Irrationality exponents of generalized Hone series

Abstract

We compute the exact irrationality exponents of certain series of rational numbers, first studied in a special case by Hone, by transforming them into suitable continued fractions.

1 Main theorem

For a real number $\alpha ,$ the irrationality exponent $\mu \left( \alpha \right)$ is defined by the infimum of the set of numbers $\mu$ for which the inequality

$$\begin{aligned} \left| \alpha -\frac{p}{q}\right| <\frac{1}{q^{\mu }} \end{aligned}$$

(1)

has only finitely many rational solutions p/q, or equivalently the supremum of the set of numbers $\mu$ for which the inequality (1) has infinitely many solutions. If $\alpha$ is irrational, then $\mu \left( \alpha \right) \ge 2$. If $\alpha$ is a real algebraic irrationality, then $\mu \left( \alpha \right) =2$ by Roth’s theorem [7]. If $\mu \left( \alpha \right) =\infty ,$ then $\alpha$ is called a Liouville number.

For every sequence $\left( u_{n}\right) _{n\ge 1}$ of nonzero numbers or indeterminates, we define $u_{0}=1$ and

$$\begin{aligned} \theta u_{k}=\frac{u_{k+1}}{u_{k}},\quad \theta ^{2}u_{k}=\theta \left( \theta u_{k}\right) =\frac{u_{k+2}u_{k}}{u_{k+1}^{2}}\quad \left( k\ge 0\right) . \end{aligned}$$

(2)

Theorem 1

Let $\left( x_{n}\right) _{n\ge 1}$ be an increasing sequence of integers and $\left( y_{n}\right) _{n\ge 1}$ be a sequence of nonzero integers such that $x_{1}>y_{1}\ge 1$

$$\begin{aligned} \frac{\theta ^{2}x_{n}-\theta ^{2}y_{n}}{x_{n}}\in {\mathbb {Z}}_{>0}\quad \left( n\ge 0\right) . \end{aligned}$$

(3)

Assume that

• (i)
  $$\begin{aligned} {}\log \left| y_{n+2}\right| =o\left( \log x_{n}\right) , \end{aligned}$$
• (ii)
  $$\begin{aligned} {}\liminf _{n\rightarrow \infty }\frac{\log x_{n+1}}{\log x_{n}}>2. \end{aligned}$$

Then the series

$$\begin{aligned} \sigma =\sum _{n=1}^{\infty }\frac{y_{n}}{x_{n}} \end{aligned}$$

(4)

is convergent and

$$\begin{aligned} \mu \left( \sigma \right) =\max \left\{ \limsup _{n\rightarrow \infty }\frac{\log x_{n+1}}{\log x_{n}}\text { },\text { }2+\frac{1}{\liminf \nolimits _{n\rightarrow \infty }\dfrac{\log x_{n+1}}{\log x_{n}}-1}\right\} . \end{aligned}$$

Remark 1

The assumption (ii) implies that

$$\begin{aligned} 2+\frac{1}{\liminf \nolimits _{n\rightarrow \infty }\dfrac{\log x_{n+1}}{\log x_{n} }-1}<3. \end{aligned}$$

Moreover, if the limit $\lambda :=\lim _{n\rightarrow \infty }\left( \log x_{n+1}/\log x_{n}\right)$ exists, then

$$\begin{aligned} \mu \left( \sigma \right) =\max \left\{ \lambda \text { },\text { }2+\frac{1}{\lambda -1}\right\} , \end{aligned}$$

and so

$$\begin{aligned} \mu \left( \sigma \right) =\left\{ \begin{array}[c]{l} \lambda \quad \text {if }\lambda \ge \dfrac{3+\sqrt{5}}{2}=2.618...,\\ 2+\dfrac{1}{\lambda -1}\quad \text {if}\quad 2<\lambda <\dfrac{3+\sqrt{5}}{2}. \end{array} \right. \end{aligned}$$

Hence, under the hypotheses of Theorem 1, $\mu \left( \sigma \right) >2$ and therefore $\sigma$ is transcendental.

Examples of series $\sigma$ satisfying the assumptions of Theorem 1 have been first given by Hone [3] in the case where $y_{n}=1$ for every positive integer n,  and later by Varona [8] in the case where $y_{n}=\left( -1\right) ^{n}.$ Both Hone and Varona computed the expansion in regular continued fraction of $\sigma$ in these special cases and succeeded in proving its transcendence by using Roth’s theorem. For more expansions in regular continued fraction, see also [4,5,6].

In this paper, we will use basically the same method and transform $\sigma$ given by (4) into a continued fraction (not regular in general) by using Lemma 2 in Sect. 3. Then we will reach our conclusion by applying a formula which gives the irrationality exponent of continued fractions under convenient assumptions (Lemma 3, also in Sect. 3).

