Open Access

The ratio log-concavity of the Cohen numbers

Journal of Inequalities and Applications20162016:278

https://doi.org/10.1186/s13660-016-1217-7

Received: 27 August 2016

Accepted: 21 October 2016

Published: 8 November 2016

Abstract

Let \(U_{n}\) denote the nth Cohen number. Some combinatorial properties for \(U_{n}\) have been discovered. In this paper, we prove the ratio log-concavity of \(U_{n}\) by establishing the lower and upper bounds for \(\frac{U_{n}}{U_{n-1}}\).

Keywords

the Cohen number log-concavity ratio log-concavity

MSC

05A20 11B83

1 Introduction

An infinite sequence \(\{a_{n}\}_{n=0}^{\infty}\) is said to be log-concave (respectively, log-convex) if for any positive integer n,
$$a_{n}^{2}\geq a_{n+1}a_{n-1}\quad\bigl( \mbox{respectively}, a_{n}^{2}\leq a_{n+1}a_{n-1} \bigr). $$
Furthermore, a positive sequence \(\{a_{n}\}_{n=0}^{\infty}\) is said to be ratio log-concave if the sequence \(\{\frac{a_{n+1}}{a_{n}}\}_{n=0}^{\infty}\) is log-concave. The aim of this paper is to prove the ratio log-concavity of the Cohen numbers. The nth Cohen number was first introduced by Cohen [1] which is defined by
$$\begin{aligned}& U_{n}=h(n)U_{n-1}+g(n)U_{n-2}\quad(n \geq2) \end{aligned}$$
(1.1)
with \(U_{0}=1\) and \(U_{1}=12\), where
$$\begin{aligned}& h(n)=\frac{3(2n-1)(3n^{2}-3n+1)(15n^{2}-15n+4)}{n^{5}} \end{aligned}$$
(1.2)
and
$$\begin{aligned}& g(n)=\frac{3(n-1)^{3}(3n-4)(3n-2)}{n^{5}}. \end{aligned}$$
(1.3)
In [2], Zudilin proved that \(D_{n} U_{n}\) is an integer where \(D_{n}\) is the least common multiple of \(1, 2, \ldots, n\). Moreover, he conjectured some stronger inclusions that were finally proved by Krattenthaler and Rivoal [3]. In particular, they proved that
$$U_{n}=\sum_{i,j} {n\choose i}^{2} {n\choose j}^{2} {n+j\choose n} {n+j-i\choose n} {2n-i \choose i}, $$
where the binomial coefficients \({a\choose b}\) are zero if \(b < 0\) or \(a < b\); see also [4].
Recently, the combinatorial properties of \(U_{n}\) were considered. Employing a criterion due to Xia and Yao [5], it is easy to prove the log-convexity of \(U_{n}\). Chen and Xia [6] proved the 2-log-convexity of \(U_{n}\), that is,
$$\bigl(U_{n-1}U_{n+1}-U_{n}^{2}\bigr) \bigl(U_{n+1}U_{n+3}-U_{n+2}^{2}\bigr) > \bigl(U_{n}U_{n+2}-U_{n+1}^{2} \bigr)^{2}. $$
In this paper, we prove the ratio log-concavity of \(U_{n}\). The main results of the paper can be stated as follows.

Theorem 1.1

The sequence \(\{U_{n}\}_{n=0}^{\infty}\) is ratio log-concave, namely, for \(n\geq2\),
$$\begin{aligned}& \frac{U_{n}^{2}}{U_{n-1}^{2}}>\frac{U_{n+1}}{U_{n}}\frac{U_{n-1}}{U_{n-2}}. \end{aligned}$$
(1.4)

2 Lower and upper bounds for \(\frac{U_{n}}{U_{n-1}}\)

In order to prove Theorem 1.1, we first establish the lower and upper bounds for \(\frac{U_{n}}{U_{n-1}}\).

