\documentclass[twoside]{article} \pagestyle{myheadings} \markboth{\hfil Oscillation criteria \hfil EJDE--2001/10} {EJDE--2001/10\hfil Jianhua Shen \& I. P. Stavroulakis \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2001}(2001), No. 10, pp. 1--15. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Oscillation criteria for delay difference equations % \thanks{ {\em Mathematics Subject Classifications:} 39A10. \hfil\break\indent {\em Key words:} Oscillation, non-oscillation, delay difference equation. \hfil\break\indent \copyright 2001 Southwest Texas State University. \hfil\break\indent Submitted January 9, 2001. Published January 23, 2001. \hfil\break\indent This work was supported by the State Scholarship Foundation (I.K.Y.), Athens, Greece, \hfil\break\indent for a postdoctoral research, and was done while the first author was visiting the \hfil\break\indent Department of Mathematics, University of Ioannina" } } \date{} % \author{ Jianhua Shen \& I. P. Stavroulakis } \maketitle \begin{abstract} This paper is concerned with the oscillation of all solutions of the delay difference equation $$x_{n+1}-x_n+p_nx_{n-k}=0, \quad n=0,1,2,\dots$$ where $\{p_n\}$ is a sequence of nonnegative real numbers and $k$ is a positive integer. Some new oscillation conditions are established. These conditions concern the case when none of the well-known oscillation conditions $$\limsup_{n\to \infty}\sum_{i=0}^kp_{n-i}>1 \quad{\rm and}\quad \liminf_{n\to \infty}\frac{1}{k}\sum_{i=1}^kp_{n-i}>\frac{k^k}{(k+1)^{k+1}}$$ is satisfied. \end{abstract} \section{Introduction} In the last few decades the oscillation theory of delay differential equations has been extensively developed. The oscillation theory of discrete analogue of delay differential equations has also attracted growing attention in the recent few years. The reader is referred to [1-5,9,10,15,16,18,20-23]. In particular, the problem of establishing sufficient conditions for the oscillation of all solutions of the delay difference equation $$x_{n+1}-x_n+p_nx_{n-k}=0,\quad n=0,1,2,\dots \eqno (1.1)$$ where $\{p_n\}$ is a sequence of nonnegative real numbers and $k$ is a positive integer, has been the subject of many recent investigations. See, for example, [2-7,9,15,16,18,20,21,23] and the references cited therein. Strong interest in (1.1) is motivated by the fact that it represents a discrete analogue of the delay differential equation $$x'(t)+p(t)x(t-\tau)=0, \quad p(t)\geq 0\,, \; \tau>0\,. \eqno (1.2)$$ By a solution of (1.1) we mean a sequence $\{x_n\}$ which is defined for $n\geq -k$ and which satisfies (1.1) for $n\geq 0$. A solution $\{x_n\}$ of (1.1) is said to be {\it oscillatory} if the terms $x_n$ of the solution are not eventually positive or eventually negative. Otherwise