The paper is organized as follows. In Sect.2, we will give examples of series generalizing both Hone and Varona series, and show how Theorem 1 applies to these series (see formula (10) below). In Sect. 3, we will state three lemmas which will be used in the proof of theorem 1. The proof of Theorem 1 will be given in Sect. 4. Finally, in Sect. 5 we will give the proof of the asymptotic estimates used in Sect. 2.

2 Applications

For any nonzero integer a and non-constant $P(X) \in {\mathbb Z}_{\ge 0}[X]$ with $P(0)=0$, we define the sequence $\{y_n\}$ by

$$\begin{aligned} y_{n}=a^{P(n)}\qquad (n\ge 0). \end{aligned}$$

Let $\varDelta P(X)=P(X+1)-P(X).$ As $P(X)\in {\mathbb {Z}}_{>0},$ we have $\varDelta P(X)\in {\mathbb {Z}}_{>0}$ and $\varDelta ^{2}P(X)\in {\mathbb {Z}}_{>0}$. Hence we see that

$$\begin{aligned} \theta ^{2}y_{n}=a^{\varDelta ^{2}P(n)}=a^{P(n+2)-2P(n+1)+P(n)}\in {\mathbb {Z}} _{>0}, \end{aligned}$$

since $P(n+2)-2P(n+1)+P(n)\equiv P(n+2)+P(n)\equiv 2P(n)\equiv 0$$\left( \mathrm{mod}~ 2\right)$ and $\varDelta ^{2}P(n)>0.$ Furthermore, let Q(X, Y) be any pseudo-polynomial defined by

$$\begin{aligned} Q(X,Y)=\sum _{i=0}^{I}\sum _{j=0}^{J}\beta _{i,j}X^{q_{i}}Y^{r_{j}}, \end{aligned}$$

where $\beta _{i,j}$ are non-negative real numbers with $\beta =\beta _{I,J}>0$, $q=q_{I}>q_{I -1}>\cdots >q_{0}\ge 0$, and $r=r_{J}>r_{J-1}>\cdots >r_{0}\ge 0$ with $q+r>0.$ Put $\alpha _{0}=1$ and

$$\begin{aligned} \alpha _{n+1}=\left\lfloor Q(x_{n},x_{n+1})\right\rfloor \quad \left( n\ge 0\right) . \end{aligned}$$

(5)

We define the sequence $\left( x_{n}\right) _{n\ge 0}$ by the recurrence relation

$$\begin{aligned} x_{n+2}x_{n}=x_{n+1}^{2}\left( \alpha _{n+1}x_{n}+\theta ^{2}y_{n}\right) \quad \left( n\ge 0\right) \end{aligned}$$

(6)

with the initial conditions

$$\begin{aligned} x_{0}=1,\quad x_{1}\in {\mathbb {Z}}_{>0}. \end{aligned}$$

(7)

We note that there exists $\xi >0$ such that

$$\begin{aligned} \alpha _{n+1}=\beta x_{n}^{q}x_{n+1}^{r}\left( 1+O(x_{n}^{-\xi })\right) . \end{aligned}$$

(8)

An easy induction shows that

$$\begin{aligned} x_{n}\ge \left( x_{1}\right) ^{2^{n-1}} \end{aligned}$$

(9)

and that $x_{n-1}$ divides $x_{n}$ for every $n\ge 1.$ By (9), the series $\sum _{n=1}^\infty y_n/x_n$ is convergent. Moreover we will prove in Sect. 5, Corollary 3, that

$$\begin{aligned} \log x_{n}{\sim }C\lambda ^{n}\qquad \text{ as }\qquad n \rightarrow \infty , \end{aligned}$$

where C is a positive constant and

$$\begin{aligned} \lambda =\lim _{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}=\frac{1}{2}\left( r+2+\sqrt{\left( r+2\right) ^{2}+4q}\right) >2, \end{aligned}$$

since $q\ge 0,$$r\ge 0$ and $q+r>0.$ Therefore, applying Theorem 1 with Remark 1, we obtain

Theorem 2

Let $(x_{n})_{n\ge 0}$ be as above. Define the number $\sigma$ by

$$\begin{aligned} \sigma =\sum _{n=1}^{\infty }\frac{y_{n}}{x_{n}}=\sum _{n=1}^{\infty }\frac{ a^{P(n)}}{x_{n}}. \end{aligned}$$

Then we have

$$\begin{aligned} \mu \left( \sigma \right) =\left\{ \begin{array}{l} \lambda \quad \mathrm{if}\quad \lambda \ge \dfrac{3+\sqrt{5}}{2} \\ 2+\dfrac{1}{\lambda -1}\quad \text {if}\quad 2<\lambda <\dfrac{3+\sqrt{5}}{2} \end{array} \right. . \end{aligned}$$

(10)

As an application, take $P(n)=n.$ Then the function

$$\begin{aligned} f(z)=\sum _{n=1}^{\infty }\frac{z^{n}}{x_{n}} \end{aligned}$$

is an entire function, and we have the following generalization of Theorem 3.4 of [6].