Lemma 2.1

For \(n\geq5\),
$$\begin{aligned}& l(n)< \frac{U_{n}}{U_{n-1}}, \end{aligned}$$
(2.1)
where
$$\begin{aligned}& l(n) = 135+78\sqrt{3}-\frac{675+390\sqrt{3}}{2n} +\frac{9{,}737\sqrt{3}+16{,}848}{48n^{2}} - \frac{3{,}497\sqrt{3}+6{,}045}{32n^{3}}. \end{aligned}$$
(2.2)

Proof

We are ready to prove Lemma 2.1 by induction on n. It is easy to check that (2.1) is true when \(n = 5\) and \(n=6\). Suppose that Lemma 2.1 holds when \(n = m \geq5\), that is,
$$\begin{aligned}& l(m)< \frac{U_{m}}{U_{m-1}}. \end{aligned}$$
(2.3)
In order to prove Lemma 2.1, it suffices to prove that this lemma holds when \(n = m + 2\), that is,
$$\begin{aligned}& l(m+2)< \frac{U_{m+2} }{U_{m+1}}. \end{aligned}$$
(2.4)
Based on (1.1) and (2.3),
$$\begin{aligned} \frac{U_{m+2}}{U_{m+1}} =& h(m+2) +g(m+2)\frac{1}{\frac{U_{m+1}}{U_{m}}} = h(m+2) +g(m+2) \frac{1}{h(m+1)+g(m+1)\frac{U_{m-1}}{U_{m}}} \\ > & h(m+2) +g(m+2) \frac{1}{h(m+1)+\frac{g(m+1)}{l(m)}}, \end{aligned}$$
(2.5)
where \(h(n)\), \(g(n)\), and \(l(n)\) are defined by (1.2), (1.3), and (2.2), respectively. Thanks to (2.5),
$$\begin{aligned}& \frac{U_{m+2}}{U_{m+1}}-l(m+2) \\& \quad >h(m+2) +g(m+2) \frac{1}{h(m+1)+\frac{g(m+1)}{l(m)}}-l(m+2) \\& \quad = \frac{13(542{,}921-313{,}428\sqrt{3})\alpha(m)}{830{,}059{,}024(m+2)^{5}\beta(m)}, \end{aligned}$$
(2.6)
where \(\alpha(m)\) and \(\beta(m)\) are defined by
$$\begin{aligned} \alpha(m) =& 121{,}396{,}132{,}260m^{9}- 257{,}880{,}671{,}236 \sqrt{3}m^{8}+675{,}703{,}429{,}830m^{8} \\ &{}+2{,}176{,}536{,}150{,}666m^{7} - 1{,}330{,}058{,}753{,}240 \sqrt{3}m^{7} +4{,}927{,}389{,}297{,}804m^{6} \\ &{}-2{,}983{,}584{,}697{,}467\sqrt{3} m^{6}-4{,}066{,}074{,}230{,}366 \sqrt{3}m^{5} \\ &{}+ 7{,}060{,}751{,}181{,}826m^{5} -3{,}674{,}684{,}488{,}924\sqrt{3}m^{4} \\ &{}+6{,}286{,}428{,}416{,}954m^{4} +3{,}481{,}214{,}050{,}452m^{3} \\ &{}-2{,}206{,}793{,}212{,}277\sqrt{3} m^{3} +1{,}169{,}733{,}232{,}808m^{2} -845{,}639{,}850{,}544\sqrt{3} m^{2} \\ &{}-187{,}272{,}537{,}764\sqrt{3}m+218{,}478{,}614{,}224m \\ &{}-18{,}263{,}322{,}480 \sqrt{3}+17{,}360{,}076{,}864 \end{aligned}$$
and
$$\begin{aligned} \beta(m) =& 864m^{8}-96m^{6}-468\sqrt{3}m^{6}+468 \sqrt{3}m^{5} -810m^{5}-2{,}206{,}269\sqrt{3}m^{4} \\ &{}+3{,}821{,}499m^{4}+6{,}665{,}166m^{3}-3{,}848{,}598 \sqrt{3} m^{3} -2{,}680{,}756\sqrt{3}m^{2} \\ &{}+4{,}642{,}290m^{2}-875{,}459\sqrt{3}m+1{,}515{,}969m+194{,}220-112{,}164\sqrt{3}. \end{aligned}$$
By (2.6) and the fact that \(\alpha(m)\beta(m)> 0\) for \(m\geq5\), we obtain (2.4). This completes the proof of Lemma 2.1 by induction. □

Lemma 2.2

For \(n\geq5\),
$$\begin{aligned}& \frac{U_{n}}{U_{n-1}}< u(n), \end{aligned}$$
(2.7)
where
$$\begin{aligned}& u(n)=135+78\sqrt{3}-\frac{675+390\sqrt{3}}{2n} +\frac{9{,}737\sqrt{3}+16{,}848}{48n^{2}} - \frac{6{,}994\sqrt{3}+6{,}045}{64n^{3}}. \end{aligned}$$
(2.8)

