\documentclass[twoside]{article}
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\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} \\
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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{}
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\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.
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\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}