Corollary 1

For every $\nu \ge \left( 3+\sqrt{5}\right) /2$, there exists infinitely many sequences $(x_{n})$ such that $\mu \left( f\left( a\right) \right) =\nu$ for every non-zero integer a.

Proof

Choose q and r such that

$$\begin{aligned} \nu =\frac{1}{2}\left( r+2+\sqrt{\left( r+2\right) ^{2}+4q}\right) \ge \frac{3+\sqrt{5}}{2}. \end{aligned}$$

There are infinitely many possible choices. Define the sequence $\left( x_{n}\right) _{n\ge 0}$ by the recurrence relation (6) and (7) with $\alpha _{n+1}=\left\lfloor x_{n}^{q}x_{n+1}^{r}\right\rfloor$. Then Theorem 2 applies, which proves Corollary 1. $\square$

Remark 2

Hone and Varona series in [3, 8] are obtained as special cases of (6) by taking $P(X)=X$ with $a=1$ and $a=-1$ respectively.

3 Lemmas

In this section, we prepare some lemmas for the proof of Theorem 1.

Lemma 1

Let $(x_{n})_{n\ge 1}$ and $(y_{n})_{n\ge 1}$ be sequences as in Theorem 1. We have

$$\begin{aligned}&\log \left| y_{n+1}\right| =o\left( \log x_{n}\right) ,\quad \log \left| y_{n}\right| =o\left( \log x_{n}\right) , \end{aligned}$$

(11)

$$\begin{aligned}&\sum _{j=1}^{n}\log \left| y_{j}\right| =o\left( \log x_{n}\right) . \end{aligned}$$

(12)

Proof

The assumption (ii) implies for large n that

$$\begin{aligned} \frac{\log \left| y_{n+1}\right| }{\log x_{n}}&=\frac{\log \left| y_{n+1}\right| }{\log x_{n-1}}\frac{\log x_{n-1}}{\log x_{n}}\le \frac{\log \left| y_{n+1}\right| }{2\log x_{n-1}},\\ \frac{\log \left| y_{n}\right| }{\log x_{n}}&=\frac{\log \left| y_{n}\right| }{\log x_{n-1}}\frac{\log x_{n-1}}{\log x_{n}}\le \frac{\log \left| y_{n}\right| }{2\log x_{n-1}}, \end{aligned}$$

which proves (11) by using (i). Now by (11) we have

$$\begin{aligned} \sum _{j=1}^{n}\log \left| y_{j}\right| =o\left( \sum _{j=1}^{n}\log x_{j}\right) \end{aligned}$$

and by (ii) there exists a constant $K>0$ such that

$$\begin{aligned} \log x_{j}\le \frac{K}{2^{n-j}}\log x_{n}\quad \left( 1\le j\le n\right) , \end{aligned}$$

which proves (12). $\square$

Lemma 2

[2, Theorem 2] Let $x_{1},$$x_{2},$$\ldots ,$$y_{1},$$y_{2},$$\ldots$ be indeterminates. Then for every $n\ge 1,$

$$\begin{aligned} \sigma _{n}=\sum _{k=1}^{n}\frac{y_{k}}{x_{k}}=\frac{a_{1}}{b_{1}} \genfrac{}{}{0.0pt}{}{{}}{+} \frac{a_{2}}{b_{2}} \genfrac{}{}{0.0pt}{}{{}}{+\cdots } \genfrac{}{}{0.0pt}{}{{}}{+} \dfrac{a_{2n}}{b_{2n}}, \end{aligned}$$

(13)

where $a_{1}=y_{1},$$b_{1}=x_{1}-y_{1},$ and for $k\ge 1$

$$\begin{aligned} a_{2k}&=\theta y_{k-1},\quad a_{2k+1}=\theta ^{2}y_{k-1},\\ \quad b_{2k}&=x_{k-1},\quad b_{2k+1}=\frac{\theta ^{2}x_{k-1}-\theta ^{2}y_{k-1}}{x_{k-1}}. \end{aligned}$$