Proof

We also prove Lemma 2.2 by induction on n. It is easy to verify that (2.7) holds for \(n = 5\) and \(n=6\). Assume that Lemma 2.2 is true for \(n = m \geq5\), that is,
$$\begin{aligned}& \frac{U_{m}}{U_{m-1}}< u(m), \end{aligned}$$
(2.9)
where \(u(m)\) is defined by (2.8). In order to prove Lemma 2.2, it suffices to prove that Lemma 2.2 is true when \(n = m + 2\), namely,
$$\begin{aligned}& \frac{U_{m+2} }{U_{m+1}}< u(m+2). \end{aligned}$$
(2.10)
Based on (1.1) and (2.9),
$$\begin{aligned} \frac{U_{m+2}}{U_{m+1}} =& h(m+2) +g(m+2)\frac{1}{\frac{U_{m+1}}{U_{m}}} \\ =& h(m+2) +g(m+2) \frac{1}{h(m+1)+g(m+1)\frac{U_{m-1}}{U_{m}}} \\ < & h(m+2)+g(m+2) \frac{1}{h(m+1)+\frac{g(m+1)}{u(m)}}, \end{aligned}$$
(2.11)
where \(h(n)\), \(g(n)\), and \(u(n)\) are defined by (1.2), (1.3), and (2.8), respectively. Thanks to (2.11),
$$\begin{aligned}& \frac{U_{m+2}}{U_{m+1}}-u(m+2) \\& \quad < h(m+2) +g(m+2) \frac{1}{h(m+1)+\frac{g(m+1)}{u(m)}}-u(m+2) \\& \quad = \frac{13(2{,}340- 1{,}351\sqrt{3})\varphi(m) }{192(m+2)^{5}\psi(m)}< 0, \end{aligned}$$
(2.12)
where \(\varphi(m)\) and \(\psi(m)\) are defined by
$$\begin{aligned} \varphi(m) =& 3{,}760{,}473{,}600m^{10}+24{,}236{,}858{,}880m^{9}- 7{,}725{,}471{,}840\sqrt{3}m^{9} \\ &{}+84{,}297{,}090{,}576m^{8}-46{,}822{,}426{,}128\sqrt{3} m^{8}-123{,}822{,}624{,}402 \sqrt{3}m^{7} \\ &{}+204{,}386{,}453{,}088m^{7}-194{,}450{,}024{,}349\sqrt{3}m^{6} +336{,}953{,}792{,}124m^{6} \\ &{}-203{,}475{,}797{,}950\sqrt{3}m^{5}+365{,}977{,}131{,}864m^{5}+260{,}362{,}891{,}056m^{4} \\ &{}-147{,}457{,}384{,}610\sqrt{3}m^{4}-73{,}568{,}487{,}135\sqrt{3}m^{3} +119{,}994{,}653{,}508m^{3}\\ &{}-24{,}169{,}674{,}728 \sqrt{3}m^{2}+34{,}436{,}526{,}528m^{2} +5{,}563{,}246{,}416m \\ &{}-4{,}710{,}672{,}460\sqrt{3}m+382{,}180{,}032-412{,}780{,}368 \sqrt{3} \end{aligned}$$
and
$$\begin{aligned} \psi(m) =& 1{,}728m^{8}-936\sqrt{3}m^{6}-192m^{6}+2{,}205{,}033{,}030m^{5} -1{,}273{,}076{,}064\sqrt{3}m^{5} \\ &{}+5{,}520{,}229{,}623m^{4}-3{,}187{,}105{,}038\sqrt{3}m^{4} -3{,}317{,}697{,}396\sqrt{3}m^{3} \\ &{}+5{,}746{,}420{,}422m^{3}-1{,}787{,}669{,}312\sqrt{3}m^{2}+3{,}096{,}333{,}090m^{2} \\ &{}+860{,}545{,}413m-496{,}836{,}418\sqrt{3}m-56{,}805{,}528\sqrt{3}+98{,}389{,}980. \end{aligned}$$
By (2.12) and the fact that \(\varphi(m)\psi(m)> 0\) for \(m\geq5\), we arrive at (2.10). This completes the proof of Lemma 2.2 by induction. □

3 Proof of Theorem 1.1

In this section, we present a proof of Theorem 1.1.

