{n}(\varepsilon-n+1)_{n}}$.\ Let $(a,b,c,\varepsilon)$ be in the domain of $\psi_{n}$ for all $n \geq 2$ with $\varepsilon a\geq b>-1$ and $1>\varepsilon\geq\,\frac{(a+1)(b+2)}{a+b+4}$ and define \[ T_{n}\equiv \,_{3}F_{2}\left(-n,a,b;a+b+2,\varepsilon -n +1;1\right). \] Theorem \ref{thm2} implies that the sequence $\{a b T_{n}\}_{n=2}^{\infty}$ is a monotone (nonincreasing) sequence.\ Now define \[S_{n} \equiv 1+ \frac{(a)_{n}(b)_{n}(-n)_{n}}{n!(a+b+2)_{n}(\varepsilon -n +1)_{n}}+ \sum_{k=1}^{n-1}\frac{(a)_{k}(b)_{k}}{k!(a+b+2)_{k}}.\] Using the definition of $\,_{3}F_{2},$ Lemma \ref{lem1}, and the fact that $\displaystyle \frac{a b (a)_{k}(b)_{k}}{k!(a+b+2)_{k}} \geq 0$ for $k=1,\ldots,n-1,$ we obtain \begin{eqnarray} a b T_{n} = a b + \frac{a b (a)_{n}(b)_{n}(-n)_{n}}{n!(a+b+2)_{n}(\varepsilon -n +1)_{n}}+ \sum_{k=1}^{n-1}\frac{a b (a)_{k}(b)_{k}(-n)_{k}}{k!(a+b+2)_{k}(\varepsilon -n +1)_{k}} \geq a b S_{n} \nonumber \end{eqnarray} \noindent {for all $n\geq 2$.} \noindent Applying Lemma \ref{lem2} with $c=a+b+2,$ we have \[ \lim_{n\rightarrow\infty} S_{n} = \,_{2}F_{1}\left(a,b;a+b+2;1\right). \] Since $a b T_{n} \geq a b S_{n}$ for all $n\geq 2,$ it follows that $\{a b T_{n}\}_{n=2}^{\infty}$ is a bounded monotone sequence.\ Thus \[ a b T_{n}\geq \lim_{n\rightarrow\infty} abT_{n} \geq \lim_{n\rightarrow\infty} a b S_{n} = a b\cdot\,_{2}F_{1}\left(a,b;a+b+2;1\right) \quad \mbox{for all } n\geq 2.\, \qquad\endproof\]%\rule{2.4mm}{2.4mm}}\] {\it Proof of Corollary {\rm\ref{cor2}}}.\ Choose $a=b=-1/2$ and $1>\varepsilon \geq 1/4$ and define \[ T_{n}\equiv \,_{3}F_{2}\left(-n,-\frac{1}{2},-\frac{1}{2};1,\varepsilon -n +1;1\right). \] It is known that (see \cite[p. 49]{R}) \[ \,_{2}F_{1}\left(-\frac{1}{2},-\frac{1}{2};1;1\right) = \frac{4}{\pi}. \] Corollary \ref{cor1} implies that \[ T_{n} \geq T_{n+1} \geq \frac{4}{\pi} \quad \mbox{for all } n\geq 2.\,\qquad\endproof\] %\rule{2.4mm}{2.4mm}}\] \section{Appendix} \subsection{Recursive relationship for Maclaurin series coefficients of Ramanujan's second estimate \protect\boldmath$A_{13}$} Writing $\beta_{n}\equiv \beta^{(13)}_{n} ,$ we have \begin{eqnarray} 3\lambda^2(10-\sqrt{4-3\lambda^2})=(A_{13}(\lambda) - 1)(10^2 -(\sqrt{4-3\lambda^2})^2)= (96+3\lambda^2)\sum_{n=1}^{\infty}\beta_{n}\lambda^{2n} \nonumber \end{eqnarray} which implies that \begin{eqnarray} 10-2\left(1-\frac{3}{4}\lambda^2\right)^{1/2}= (32 +\lambda^2)\sum_{n=1}^{\infty}\beta_{n}\lambda^{2n-2}.\ \nonumber \end{eqnarray} Applying $(1-x)^{q} = \sum_{n=0}^{\infty}\frac{(-q)_{n}}{n!}x^{n}$ and simplifying yields \begin{eqnarray} 8 -2\sum_{n=1}^{\infty}\frac{(-1/2)_{n}(3/4)^n}{n!}\lambda^{2n} = 32\beta_{1} + \sum_{n=1}^{\infty}(32\beta_{n+1}+\beta{n})\lambda^{2n}.\ \nonumber \end{eqnarray} Thus $\beta_{0}=1,$ $\beta_{1}=1/4,$ $\beta_{2}=1/64,$ and \begin{eqnarray} \beta_{n+1} = \frac{-(-1/2)_{n}(3/4)^n}{16\cdot n!} -\frac{\beta_{n}}{32} \quad \mbox{for all} \ n\geq 1.