Lemma 3

[1, Corollary 4] Let an infinite continued fraction

$$\begin{aligned} \alpha =\frac{a_{1}}{b_{1}} \genfrac{}{}{0.0pt}{}{{}}{+} \frac{a_{2}}{b_{2}} \genfrac{}{}{0.0pt}{}{{}}{+} \dfrac{a_{3}}{b_{3}} \genfrac{}{}{0.0pt}{}{{}}{+} \cdots \end{aligned}$$

be convergent, where $a_{n},$$b_{n}$$\left( n\ge 1\right)$ are nonzero rational integers. Assume that

• (I)
  $$\begin{aligned} \sum _{n=1}^{+\infty }\left| \frac{a_{n+1}}{b_{n}b_{n+1} }\right| <\infty , \end{aligned}$$
• (II)
  $$\begin{aligned} \lim _{n\rightarrow +\infty }\frac{\log \left| a_{n} \right| }{\log \left| b_{n}\right| }=0. \end{aligned}$$

Then $\alpha$ is irrational and

$$\begin{aligned} \mu \left( \alpha \right) =2+\limsup _{n\rightarrow +\infty }\frac{\log \left| b_{n+1}\right| }{\log \left| b_{1}b_{2}\cdots b_{n}\right| }. \end{aligned}$$

(14)

4 Proof of Theorem 1

The assumptions (i), (ii) and (11) imply that

$$\begin{aligned} \max \left( \left| y_{k}\right| ,\left| y_{k+1}\right| ,\left| y_{k+2}\right| \right) \le x_{k}^{\varepsilon } \end{aligned}$$

(15)

and that $x_{k+1}\ge x_{k}^{2}$$\left( k\ge k_{0}\left( \varepsilon \right) \right)$ for any $\varepsilon \in (0,1)$. We have

$$\begin{aligned} \sum _{k=k_{0}}^{\infty }\frac{\left| y_{k}\right| }{x_{k}}\le \sum _{k=k_{0}}^{\infty }\frac{1}{x_{k}^{1-\varepsilon }}\le \sum _{k=0}^{\infty }\frac{1}{x_{k_{0}}^{\left( 1-\varepsilon \right) 2^{k}}}, \end{aligned}$$

and hence the series (4) is absolutely convergent. Using Lemma 2, we get the continued fraction expansion of $\sigma =\lim _{n\rightarrow \infty }\sigma _{n}.$ To apply Lemma 3 we transform this to a continued fraction with integral partial numerators and denominators by using the formula

$$\begin{aligned} \frac{a_{1}}{b_{1}} \genfrac{}{}{0.0pt}{}{{}}{+} \frac{a_{2}}{b_{2}} \genfrac{}{}{0.0pt}{}{{}}{+} \dfrac{a_{3}}{b_{3}} \genfrac{}{}{0.0pt}{}{{}}{+} \cdots =\frac{r_{1}a_{1}}{r_{1}b_{1}} \genfrac{}{}{0.0pt}{}{{}}{+} \frac{r_{2}r_{1}a_{2}}{r_{2}b_{2}} \genfrac{}{}{0.0pt}{}{{}}{+} \dfrac{r_{2}r_{3}a_{3}}{r_{3}b_{3}} \genfrac{}{}{0.0pt}{}{{}}{+} \cdots . \end{aligned}$$

By taking $r_{2k}=1$ and $r_{2k+1}=y_{k}^{2}$$(k\ge 1),$ we obtain the expansion

$$\begin{aligned} \sigma =\frac{a_{1}}{b_{1}} \genfrac{}{}{0.0pt}{}{{}}{+} \frac{a_{2}}{b_{2}} \genfrac{}{}{0.0pt}{}{{}}{+\cdots } \genfrac{}{}{0.0pt}{}{{}}{+} \dfrac{a_{n}}{b_{n}} \genfrac{}{}{0.0pt}{}{{}}{+\cdots } , \end{aligned}$$

where $a_{1}=y_{1},$$b_{1}=x_{1}-y_{1}\ge 1,$ and for $k\ge 1$

$$\begin{aligned} a_{2k}&=y_{k-1}y_{k},\quad a_{2k+1}=y_{k-1}y_{k+1}, \end{aligned}$$

(16)

$$\begin{aligned} \quad b_{2k}&=x_{k-1},\quad b_{2k+1}=y_{k}^{2}\frac{\theta ^{2} x_{k-1}-\theta ^{2}y_{k-1}}{x_{k-1}}\ge 1. \end{aligned}$$

(17)