the solution is called {\it non-oscillatory}. In 1989, Erbe and Zhang [9] and Ladas, Philos and Sficas [16] studied the oscillation of (1.1) and proved that all solutions oscillate if $$\limsup_{n\to \infty}\sum_{i=0}^kp_{n-i}>1, \eqno (1.3)$$ or $$\liminf_{n\to \infty}p_n>\frac{k^k}{(k+1)^{k+1}}, \eqno (1.4)$$ or $$\liminf_{n\to \infty}\frac{1}{k}\sum_{i=1}^kp_{n-i}>\frac{k^k}{(k+ 1)^{k+1}}. \eqno (1.5)$$ Observe that (1.5) improves (1.4). It is interesting to establish sufficient conditions for the oscillation of all solutions of (1.1) when (1.3) and (1.5) are not satisfied. (For (1.2), this question has been investigated by many authors, see, for example, [8,11-14,19] and the references cited therein). In 1993, Yu, Zhang and Qian [23] and Lalli and Zhang [18] derived some results in this direction. Unfortunately, the main results in [23,18] are not correct. This is because these results are based on a false discrete version of Koplatadze-Chanturia Lemma (a counterexample is given in [5]). In 1998 Domshlak [4], studied the oscillation of all solutions and the existence of non-oscillatory solution of (1.1) with $r$ -periodic positive coefficients $\{p_n\}, p_{n+r}=p_n$. It is very important that in the following cases where $\{r=k\},\{r=k+1\},\{r=2\},\{k=1,r=3\}$ and $\{k=1,r=4\}$ the results obtained are stated in terms of necessary and sufficient conditions, and their checking is very easy. Following this historical (and chronological) review we also mention that in the case where $$\frac{1}{k}\sum_{i=1}^k p_{n-i}\geq \frac{k^k}{(k+1)^{k+1}} \quad{\rm and}\quad \lim_{n\to \infty}\frac{1}{k}\sum_{i=1}^kp_{n-i}= \frac{k^k}{(k+1)^{k+1}},$$ the oscillation of (1.1) has been studied in 1994 by Domshlak [3] and in 1998 by Tang [21] (see also Tang and Yu [22]). In a case when $p_n$ is asymptotically close to one of the periodic critical states, optimal results about oscillation properties of the equation $$x_{n+1}-x_n+p_nx_{n-1}=0$$ were obtained by Domshlak in 1999 [6] and in 2000 [7]. The aim of this paper is to use some new techniques and improve the methods previously used to obtain new oscillation conditions for (1.1). Our results are based on two new lemmas established in section 2. For convenience, we will assume that inequalities about values of sequences are satisfied eventually for all large $n$. \section{Some new lemmas} \paragraph{Lemma 2.1} {\it Let the number $h\geq 0$ be such that for large $n$, $$\frac{1}{k}\sum_{i=1}^kp_{n-i}\geq h\,. \eqno (2.1)$$ Assume that {\rm (1.1)} has an eventually positive solution $\{x_n\}$. Then $h\leq k^k/(k+1)^{k+1}$ and $$\limsup_{n\to \infty}\frac{x_n}{x_{n-k}}\leq [d(h)]^k, \eqno (2.2)$$ where $d(h)$ is the greater real root of the algebraic equation $$d^{k+1}-d^k+h=0 \eqno (2.3)$$ on the interval [0,1].