\ \nonumber \end{eqnarray} Letting $\displaystyle \phi_{n} \equiv - \frac{(-1/2)_{n}(3/4)^n}{16\cdot n!},$ we obtain \begin{eqnarray} \beta_{n+1} = \phi_{n} - 2^{-5}\beta_{n} \quad \mbox{for all } n\geq 2.\ \nonumber \,\qquad\endproof %$\rule{2.4mm}{2.4mm}$} \nonumber \end{eqnarray} \subsection{Recursive relationship for Maclaurin series coefficients of Sipos $\mbox{and}$ Ekwall's estimate \protect\boldmath$A_{3}$} Writing $\beta_{n}\equiv \beta^{(3)}_{n} $ and using the Cauchy product, we have \begin{eqnarray} 2=A_{3}(\lambda)(1+\sqrt{1-\lambda^2})=\sum_{n=0}^{\infty}\beta_{n}\lambda^{ 2n} + \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1/2)_{n-k}}{(n-k)!}\beta_{k}\lambda ^{2n}.\ \nonumber \end{eqnarray} Thus $\beta^{(3)}_{0}=1, \ \beta^{(3)}_{1}=1/4, \ \beta^{(3)}_{2}=1/8, \quad \mbox{and} $ \begin{eqnarray} \beta^{(3)}_{n} = \frac{-1}{2}\sum_{k=0}^{n-1}\frac{(-1/2)_{n-k}}{(n-k)!}\beta^{(3)}_{k} \quad \mbox{for all } n\geq 2.\, \nonumber \qquad\endproof%$\rule{2.4mm}{2.4mm}$} \nonumber \end{eqnarray} \subsection{Establishing inequality (\ref{eqw3.1})} Let $\phi_{n} \equiv \frac{-(-1/2)_{n}(3/4)^n}{16\cdot n!},$ $\gamma_{n} \equiv \beta^{(11)}_{n} = \frac{-(-1/2)_{n}}{n!2^{n+1}},$ and $n\geq 4$. Inequality (\ref{eqw3.1}) claims that $\displaystyle \gamma_{n}\leq \phi_{n-1} - 2^{-5}\phi_{n-2} + 2^{-10}\gamma_{n-2}$. Direct calculation reveals that the desired inequality holds for $n=4,\dots,7$. Now suppose that $n \geq 7$.\ Since $\gamma_{n-2}>0,$ we have \begin{eqnarray*} \gamma_{n}&-& \phi_{n-1} + 2^{-5}\phi_{n-2} - 2^{-10}\gamma_{n-2} < \gamma_{n}- \phi_{n-1} + 2^{-5}\phi_{n-2} \nonumber \\ &=& \frac{-(-1/2)_{n}}{n!2^{n+1}} +\frac{(-1/2)_{n-1}(3/4)^{n-1}}{16\cdot (n-1)!} -2^{-9}\frac{(-1/2)_{n-2}(3/4)^{n-2}}{ (n-2)!} \nonumber \\ &=& -2^{-9}\frac{(-1/2)_{n-2}(3/4)^{n-2}}{(n-2)!}\cdot\left[ \frac{2^9(4/3)^{n-2}(n-3/2)(n-5/2)}{n(n-1)2^{n+1}} - \frac{2^5(3/4)(n-5/2)}{(n-1)}+1\right] \nonumber \\ &=& -2^{-9}\frac{(-1/2)_{n-2}(3/4)^{n-2}}{(n-2)!}\cdot\left[ \frac{2^{n+4}(n-3/2)(n-5/2)}{3^{n-2}n(n-1)} - \frac{24(n-5/2)}{(n-1)}+1\right]. \end{eqnarray*} Since $\frac{(n-5/2)}{(n-1)} \geq \frac{1}{2},$ it follows that \begin{eqnarray} \frac{2^{n+4}(n-3/2)(n-5/2)}{3^{n-2}n(n-1)} - \frac{24(n-5/2)}{(n-1)}+1 \leq \frac{2^{n+4}(n-3/2)(n-5/2)}{3^{n-2}n(n-1)} - 11.\ \nonumber \end{eqnarray} Thus \begin{eqnarray} \gamma_{n}&-& \phi_{n-1} + 2^{-5}\phi_{n-2} - 2^{-10}\gamma_{n-2}\nonumber \\ &<& -2^{-5}\cdot9\frac{(-1/2)_{n-2}(3/4)^{n-2}}{(n-2)!}\cdot\left[ \frac{2^{n}(n-3/2)(n-5/2)}{3^{n}n(n-1)} - \frac{11\cdot2^{-4}}{3^2}\right] \nonumber \\ &\leq& \frac{-9(-1/2)_{n-2}(3/4)^{n-2}}{32(n-2)!}\cdot\left[ \left(\frac{2}{3}\right)^{n} - \frac{11}{144}\right] \nonumber \\ &\leq& \frac{-9(-1/2)_{n-2}(3/4)^{n-2}}{32(n-2)!}\cdot\left[\left(\frac{2}{3}\right )^{7} - \frac{11}{144}\right] <0.\ \nonumber \end{eqnarray} Hence the claim in (\ref{eqw3.1}) is established.\, \qquad\endproof%$\rule{2.4mm}{2.4mm}$ \subsection{Establishing inequality (\ref{eqw3.2})} Let $\textstyle \phi_{n} \equiv - \frac{(-1/2)_{n}(3/4)^n}{16\cdot n!