We verify the conditions (I) and (II) in Lemma 3. First for (I), we have

$$\begin{aligned} \sum _{n=1}^{+\infty }\left| \frac{a_{n+1}}{b_{n}b_{n+1}}\right| \le \sum _{k=1}^{+\infty }\left| \frac{a_{2k+1}}{b_{2k}b_{2k+1}}\right| +\sum _{k=1}^{+\infty }\left| \frac{a_{2k}}{b_{2k-1}b_{2k}}\right| , \end{aligned}$$

where

$$\begin{aligned} \left| \frac{a_{2k+1}}{b_{2k}b_{2k+1}}\right|&\le \frac{\left| y_{k-1}y_{k+1}\right| }{x_{k-1}y_{k}^{2}}\le \frac{\left| y_{k-1}y_{k+1}\right| }{x_{k-1}},\\ \left| \frac{a_{2k}}{b_{2k-1}b_{2k}}\right|&\le \frac{\left| y_{k-1}y_{k}\right| }{y_{k-1}^{2}x_{k-1}}\le \frac{\left| y_{k-1} y_{k}\right| }{x_{k-1}}, \end{aligned}$$

noting that $b_{2k+1}\ge y_{k}^{2}$ by (3). By using (15) we deduce similarly as above that

$$\begin{aligned} \max \left\{ \sum _{k=k_{0}+1}^{+\infty }\frac{\left| y_{k-1}y_{k+1} \right| }{x_{k-1}},\sum _{k=k_{0}+1}^{+\infty }\frac{\left| y_{k-1} y_{k}\right| }{x_{k-1}}\right\} \le \sum _{k=k_{0}+1}^{+\infty }\frac{1}{x_{k-1}^{1-2\varepsilon }} \end{aligned}$$

and (I) follows. Now we prove that (II) holds. We have by (i) and (11)

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{\log \left| a_{2k}\right| }{\log b_{2k} }=\lim _{k\rightarrow \infty }\frac{\log \left| y_{k-1}\right| +\log \left| y_{k}\right| }{\log x_{k-1}}=0 \end{aligned}$$

and

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{\log \left| a_{2k+1}\right| }{\log b_{2k+1}}=\lim _{k\rightarrow \infty }\frac{\log \left| y_{k-1}\right| +\log \left| y_{k+1}\right| }{2\log \left| y_{k}\right| +\log \left( \theta ^{2}x_{k-1}-\theta ^{2}y_{k-1}\right) -\log x_{k-1}}.\nonumber \\ \end{aligned}$$

(18)

Here, since $\theta ^{2}x_{k-1}-\theta ^{2}y_{k-1}\ge x_{k-1}$ by (3), we have

$$\begin{aligned} \frac{\left| \theta ^{2}y_{k-1}\right| }{\theta ^{2}x_{k-1}}\le \frac{\left| \theta ^{2}y_{k-1}\right| }{x_{k-1}+\theta ^{2}y_{k-1} }. \end{aligned}$$

(19)

However, by (11) we can write for every $\varepsilon \in (0,1)$

$$\begin{aligned} \log \frac{\left| \theta ^{2}y_{k-1}\right| }{x_{k-1}}&=\log \left| y_{k+1}\right| +\log \left| y_{k-1}\right| -2\log \left| y_{k}\right| -\log x_{k-1}\\&=-\log x_{k-1}+o\left( \log x_{k-1}\right) \le -\varepsilon \log x_{k-1}\quad \left( k\rightarrow \infty \right) , \end{aligned}$$

which implies

$$\begin{aligned} \frac{\left| \theta ^{2}y_{k-1}\right| }{x_{k-1}}\le \frac{1}{x_{k-1}^{\varepsilon }}\quad \left( k\rightarrow \infty \right) . \end{aligned}$$

Therefore from (19) we see that for every $\varepsilon \in (0,1)$

$$\begin{aligned} \frac{\left| \theta ^{2}y_{k-1}\right| }{\theta ^{2}x_{k-1}}\le \frac{\left| \theta ^{2}y_{k-1}\right| }{x_{k-1}+\theta ^{2}y_{k-1}} \le \frac{2}{x_{k-1}^{\varepsilon }}\rightarrow 0\quad \left( k\rightarrow \infty \right) , \end{aligned}$$

(20)

which yields

$$\begin{aligned} \log \left( \theta ^{2}x_{k-1}-\theta ^{2}y_{k-1}\right) =\log \theta ^{2} x_{k-1}+o(1). \end{aligned}$$

Hence we get from (11) and (18)