Lemma 3.1

For \(n\geq5\),
$$\begin{aligned}& \frac{U_{n+1}U_{n-1}}{U_{n}^{2}}< f(n), \end{aligned}$$
(3.1)
where
$$\begin{aligned}& f(n)= \frac{(144n^{2}-216n+298-39\sqrt{3})(2n-1)^{2}}{ (144n^{2}-504n+658-39\sqrt{3})(2n+1)^{2}}. \end{aligned}$$
(3.2)

Proof

Let \(h(n)\) and \(g(n)\) be defined by (1.2) and (1.3), respectively. It is easy to verify that, for \(n\geq5\),
$$\begin{aligned}& h^{2}(n+1)+4f(n)g(n+1)=\frac{3a(n)}{ (144n^{2}-504n+658- 39\sqrt{3})(n+1)^{10}(2n+1)^{2}}>0, \end{aligned}$$
(3.3)
where
$$\begin{aligned} a(n) =& 14{,}017{,}536n^{14} +35{,}043{,}840n^{13}-3{,}796{,}416 \sqrt{3}n^{12} +3{,}796{,}416n^{12} \\ &{}-22{,}767{,}264\sqrt{3}n^{11} -35{,}430{,}048n^{11}-63{,}335{,}220 \sqrt{3}n^{10}+136{,}740{,}296n^{10} \\ &{}-108{,}042{,}324\sqrt{3}n^{9}+585{,}572{,}912n^{9}+1{,}001{,}472{,}846n^{8} -125{,}838{,}297\sqrt{3}n^{8} \\ &{}-105{,}379{,}248\sqrt{3}n^{7}+1{,}047{,}661{,}216n^{7} -65{,}023{,}608\sqrt{3}n^{6}+749{,}372{,}512n^{6} \\ &{}-29{,}770{,}650\sqrt{3}n^{5}+381{,}561{,}324n^{5} +139{,}393{,}232n^{4}-10{,}033{,}296\sqrt{3}n^{4} \\ &{}+35{,}929{,}568n^{3}-2{,}426{,}892\sqrt{3}n^{3} -399{,}789\sqrt{3}n^{2}+6{,}231{,}942n^{2} \\ &{}+654{,}864n -40{,}248\sqrt{3}n+31{,}584-1{,}872\sqrt{3}. \end{aligned}$$
Moreover, it is easy to check that, for \(n\geq0\),
$$\begin{aligned}& 2f(n)l(n)-h(n+1) \\& \quad =\frac{\sqrt{3}b(n)}{48n^{3}( 144 n^{2}-504n+658+39 \sqrt{3} ) (2n+1)^{2}(n+1)^{5}}>0 \end{aligned}$$
(3.4)
and
$$\begin{aligned}& \bigl(2f(n)l(n)-h(n+1) \bigr)^{2} - \bigl(h^{2}(n+1)+4f(n)g(n+1) \bigr) \\& \quad =\frac{(1{,}351+780 \sqrt{3})(2n-1)^{2}(144n^{2}-216n+298-39 \sqrt{3})c(n)}{ 384 n^{6} ( 144 n^{2}-504n+658+39 \sqrt{3} ) (2n+1)^{4}(n+1)^{5}}>0, \end{aligned}$$
(3.5)
where