} \vskip 0.2cm \noindent {\it Proof}. Since (1.5) implies that all solutions of (1.1) oscillate, but (1.1) has an eventually positive solution, from (2.1), it follows that $h\leq k^k/(k+1)^{k+1}$ must hold. We now prove (2.2). To this end, we let $$w_n=\frac{1}{k}\sum_{i=1}^k\frac{x_{n-i}}{x_{n-i-1}}. \eqno (2.4)$$ and first prove that $\limsup_{n\to \infty}w_n\leq d(h)$. From (1.1), it follows that $\{x_n\}$ is eventually decreasing and so for large $n$, we have $x_{n-i-1}\geq x_{n-i}$ for $i=1,2,\dots ,k$. This implies that $$w_n=\frac{1}{k}\sum_{i=1}^k\frac{x_{n-i}}{x_{n-i-1}}\leq 1:=d_1. \eqno (2.5)$$ Thus, $\limsup_{n\to \infty}w_n\leq d(h)$ holds for $h=0$ because of $d(0)=1$. We now consider the case when $0d_2>\cdots >d_m>d_{m+1}>0,m=1,2,\dots$. Therefore, the limit $\lim_{m\to \infty}d_m=d$ exists and satisfies (2.3). Since (2.7) holds for all $m=1,2,\dots ,\{d_m\}$ is decreasing and $d(h)$ is the greater real root of the equation (2.3), it follows that $\limsup_{n\to \infty}w_n\leq d(h)$ holds. Finally, using the Arithmetic-Geometric Mean Inequality , we have $$\limsup_{n\to \infty}\left(\frac{x_{n-1}}{x_{n-k-1}}\right)^{1/k}\leq \limsup_{n\to \infty}\frac{1}{k}\sum_{i= 1}^k\frac{x_{n-i}}{x_{n-i-1}}\leq d(h).$$ This implies (2.2). The proof is complete. \vskip 0.1cm We describe by the following proposition and remark the number $d(h)$. \paragraph{Proposition 2.1} {\it For {\rm (2.3)}, the following statements hold true: \begin{description} \item{\rm (i)} If $h=0$, then {\rm (2.3)} has exactly two different real roots $d_1=0$ and $d_2=1$. \item{\rm (ii)} If $0\sum_{j=0}^{k-1}p_{n+j}x_{n+j-k}\geq \left(\sum_{j=0}^{k-1}p_{n+ j}\right)x_{n-1}, $$and so for i=1,2,\dots ,k, we have$$ \frac{x_{n-i}}{x_{n-i+1}}\left(\sum_{j=1}^kp_{n-i+j}\right)<1. $$Summing the last inequality from i=1 to i=k, we obtain$$ \overline{w}_n=\frac{1}{k}\sum_{i=1}^k\frac{x_{n-i}}{x_{n-i+1}}\left( \sum_{j=1}^kp_{n-i+j}\right)<1:=d_1. \eqno (2.14) $$Thus (2.12) holds for M=0 because of \overline{d}(0)=1. We now consider the case when 0\frac{M^k}{d_1^k}+\sum_{j=1}^kp_{n-i+j}\frac{x_{n-i}}{x_{n-i+1}}.$$ Summing the last inequality from$i=1$to$i=k$, we obtain $$k>\frac{kM^k}{d_1^k}+\sum_{i=1}^k\frac{x_{n-i}}{x_{n-i+ 1}}\left(\sum_{j=1}^kp_{n-i+j}\right).$$ Thus $$\overline{w}_n=\frac{1}{k}\sum_{i=1}^k\frac{x_{n-i}}{x_{n-i+ 1}}\left(\sum_{j=1}^kp_{n-i+j}\right)<1-\frac{M^k}{d_1^k}:=d_2. \eqno (2.17)$$ Using the inequality (2.17) and repeating the above arguments, we have $$\overline{w}_n<1-\frac{M^k}{d_2^k}:=d_3.