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{\log \left| a_{2k+1}\right| }{\log b_{2k+1}}=\lim _{k\rightarrow \infty }\frac{\log \left| y_{k-1}\right| +\log \left| y_{k+1}\right| }{2\log \left| y_{k}\right| +\log x_{k+1}-2\log x_{k}+o(1)}=0, \end{aligned}$$

and (II) is ensured. Now we compute the right-hand side of (14). We have by (17)

$$\begin{aligned} b_{1}b_{2}\cdots b_{2k+1}&=\left( x_{1}-y_{1}\right) \prod _{j=0} ^{k-1}y_{j+1}^{2}\left( \theta ^{2}x_{j}-\theta ^{2}y_{j}\right) \\&=\left( x_{1}-y_{1}\right) \theta ^{2}x_{0}\theta ^{2}x_{1}\cdots \theta ^{2}x_{k-1}\prod _{j=0}^{k-1}y_{j+1}^{2}\left( 1-\frac{\theta ^{2}y_{j}}{\theta ^{2}x_{j}}\right) \\&=\frac{x_{1}-y_{1}}{x_{1}}\frac{x_{k+1}}{x_{k}}\prod _{j=0}^{k-1}y_{j+1} ^{2}\left( 1-\frac{\theta ^{2}y_{j}}{\theta ^{2}x_{j}}\right) . \end{aligned}$$

Now the infinite product

$$\begin{aligned} \prod _{j=0}^{\infty }\left( 1-\frac{\theta ^{2}y_{j}}{\theta ^{2}x_{j}}\right) \end{aligned}$$

is convergent by (20) and (9). Hence we get by (12)

$$\begin{aligned} \log \left( b_{1}b_{2}\cdots b_{2k+1}\right)&=\log x_{k+1}-\log x_{k}+2\sum _{j=1}^{k}\log \left| y_{j}\right| +O(1)\nonumber \\&=\log x_{k+1}-\log x_{k}+o\left( \log x_{k}\right) . \end{aligned}$$

(21)

Therefore

$$\begin{aligned} \limsup _{k\rightarrow +\infty }\frac{\log b_{2k+2}}{\log \left( b_{1}b_{2}\cdots b_{2k+1}\right) }=\limsup _{k\rightarrow +\infty }\frac{\log x_{k}}{\log x_{k+1}-\log x_{k}}. \end{aligned}$$

(22)

Furthermore, we have by (21)

$$\begin{aligned} \log \left( b_{1}b_{2}\cdots b_{2k}\right) =\log \left( b_{1}b_{2}\cdots b_{2k-1}\right) +\log b_{2k}=\log x_{k}+o\left( \log x_{k}\right) \end{aligned}$$

and

$$\begin{aligned} \log b_{2k+1}=\log \left( b_{1}b_{2}\cdots b_{2k+1}\right) -\log \left( b_{1}b_{2}\cdots b_{2k}\right) =\log x_{k+1}-2\log x_{k}+o\left( \log x_{k}\right) . \end{aligned}$$

Hence

$$\begin{aligned} \limsup _{k\rightarrow +\infty }\frac{\log b_{2k+1}}{\log \left( b_{1}b_{2}\cdots b_{2k}\right) }=\limsup _{n\rightarrow +\infty }\frac{\log x_{k+1}}{\log x_{k} }-2. \end{aligned}$$

(23)

Therefore, it follows from (14), (22) and (23) that

$$\begin{aligned} \mu \left( \sigma \right) =\max \left\{ 2+\frac{1}{\liminf \nolimits _{k\rightarrow \infty }\dfrac{\log x_{k+1}}{\log x_{k}}-1}\text { },\text { }\limsup _{k\rightarrow \infty }\frac{\log x_{k+1}}{\log x_{k}}\right\} \end{aligned}$$

and the proof of Theorem 1 is completed.

5 Asymptotic behaviour

We now study the asymptotic behaviour of sequences $\left( x_{n}\right) _{n\ge 1}$ satisfying (6) and (7). We follow basically the method indicated in [3]. Let $\left( u_{n}\right) _{n\ge 0}$ be any sequence of complex numbers satisfying the recurrence relation

$$\begin{aligned} u_{n+2}-Au_{n+1}+Bu_{n}=\tau _{n}\quad \left( n\ge 0\right) , \end{aligned}$$

(24)

where A and B are complex numbers with $A^{2}-4B\ne 0$ and $\tau _{n}$ is a function of n,  $u_{n}$ and $u_{n+1}$. As $A^{2}-4B\ne 0,$ the equation

$$\begin{aligned} x^{2}-Ax+B=0. \end{aligned}$$

(25)

has two distinct roots $\lambda$ and $\nu ,$ with $\lambda \ne \nu .$ Morever, at least one of these roots is not zero, and we can assume without loss of generality that $\lambda \ne 0.$