$$\begin{aligned} b(n) =& 4{,}313{,}088n^{12}-10{,}048{,}896n^{10} -1{,}168{,}128\sqrt{3}n^{10}+2{,}134{,}080\sqrt{3}n^{9} \\ &{}+18{,}944{,}640n^{9}+25{,}025{,}104n^{8}-18{,}411{,}096\sqrt{3}n^{8}-23{,}385{,}908n^{7} \\ &{}-52{,}175{,}844\sqrt{3}n^{7}-856{,}596n^{6}-24{,}604{,}446\sqrt{3} n^{6}+30{,}202{,}523n^{5} \\ &{}+8{,}838{,}489\sqrt{3}n^{5}-10{,}242{,}333\sqrt{3} n^{4}-14{,}907{,}529n^{4}-17{,}776{,}278n^{3} \\ &{}-10{,}403{,}322\sqrt{3} n^{3}+6{,}462{,}024\sqrt{3}n^{2}+11{,}237{,}434n^{2}+2{,}501{,}109\sqrt{3} n\\ &{}+4{,}336{,}111n-2{,}419{,}053-1{,}392{,}261\sqrt{3} \end{aligned}$$
and
$$\begin{aligned} c(n) =&59{,}719{,}680n^{12} -106{,}074{,}695{,}808n^{11}+61{,}088{,}601{,}600 \sqrt{3}n^{11} \\ &{}-46{,}198{,}518{,}912\sqrt{3}n^{10}+80{,}016{,}457{,}536n^{10}+143{,}774{,}162{,}864n^{9} \\ &{}-82{,}704{,}126{,}024 \sqrt{3} n^{9}+283{,}349{,}090{,}856\sqrt{3}n^{8}-491{,}163{,}569{,}992n^{8} \\ &{}-1{,}030{,}144{,}421{,}232n^{7}+594{,}612{,}735{,}990 \sqrt{3} n^{7}+478{,}140{,}044{,}616\sqrt{3}n^{6} \\ &{}-827{,}484{,}760{,}322n^{6}+199{,}910{,}945{,}130\sqrt{3} n^{5}-346{,}327{,}459{,}907n^{5}\\ &{}-74{,}608{,}485{,}009n^{4}+42{,}928{,}060{,}188\sqrt{3}n^{4}-6{,}144{,}385{,}390n^{3} \\ &{}+3{,}678{,}965{,}082\sqrt{3} n^{3}-499{,}917{,}028n^{2}+293{,}762{,}352\sqrt{3}n^{2}-191{,}636{,}367n\\ &{}+71{,}139{,}198 \sqrt{3}n-55{,}991{,}052\sqrt{3}+115{,}716{,}159. \end{aligned}$$
It follows from (3.3)-(3.5) that, for \(n\geq0\),
$$2f(n)l(n)-h(n+1)>\sqrt{h^{2}(n+1)+4f(n)g(n+1)} $$
and thus
$$\begin{aligned}& l(n)> \frac{h(n+1)+\sqrt{h^{2}(n+1)+4f(n)g(n+1)}}{2f(n)}. \end{aligned}$$
(3.6)
In view of (2.1) and (3.6),
$$\begin{aligned}& \frac{U_{n}}{U_{n-1}} >\frac{h(n+1)+\sqrt{h^{2}(n+1)+4f(n)g(n+1)}}{2f(n)}, \end{aligned}$$
(3.7)
which implies that, for \(n\geq5\),
$$\begin{aligned}& f(n) \biggl(\frac{U_{n}}{U_{n-1}} \biggr)^{2}- h(n+1) \frac{U_{n}}{U_{n-1}}-g(n+1)>0. \end{aligned}$$
(3.8)
Thanks to (1.1),
$$\begin{aligned}& f(n)U_{n}^{2}-U_{n-1}U_{n+1}=U_{n}^{2} \biggl(f(n) \biggl(\frac{ U_{n}}{U_{n-1}} \biggr)^{2}- h(n+1) \frac{U_{n}}{ U_{n-1}}-g(n+1) \biggr). \end{aligned}$$
(3.9)
Lemma 3.1 follows from (3.8) and (3.9). This completes the proof. □