$$ Following this iterative procedure, by induction, we have $$\overline{w}_n<1-\frac{M^k}{d_m^k}:=d_{m+1}, m=1,2,\dots \eqno (2.18)$$ Now (2.12) follows from similar proof as in Lemma 2.1. Next, using the Arithmetic-Geometric Mean Inequality in (2.12) we have $$\limsup_{n\to \infty}\Big(\frac{x_{n-k}}{x_n}\prod_{i=1}^k\sum_{j =1}^kp_{n-i+ j}\Big)^{1/k}\leq \limsup_{n\to \infty}\frac{1}{k}\sum_{i=1}^k\frac{x_{n-i}}{x_{n-i+ 1}}\Big(\sum_{j=1}^kp_{n-i+j}\Big)\leq \overline{d}(M),$$ which leads to (2.9). The proof is complete. Observe that the number$M$in Lemma 2.2 satisfies $$0\leq M^k\leq \left(\frac{k^{k+1}}{(k+1)^{k+1}}\right)^k\leq \frac{k^k}{(k+1)^{k+1}},$$ and the last equality holds if and only if$k=1$. Thus, from Proposition 2.1, we have the following conclusion about the equation (2.10). \paragraph{Proposition 2.2} {\it For {\rm (2.10)}, the following statements hold true: \begin{description} \item{\rm (i)} If$M=0$, then {\rm (2.10)} has exactly two different real roots$d_1=0$and$d_2=1$. \item{\rm (ii)} If$k\neq 1$and$01, \eqno (3.2) $$where \overline{d}(k\mu) and d(\mu) are the greater real roots of the equations$$ d^{k+1}-d^k+(k\mu)^k=0 \eqno (3.3) $$and$$ d^{k+1}-d^k+\mu=0, \eqno (3.4) $$respectively. Then all solutions of {\rm (1.1)} oscillate.} \vskip 0.1cm \noindent {\it Proof}. Assume, for the sake of contradiction, that (1.1) has an eventually positive solution \{x_n\}. We consider the two possible cases: CASE 1. \mu=0. In this case we have \overline{d}(k\mu)=d(\mu)=1. From (1.1), we have$$ x_{n-i}=x_{n-i+1}+p_{n-i}x_{n-k-i}, \quad i=1,2,\dots ,k\,. $$Summing both sides of the above equality from i=1 to i=k leads to$$ x_{n-k}=x_n+\sum_{i=1}^kp_{n-i}x_{n-k-i}. \eqno (3.5) $$From (1.1), for any positive integer j, we have$$ x_{n-k-j}=x_{n-k-j+1}+p_{n-k-j}x_{n-k-j-k}. \eqno (3.6) $$Substituting (3.6) for j=i into (3.5), we have$$ x_{n-k}=x_n+\sum_{i=1}^kp_{n-i}x_{n-k-i+1}+\sum_{i=1}^kp_{n-i}p_{n-k- i}x_{n-i-2k}. $$Substituting (3.6) for j=i+k into the last equality, we have \begin{eqnarray*} x_{n-k}&=&x_n+\sum_{i=1}^kp_{n-i}x_{n-k-i+1}+\sum_{i=1}^kp_{n-i}p_{n-k- i}x_{n-2k-i+1}\\ &&+\sum_{i=1}^kp_{n-i}p_{n-k-i}p_{n-2k-i}x_{n-i-3k}. \end{eqnarray*} By induction, it is easy to prove that \begin{eqnarray*} x_{n-k}&=&x_n+\sum_{i=1}^kp_{n-i}x_{n-k-i+1}+\sum_{i=1}^kp_{n-i}p_{n-k- i}x_{n-2k-i+1}\\ &&+\sum_{i=1}^kp_{n-i}p_{n-k-i}p_{n-2k-i}x_{n-3k-i+1}+\cdots \\ &&+\sum_{i=1}^kp_{n-i}p_{n-k-i}\cdots p_{n-lk-i}x_{n-(l+1)k-i+1}\\ &&+\sum_{i=1}^kp_{n-i}p_{n-k-i}\cdots p_{n-(l+1)k-i}x_{n-i-(l+2)k}\,. \end{eqnarray*} Removing the last term of the last equality, we have$$ x_{n-k}\geq x_n+\sum_{i=1}^kp_{n-i}x_{n-k-i+1}+\sum_{m=0}^{l-1}\sum_{i= 1}^kx_{n-(m+2)k-i+1}\prod_{j=0}^{m+1}p_{n-jk-i}. \eqno (3.7) $$In the proof of Lemma 2.2, we have (2.14) holds. Using the Arithmetic-Geometric Mean Inequality in (2.14), we have$$ \left(\frac{x_{n-k}}{x_n}\prod_{i=1}^k\sum_{j=1}^kp_{n-i+j}\right)^{1/k} <1, $$and so$$ x_n>\left(\prod_{i=1}^k\sum_{j=1}^kp_{n-i+ j}\right)x_{n-k}. \eqno (3.8) $$Substituting (3.8) into (3.7) and using the fact that \{x_n\} is eventually decreasing, we have$$ x_{n-k}>\left(\sum_{i=1}^kp_{n-i}+ \prod_{i=1}^k\sum_{j=1}^kp_{n-i+j} +\sum_{m=0}^{l-1}\sum_{i=1}^k\prod_{j=0}^{m+1}p_{n- jk-i}\right)x_{n-k}. $$Dividing both sides of the last inequality by x_{n-k}, and taking the limit superior as n\to \infty, we have$$ 1\geq \limsup_{n\to \infty}\left\{\sum_{i=1}^kp_{n-i}+ \prod_{i=1}^k\sum_{j=1}^kp_{n-i+j} +\sum_{m=0}^{l-1}\sum_{i=1}^k\prod_{j=0}^{m+1}p_{n- jk-i}\right\}. $$This contradicts (3.2). CASE 2. 0<\mu\leq k^k/(k+1)^{k+1}. In this case, for any \eta\in (0,\mu), we have$$ \frac{1}{k}\sum_{i=1}^kp_{n-i}\geq \mu-\eta. \eqno (3.9) $$From (3.7), we have$$ x_{n-k}\geq x_n +\sum_{i=1}^kp_{n-i}x_{n-k}+\sum_{m=0}^{l-1}x_{n-(m+2)k}\sum_{i= 1}^k\prod_{j=0}^{m+1}p_{n-jk-i}.\eqno (3.10) $$By Lemma 2.2, we have$$ x_n\geq \{[\overline{d}(k(\mu-\eta))]^{-k}-\eta\}\prod_{i= 1}^k\sum_{j=1}^kp_{n-i +j}x_{n-k}, \eqno (3.11) $$where \overline{d}(k(\mu-\eta)) is the greater real root of the equation$$ d^{k+1}-d^k+k^k(\mu-\eta)^k=0. \eqno (3.12) $$By Lemma 2.1, we have$$ x_{n-(m+2)k}\geq \{[d(\mu-\eta)]^{-(m+1)k}-\eta\}x_{n- k}, \eqno (3.13) $$where d(\mu-\eta) is the greater real root of the equation$$ d^{k+1}-d^k+(\mu-\eta)=0. \eqno (3.14) $$Now substituting (3.11) and (3.13) into (3.10), we obtain \begin{eqnarray*} x_{n-k}&\geq&\sum_{i=1}^kp_{n-i}x_{n-k}+\{[\overline{d}(k(\mu-\eta))]^{- k}-\eta\}\prod_{i=1}^k\sum_{j=1}^kp_{n-i+j}x_{n-k}\\ &&+\sum_{m=0}^{l-1}\{[d(\mu-\eta)]^{-(m+1)k}-\eta\}\sum_{i= 1}^k\prod_{j=0}^{m+ 1}p_{n-jk-i}x_{n-k}. \end{eqnarray*} Dividing both sides of the last inequality by x_{n-k} then taking the limit superior as n\to \infty, we have$$ 1\geq \limsup_{n\to \infty}\left\{\sum_{i=1}^kp_{n-i}+ \{[\overline{d}(k(\mu-\eta))]^{- k}-\eta\}\prod_{i=1}^k\sum_{j=1}^kp_{n-i+j}\right.  \left.+\sum_{m=0}^{l-1}\{[d(\mu-\eta)]^{-(m+1)k}-\eta\}\sum_{i= 1}^k\prod_{j=0}^{m+ 1}p_{n-jk-i}\right\}. $$Letting \eta\to 0, we have \overline{d}(k(\mu-\eta))\to \overline{d}( k\mu) and d(\mu-\eta)\to d(\mu), so that the last inequality contradicts (3.2). The proof is now complete. \vskip 0.1cm Notice that when k=1, from Remark 2.1 and Remark 2.2, we have d(\mu)=\overline{d}(\mu)=(1+\sqrt{1-4\mu})/2, so condition (3.2) reduces to$$ \limsup_{n\to \infty}\left\{Cp_n+p_{n-1}+\sum_{m=0}^{l-1}C^{m+1}\prod_{j= 0}^{m+1}p_{n-j-1}\right\}>1, \eqno (3.15) $$where C=2/(1+\sqrt{1-4\mu}), \mu=\liminf_{n\to \infty}p_n. Therefore, from Theorem 3.1, we have the following corollary. \paragraph{Corollary 3.1} {\it Assume that 0\leq \mu\leq 1/4 and that {\rm (3.15)} holds. Then all solutions of the equation$$ x_{n+1}-x_n+p_nx_{n-1}=0 \eqno (3.16) $$oscillate.