Theorem 3

Assume that $\left( u_{n}\right) _{n\ge 0}$ satisfies (24). Let $\lambda$ and $\nu ,$ with $\lambda \ne \nu$ and $\lambda \ne 0,$ be the roots of (25). Then for every $n\ge 1$

$$\begin{aligned} u_{n}&=\frac{1}{\lambda -\nu }\left( \left( u_{1}-\nu u_{0}\right) \lambda ^{n}+\left( \lambda u_{0}-u_{1}\right) \nu ^{n}+\sum _{k=1}^{n-1} \tau _{k-1}\left( \lambda ^{n-k}-\nu ^{n-k}\right) \right) \quad \left( B\ne 0\right) , \end{aligned}$$

(26)

$$\begin{aligned} u_{n}&=\frac{1}{\lambda }\left( u_{1}\lambda ^{n}+\sum _{k=1}^{n-1} \tau _{k-1}\lambda ^{n-k}\right) \quad \left( B=0\right) . \end{aligned}$$

(27)

Proof

First we assume that $B\ne 0,$ which implies $\nu \ne 0.$ For every $n\ge 0,$ let

$$\begin{aligned} v_{n}=\sum _{k=1}^{n}\tau _{k-1}\left( \lambda ^{n-k}-\nu ^{n-k}\right) . \end{aligned}$$

We have for $n\ge 0$

$$\begin{aligned} v_{n+1}&=\sum _{k=1}^{n+1}\tau _{k-1}\left( \lambda ^{n+1-k}-\nu ^{n+1-k}\right) =\sum _{k=1}^{n}\tau _{k-1}\left( \lambda ^{n-k+1}-\nu ^{n-k+1}\right) ,\\ v_{n+2}&=\sum _{k=1}^{n+1}\tau _{k-1}\left( \lambda ^{n-k+2}-\nu ^{n-k+2}\right) =\left( \lambda -\nu \right) \tau _{n}+\sum _{k=1}^{n} \tau _{k-1}\left( \lambda ^{n-k+2}-\nu ^{n-k+2}\right) . \end{aligned}$$

Therefore, for every $n\ge 0,$ the sequence

$$\begin{aligned} w_{n}=\frac{1}{\lambda -\nu }\left( \left( u_{1}-\nu u_{0}\right) \lambda ^{n}+\left( \lambda u_{0}-u_{1}\right) \nu ^{n}+v_{n}\right) \end{aligned}$$

satisfies $w_{n+2}-Aw_{n+1}+Bw_{n}=\tau _{n}.$ Moreover $w_{0}=u_{0}$ and $w_{1}=u_{1}$ since $v_{0}=v_{1}=0.$ Hence $w_{n}=u_{n}$ for every $n\ge 0$, which proves Theorem 3 when $B\ne 0$ since

$$\begin{aligned} v_{n}=\sum _{k=1}^{n-1}\tau _{k-1}\left( \lambda ^{n-k}-\nu ^{n-k}\right) \quad \left( n\ge 1\right) . \end{aligned}$$

Letting $B\rightarrow 0$ in (26), we obtain (27). $\square$

Corollary 2

With the notations of Theorem 3, assume that $\left| \nu \right| <\left| \lambda \right|$ and that $\left| \lambda \right| >1.$ Assume moreover that $\tau _{n}$ is bounded. Then

$$\begin{aligned} u_{n}&=C\lambda ^{n}+O\left( 1\right) \quad \mathrm{if}\quad \left| \nu \right| <1, \end{aligned}$$

(28)

$$\begin{aligned} u_{n}&=C\lambda ^{n}+O\left( n\right) \quad \mathrm{if}\quad \left| \nu \right| =1, \end{aligned}$$

(29)

$$\begin{aligned} u_{n}&=C\lambda ^{n}+D\nu ^{n}+O\left( 1\right) \quad \mathrm{if} \quad \left| \nu \right| >1, \end{aligned}$$

(30)

where

$$\begin{aligned} C=\frac{1}{\lambda -\nu }\left( u_{1}-\nu u_{0}+\sum _{k=1}^{\infty }\tau _{k-1}\lambda ^{-k}\right) \end{aligned}$$

and, in the case where $\left| \nu \right| >1,$

$$\begin{aligned} D=\frac{1}{\lambda -\nu }\left( \lambda u_{0}-u_{1}-\sum _{k=1}^{\infty } \tau _{k-1}\nu ^{-k}\right) . \end{aligned}$$

Proof

First assume that $\nu \ne 0.$ By (26) we have

$$\begin{aligned} u_{n}=\frac{u_{1}-\nu u_{0}}{\lambda -\nu }\lambda ^{n}+\frac{\lambda u_{0} -u_{1}}{\lambda -\nu }\nu ^{n}+\frac{\lambda ^{n}}{\lambda -\nu }\sum _{k=1}^{n} \tau _{k-1}\lambda ^{-k}-\frac{\nu ^{n}}{\lambda -\nu }\sum _{k=1}^{n}\tau _{k-1} \nu ^{-k}. \end{aligned}$$

