The ratio log-concavity of the Cohen numbers

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 $\frac{U_{n}}{U_{n-1}}$.

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. □