},$ $\textstyle \alpha_{n} \equiv (\frac{(-1/2)_{n}}{n!})^2,$ and $n\geq 4$. Inequality (\ref{eqw3.2}) claims that $\displaystyle \phi_{n} - 2^{-5}\phi_{n-1} + 2^{-10}\alpha_{n-1} \leq \alpha_{n+1}$.\ Note that \begin{eqnarray*} &\phi&_{n} - 2^{-5}\phi_{n-1} + 2^{-10}\alpha_{n-1} - \alpha_{n+1}\nonumber\\ &=& -\frac{(-1/2)_{n}(3/4)^n}{16\cdot n!}+2^{-5}\frac{(-1/2)_{n-1}(3/4)^{n-1}}{16\cdot (n-1)!} + 2^{-10}\left(\frac{(-1/2)_{n-1}}{(n-1)!}\right)^2 - \left(\frac{(-1/2)_{n+1}}{(n+1)!}\right)^2 \\ &=& \frac{(-1/2)_{n-1}}{(n-1)!}\left\{ \frac{-(n-3/2)3^n}{n2^{2n+4}} + \frac{3^{n-1}}{2^{2n+7}} +\frac{(-1/2)_{n-1}}{2^{10}(n-1)!}- \frac{(n-3/2)(n-1/2)(-1/2)_{n+1}}{n(n+1)\cdot(n+1)!}\right\} \\ &=& \frac{(-1/2)_{n-1}}{(n-1)!}\left\{ \frac{3^{n-1}}{2^{2n+4}}\left[\frac{3(3/2-n)}{n}+ \frac{1}{2^3}\right] + \frac{(-1/2)_{n-1}}{(n-1)!}\left[ \frac{1}{2^{10}}-\frac{(n-3/2)^2(n-1/2)^2}{n^2(n+1)^2}\right] \right\} \\ &=& \frac{(-1/2)_{n-1}}{(n-1)!}\left\{\frac{U(n)}{V(n)} +1\right\}V(n) , \end{eqnarray*} where $$\displaystyle U(n) \equiv \frac{3^{n-1}}{2^{2n+4}}\left[\frac{3(3/2-n)}{n}+ \frac{1}{2^3}\right]$$ and $$ \displaystyle V(n) \equiv \frac{(-1/2)_{n-1}}{(n-1)!}\left[\frac{1}{2^{10}}-\frac{(n-3/2)^2(n-1/2)^2}{ n^2(n+1)^2}\right]. $$ \noindent It follows that $V(n) > 0$.\ Now let $W(n) \equiv U(n)/V(n)$.\ Since $(-1/2)_{n-1} < 0,$ we will be finished if we can show that $W(n)+1 > 0$ for all $n\geq 4$.\ Direct calculation again yields $W(4) +1>0$.\ For $n\geq4,$ it is easy to check that \begin{eqnarray} W(n+1) -W(n) &=& \frac{ \frac{3^{n}}{2^{2n+6}}\left[\frac{3(1/2-n)}{n+1}+ \frac{1}{2^3}\right]}{ \frac{(-1/2)_{n}}{n!}\left[ \frac{1}{2^{10}}-\frac{(n-1/2)^2(n+1/2)^2}{(n+1)^2(n+2)^2}\right]} \nonumber \\ &-&\frac{ \frac{3^{n-1}}{2^{2n+4}}\left[\frac{3(3/2-n)}{n}+ \frac{1}{2^3}\right]}{ \frac{(-1/2)_{n-1}}{(n-1)!}\left[ \frac{1}{2^{10}}-\frac{(n-3/2)^2(n-1/2)^2}{n^2(n+1)^2}\right] } \nonumber \\ &=& \left\{\frac{\frac{3^{n-1}}{2^{2n+4}}}{\frac{(-1/2)_{n-1}}{(n-1) !}}\right\} \left\{\frac{3n}{4(n-3/2)}Z(n+1) - Z(n) \right\}, \label{eq77} \end{eqnarray} where $Z(n) \equiv [\frac{3(3/2-n)}{n}+ \frac{1}{2^3}]/ [\frac{1}{2^{10}}-\frac{(n-3/2)^2(n-1/2)^2}{n^2(n+1)^2}].$ Direct calculation reveals that the expression in (\ref{eq77}) is nonnegative for $n=4$ and $n=5$. 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Van der Jeugt}, eds., Allied Publishers, New Delhi, 1998. \end{thebibliography} %{\small{\sc Department of Mathematics, Texas Tech University, Lubbock, %Texas, 79409} %{\em E-mail address}: {\tt barnard@math.ttu.edu } %\vspace{.1in} %{\sc Department of Mathematics, Texas Tech University, Lubbock, Texas, 79409} %{\em E-mail address}: {\tt pearce@math.ttu.edu } %\vspace{.1in} %{\sc Department of Mathematics, Southwestern University, Georgetown, Texas, %78626} %{\em E-mail address}: {\tt richards@southwestern.edu } %} \end{document}