We observe that

$$\begin{aligned} \sum _{k=1}^{n}\tau _{k-1}\lambda ^{-k}=\sum _{k=1}^{\infty }\tau _{k-1}\lambda ^{-k}-\sum _{k=n+1}^{\infty }\tau _{k-1}\lambda ^{-k}=\sum _{k=1}^{\infty } \tau _{k-1}\lambda ^{-k}+O\left( \lambda ^{-n}\right) , \end{aligned}$$

and the same equality holds with $\lambda$ replaced by $\nu$ if $\left| \nu \right| >1,$ which proves (30). On the other hand, if $\left| \nu \right| <1,$

$$\begin{aligned} \left| \frac{\nu ^{n}}{\lambda -\nu }\sum _{k=1}^{n}\tau _{k-1}\nu ^{-k}\right| \le \frac{M\left( 1-\left| \nu \right| ^{n}\right) }{\left| \lambda -\nu \right| \left( 1-\left| \nu \right| \right) }=O(1), \end{aligned}$$

(31)

where $M=\max _{k\in {\mathbb {N}}}\left| \theta _{k}\right|$, which proves (28). Finally, if $\left| \nu \right| =1,$

$$\begin{aligned} \left| \frac{\nu ^{n}}{\lambda -\nu }\sum _{k=1}^{n}\tau _{k-1}\nu ^{-k}\right| \le \frac{Mn}{\left| \lambda -\nu \right| }=O(n), \end{aligned}$$

(32)

which proves (29). When $\nu =0$ one argues the same way by using (27) in place of (26). $\square$

Now we can give an asymptotic expansion of the sequences $x_{n}$ defined by (7) and (6).

Corollary 3

Let $(x_{n})_{n\ge 0}$ be defined by (6) with (7). Define

$$\begin{aligned} \lambda&=\frac{1}{2}\left( r+2+\sqrt{\left( r+2\right) ^{2}+4q}\right) ,\\ \nu&=\frac{1}{2}\left( r+2-\sqrt{\left( r+2\right) ^{2}+4q}\right) . \end{aligned}$$

Then $\lambda \ge 1+\sqrt{q+1},$$\left| \nu \right| <\lambda$ and

$$\begin{aligned} \left\{ \begin{array} [c]{l} \log x_{n}=C\lambda ^{n}+O\left( 1\right) \quad \mathrm{if}\quad q<r+3,\\ \log x_{n}=C\lambda ^{n}+O\left( n\right) \quad \mathrm{if}\quad q=r+3,\\ \log x_{n}=C\lambda ^{n}+D\nu ^{n}+O\left( 1\right) \quad \mathrm{if}\quad q>r+3, \end{array} \right. \end{aligned}$$

where C and D are constants.

Proof

By (8) we can find $\xi >0$ such that

$$\begin{aligned} \alpha _{n+1}x_{n}+\theta ^{2}y_{n}=\beta x_{n}^{q+1}x_{n+1}^{r}\left( 1+h_{n}\right) ,\quad h_{n}=O\left( x_{n}^{-\xi }\right) . \end{aligned}$$

Taking the logarithms in (6) yields

$$\begin{aligned} \log x_{n+2}-(r+2)\log x_{n+1}-q\log x_{n}=\log \beta +\log (1+h_{n}). \end{aligned}$$

With the notations of Corollary 1, define $u_{n}=\log x_{n},$$A=r+2,$$B=-q$, and $\tau _{n}=\log \beta +\log (1+h_{n}).$ Then $\tau _{n}$ is bounded and

$$\begin{aligned} \lambda&=\frac{1}{2}\left( r+2+\sqrt{\left( r+2\right) ^{2}+4q}\right) \ge 1+\sqrt{q+1},\\ \nu&=\frac{1}{2}\left( r+2-\sqrt{\left( r+2\right) ^{2}+4q}\right) \le 0. \end{aligned}$$

Hence $\lambda \ge 2$ and $q<\lambda ^{2}$, which implies

$$\begin{aligned} \left| \nu \right| =\frac{q}{\lambda }<\lambda . \end{aligned}$$

Moreover

$$\begin{aligned} \left| \nu \right|<1\Leftrightarrow \sqrt{\left( r+2\right) ^{2} +4q}<r+4\Leftrightarrow q<r+3. \end{aligned}$$

Therefore Corollary 2 applies, which proves Corollary 3. $\square$