} \vskip 0.1cm A condition obtained from (3.15) and whose checking is more easy is given in next corollary. \paragraph{\bf Corollary 3.2} {\it Assume that 0\leq \mu\leq 1/4 and that$$ \limsup_{n\to \infty}p_n>\left(\frac{1+\sqrt{1-4\mu}}{2}\right)^2. \eqno (3.17) $$Then all solutions of {\rm (3.16)} oscillate.} \vskip 0.1cm \noindent {\it Proof}. When \mu=0, by condition (1.3), all solutions of (3.17) oscillate. For the case when 0<\mu\leq 1/4, by Theorem 3.1, it suffices to prove that (3.17) implies (3.15). Notice$$ \frac{1+\sqrt{1-4\mu}}{2}=1-\frac{\mu}{1-C\mu}, $$by (3.17) and \mu=\liminf_{n\to \infty}p_n, there exists \varepsilon\in (0,\mu) such that p_n\geq \mu-\varepsilon and$$ C\limsup_{n\to \infty}p_n>1-\frac{\mu-\varepsilon}{1- C(\mu-\varepsilon)}. $$The last inequality, in view of the fact that [C(\mu-\varepsilon)]^m\to 0 as m\to \infty, implies that for some sufficiently large integer l>1 \begin{eqnarray*} C\limsup_{n\to \infty}p_n&>&1-\frac{(\mu-\varepsilon)\{1 -[C(\mu-\varepsilon)]^{l+1}\}}{1- C(\mu-\varepsilon)}\\ &=&1-(\mu-\varepsilon)-C(\mu-\varepsilon)^2-\cdots - C^l(\mu-\varepsilon)^{l+1}, \end{eqnarray*} which leads to (3.15), because$$ p_{n-1}+\sum_{m=0}^{l-1}C^{m+1}\prod_{j=0}^{m+1}p_{n-j-1}\geq (\mu-\varepsilon)+C(\mu-\varepsilon)^2+\cdots + C^l(\mu-\varepsilon)^{l+1}. $$The proof is complete. Observe that when \mu=1/4, condition (3.17) reduces to \limsup_{n\to \infty}p_n>1/4, which can not be improved in the sense that the lower bound 1/4 can not be replaced by a smaller number. Indeed, by Theorem 2.3 in [9], we see that (3.16) has a non-oscillatory solution if p_n\leq 1/4 for large n. Note, however, that even in the critical state \lim_{n\to \infty}p_n=1/4 (3.16) can be either oscillatory or non-oscillatory. For example, if p_n=\frac{1}{4}+\frac{c}{n^2} then (3.16) will be oscillatory in case c>1/4 and non-oscillatory in case c<1/4 (the Kneser-like theorem, [3]). \paragraph{Example.} Consider the equation$$ x_{n+1}-x_n+\left(\frac{1}{4}+a\sin^4\frac{n\pi}{8}\right)x_{n-1}=0, $$where a>0 is a constant. It is easy to see that$$ \liminf_{n\to \infty}p_n=\liminf_{n\to \infty}\left(\frac{1}{4} +a\sin^4\frac{n\pi}{8}\right)=\frac{1}{4},  \limsup_{n\to \infty}p_n=\limsup_{n\to \infty}\left(\frac{1}{4} +a\sin^4\frac{n\pi}{8}\right)=\frac{1}{4}+a. $$Therefore, by Corollary 3.2, all solutions of the equation oscillate. However, none of the conditions (1.3)-(1.5) and those appear in [4,20,23] is satisfied. \vskip 0.1cm The following corollary concerns the case when k>1. \paragraph{Corollary 3.3} {\it Assume that 0\leq \mu\leq k^k/(k+1)^{k +1} and that$$ \limsup_{n\to \infty}\sum_{i=1}^kp_{n-i}>1-[\overline{d}(k\mu)]^{-k}( k\mu)^{k}-\frac{k[d(\mu)]^{-k}\mu_{\ast}^2}{1-[ d(\mu)]^{-k}\mu_{\ast}},\eqno (3.18) $$where \mu_{\ast}=\liminf_{n\to \infty}p_n, and \overline{d}(k\mu), d (\mu) are as in Theorem {\rm 3.1}. Then all solutions of {\rm (1.1)} oscillate.