Lemma 3.2

For \(n\geq5\),
$$\begin{aligned}& \frac{U_{n+1}U_{n-1}}{U_{n}^{2}}>f(n+1), \end{aligned}$$
(3.10)
where \(f(n)\) is defined by (3.2).

Proof

It is easy to check that, for \(n\geq5\),
$$\begin{aligned}& h^{2}(n+1)+4f(n+1)g(n+1) \\& \quad =\frac{3(2n+1)^{2} d(n)}{ (n+1)^{10} (144n^{2}-216n+298-39\sqrt{3})(2n+3)^{2}}>0 \end{aligned}$$
(3.11)
and
$$\begin{aligned}& 2f(n+1) l(n) -h(n+1) \\& \quad =\frac{\sqrt{3}(2n+1)e(n)}{48n^{3}(144n^{2}-216n+298-39\sqrt {3})(2n+3)^{2}(n+1)^{5}}>0, \end{aligned}$$
(3.12)
where
$$\begin{aligned} d(n) =& 3{,}504{,}384n^{12}+19{,}274{,}112n^{11} +45{,}630{,}000n^{10}- 949{,}104\sqrt{3}n^{10} \\ &{}-6{,}640{,}920 \sqrt{3}n^{9}+71{,}887{,}752n^{9}+109{,}395{,}878n^{8} -20{,}342{,}049\sqrt{3}n^{8} \\ &{}+166{,}770{,}736n^{7}-36{,}215{,}400 \sqrt{3}n^{7}-41{,}834{,}832\sqrt{3} n^{6}+203{,}641{,}520n^{6} \\ &{}-32{,}969{,}274\sqrt{3}n^{5}+177{,}081{,}116n^{5}+106{,}446{,}904n^{4}-18{,}035{,}004\sqrt{3}n^{4} \\ &{}-6{,}785{,}376\sqrt{3}n^{3}+43{,}438{,}424n^{3}+11{,}552{,}574n^{2}-1{,}684{,}917\sqrt{3}n^{2} \\ &{}-249{,}912\sqrt{3}n+1{,}816{,}272n+128{,}736-16{,}848 \sqrt{3}, \\ e(n) =&2{,}156{,}544n^{11}+7{,}547{,}904n^{10}+9{,}532{,}224n^{9} -584{,}064\sqrt{3}n^{9} \\ &{}-6{,}029{,}856\sqrt{3}n^{8}+6{,}413{,}472n^{8}-17{,}305{,}116 \sqrt{3}n^{7} +3{,}738{,}488n^{7} \\ &{}-849{,}238n^{6}-23{,}508{,}972\sqrt{3}n^{6}-7{,}884{,}539n^{5}-19{,}078{,}677 \sqrt{3}n^{5} \\ &{}-15{,}177{,}045n^{4}-13{,}428{,}969\sqrt{3}n^{4}-21{,}962{,}902n^{3}-13{,}234{,}830 \sqrt{3}n^{3} \\ &{}-11{,}506{,}974\sqrt{3}n^{2}-20{,}028{,}320n^{2}-5{,}355{,}441\sqrt{3}n -9{,}314{,}279n\\ &{}-1{,}663{,}701-957{,}021 \sqrt{3}. \end{aligned}$$
By (3.11) and (3.12),
$$-\sqrt{h^{2}(n+1)+4f(n+1)g(n+1)}< 2f(n+1)l(n)-h(n+1) $$
and thus
$$\begin{aligned}& \frac{h(n+1)-\sqrt{h^{2}(n+1)+4f(n+1)g(n+1)}}{2f(n+1)}< l(n). \end{aligned}$$
(3.13)
Furthermore, it is easy to check that, for \(n\geq0\),
$$\begin{aligned}& 2f(n+1)u(n) -h(n+1) \\& \quad = \frac{\sqrt{3}(2n+1)r(n) }{96n^{3}(144n^{2}-216n+298-39\sqrt{3})(2n+3)^{2}(n+1)^{5}}>0 \end{aligned}$$
(3.14)
and
$$\begin{aligned}& \bigl(h^{2}(n+1)+4f(n+1)g(n+1) \bigr)- \bigl(2f(n+1)u(n) -h(n+1) \bigr)^{2} \\& \quad=\frac{(32{,}424+13{,}481\sqrt{3})(144n^{2}+72n+226- 39\sqrt{3})(2n+1)^{2}s(n)}{1{,}554{,}750{,}544{,}896n^{6}(2n + 3)^{4}(n + 1)^{5}(144n^{2}-216n+298-39\sqrt{3})^{2}}>0, \end{aligned}$$
(3.15)
where
$$\begin{aligned} r(n) = & 4{,}313{,}088n^{11}+15{,}095{,}808n^{10} +19{,}064{,}448n^{9}-1{,}168{,}128\sqrt{3}n^{9} \\ &{}-10{,}318{,}752\sqrt{3}n^{8}+12{,}826{,}944n^{8} -24{,}164{,}472\sqrt{3}n^{7}+7{,}476{,}976n^{7} \\ &{}-3{,}113{,}006n^{6}-17{,}735{,}964\sqrt{3}n^{6} -23{,}548{,}993n^{5}+13{,}865{,}916\sqrt{3}n^{5} \\ &{}-48{,}035{,}715n^{4}+37{,}763{,}112\sqrt{3}n^{4} -65{,}143{,}754n^{3}+29{,}313{,}600\sqrt{3}n^{3} \\ &{}+8{,}226{,}612\sqrt{3}n^{2}-54{,}201{,}940n^{2}-712{,}452\sqrt{3}n-23{,}579{,}413n\\ &{}-4{,}034{,}667-547{,}872 \sqrt{3}, \\ s(n) =& 10{,}074{,}783{,}530{,}926{,}080n^{12}-7{,}734{,}966{,}108{,}060{,}672 \sqrt{3}n^{11} \\ &{}+26{,}936{,}680{,}137{,}670{,}656 n^{11}+26{,}619{,}191{,}006{,}206{,}464n^{10}\\ &{}-23{,}315{,}086{,}010{,}211{,}840 \sqrt{3}n^{10} -33{,}744{,}534{,}523{,}380{,}928\sqrt{3}n^{9} \\ &{}+41{,}557{,}278{,}922{,}739{,}904n^{9}-43{,}703{,}007{,}806{,}729{,}856 \sqrt{3}n^{8} \\ &{}+98{,}524{,}492{,}003{,}096{,}704n^{8}+111{,}734{,}337{,}079{,}096{,}644n^{7}\\ &{}-47{,}522{,}128{,}660{,}057{,}480 \sqrt{3}n^{7} -24{,}419{,}890{,}979{,}639{,}584\sqrt{3}n^{6} \\ &{}+24{,}142{,}153{,}127{,}323{,}080n^{6}-69{,}518{,}699{,}851{,}789{,}311n^{5} \\ &{}+24{,}355{,}619{,}096{,}164{,}226\sqrt{3}n^{5}-90{,}260{,}471{,}288{,}625{,}639n^{4}\\ &{}+57{,}503{,}789{,}690{,}970{,}194 \sqrt{3} n^{4}-66{,}508{,}795{,}463{,}791{,}122n^{3} \\ &{}+46{,}318{,}086{,}975{,}242{,}316\sqrt{3}n^{3} +17{,}867{,}770{,}385{,}080{,}772\sqrt{3}n^{2} \\ &{}-35{,}327{,}233{,}535{,}891{,}622n^{2}+2{,}975{,}211{,}060{,}929{,}562\sqrt{3}n \\ &{}-11{,}558{,}904{,}059{,}737{,}827n+111{,}646{,}193{,}915{,}178\sqrt{3}\\ &{}-1{,}615{,}722{,}383{,}317{,}419. \end{aligned}$$
Combining (3.11), (3.14), and (3.15) yields
$$\begin{aligned}& u(n)< \frac{h(n+1)+\sqrt{h^{2}(n+1)+4f(n+1)g(n+1)}}{2f(n+1)}. \end{aligned}$$
(3.16)
It follows from (2.1), (2.7), (3.13), and (3.16) that, for \(n\geq5\),
$$\begin{aligned}& \frac{h(n+1)-\sqrt{h^{2}(n+1)+4f(n+1)g(n+1)}}{2f(n+1)} \\& \quad< l(n) < \frac{P_{n}}{P_{n-1}} < u(n) < \frac{h(n+1)+\sqrt{h^{2}(n+1)+4f(n+1) g(n+1)}}{2f(n+1)}, \end{aligned}$$
which yields
$$\begin{aligned}& f(n+1) \biggl(\frac{U_{n}}{U_{n-1}} \biggr)^{2} -h(n+1) \frac{U_{n}}{U_{n-1}}-g(n+1)< 0. \end{aligned}$$
(3.17)
In view of (1.1),
$$\begin{aligned}& f(n+1)U_{n}^{2}-U_{n-1}U_{n+1}=P_{n}^{2} \biggl(f(n+1) \biggl(\frac{U_{n}}{U_{n-1}} \biggr)^{2} -h(n+1) \frac{U_{n}}{U_{n-1}}-g(n+1) \biggr). \end{aligned}$$
(3.18)
Lemma 3.2 follows from (3.17) and (3.18). This completes the proof. □

Now, we turn to the proof of Theorem 1.1.

Proof of Theorem 1.1

Replacing n by \(n-1\) in (3.10), we deduce that, for \(n \geq6\),
$$\begin{aligned}& \frac{U_{n}U_{n-2}}{U_{n-1}^{2}}>f(n). \end{aligned}$$
(3.19)
In view of (3.1) and (3.19), we deduce that, for \(n\geq 6\),
$$\begin{aligned}& \frac{U_{n}^{2} }{U_{n-1}^{2}} >\frac{U_{n+1} }{U_{n} }\frac{U_{n-1}}{ U_{n-2}}. \end{aligned}$$
(3.20)
It is easy to verify that (3.20) also holds for \(2\leq n \leq 5\). This completes the proof of Theorem 1.1. □

Declarations

Acknowledgements

This work was supported by the National Science Foundation of China (11526136).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Statistics and Information, Shanghai University of International Business and Economics
(2)
School of Mathematics, Nanjing Normal University, Taizhou College

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© Liu and Jin 2016