} \vskip 0.1cm \noindent {\it Proof}. If \mu=0 (then \mu_{\ast}=0 ), then, by (1.3), all solutions of (1.1) oscillate. If \mu_{\ast}=0,\mu>0, then (3.18) reduces to$$ \limsup_{n\to \infty}\sum_{i=1}^kp_{n-i}>1-[\overline{d}(k\mu)]^{-k}( k\mu)^{k}.\eqno (3.19) $$From (3.1) and (3.19), for some sufficiently small \eta\in (0,\mu) we have$$ \frac{1}{k}\sum_{i=1}^kp_{n-i}\geq \mu-\eta,\quad \limsup_{n\to \infty}\sum_{i=1}^kp_{n-i}>1-[\overline{d}(k\mu)]^{-k}( k(\mu-\eta))^{k}.\eqno (3.20) $$Thus, we obtain$$ [\overline{d}(k\mu)]^{-k}\prod_{i=1}^k\sum_{j=1}^kp_{n-i+j}\geq [\overline{d}(k\mu)]^{-k}(k(\mu-\eta))^{k}. $$From this and the second inequality of (3.20), we see that (3.2) holds. By Theorem 3.1, all solutions of (1.1) oscillate. We now consider the case when 0<\mu_{\ast}\leq k^k/(k+ 1)^{k+1}. By Theorem 3.1, it suffices to prove that condition (3.18) implies condition (3.2). From (3.18), it follows that, for some sufficiently small \eta\in (0,\mu_{\ast}) we have$$ \limsup_{n\to \infty}\sum_{i=1}^kp_{n-i}>1-[\overline{d}(k\mu)]^{-k}( k(\mu-\eta))^{k}-\frac{k[d(\mu)]^{-k}(\mu_{\ast}-\eta)^2}{1-[ d(\mu)]^{-k}(\mu_{\ast}-\eta)}.  This, in view of the fact that $[[d(\mu)]^{-k}(\mu_{\ast}-\eta)]^m\to 0$ as $m\to \infty$, implies that for some sufficiently large integer $l>1$ \begin{eqnarray*} \limsup_{n\to \infty}\sum_{i=1}^kp_{n-i}&>&1-[\overline{d}(k\mu)]^{-k}( k(\mu-\eta))^{k}\\ &&-\frac{k(\mu_{\ast}-\eta)^2[d(\mu)]^{-k}\{1-[ [d(\mu)]^{-k}(\mu_{\ast}-\eta)]^l\}}{1-[d(\mu)]^{-k}(\mu_{\ast}-\eta)}\\ &=&1-[\overline{d}(k\mu)]^{-k}(k(\mu-\eta))^{k}-k(\mu_{\ast}-\eta)^2[ d(\mu)]^{-k}\\ &&\times \{1+[d(\mu)]^{-k}(\mu_{\ast}-\eta)+[d(\mu)]^{-2k}(\mu_{\ast} -\eta)^2\\ &&+\cdots +[d(\mu)]^{-(l-1)k}(\mu_{\ast}-\eta)^{l-1}\}. \end{eqnarray*} This leads to (3.2) because \begin{eqnarray*} \lefteqn{ [\overline{d}(k\mu)]^{-k}\prod_{i=1}^k\sum_{j=1}^kp_{n-i+j}+\sum_{m= 0}^{l-1}[d(\mu)]^{-(m+1)k}\sum_{i=1}^k\prod_{j=0}^{m+1}p_{n-jk-i} }\\ &\geq&[\overline{d}(k\mu)]^{-k}(k(\mu-\eta))^{k}+k(\mu_{\ast}-\eta)^2[ d(\mu)]^{-k}+k(\mu_{\ast}-\eta)^3[d(\mu)]^{-2k}\\ &&+\cdots +k(\mu_{\ast}-\eta)^{l+1}[d(\mu)]^{-lk}. \end{eqnarray*} The proof is complete. \begin{thebibliography}{00} \bibitem{a1} R.P. 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Appl.}, {\bf 177} (1993), 432-444. \end{thebibliography} \noindent{\sc Jianhua Shen} \\ Department of Mathematics, Hunan Normal University \\ Changsha, Hunan 410081, China \\ e-mail: jhsh@public.cs.hn.cn \smallskip \noindent{\sc Ioannis P. Stavroulakis} \\ Department of Mathematics, University of Ioannina \\ 451 10 Ioannina, Greece \\ e-mail: ipstav@cc.uoi.gr \end{document}