In this section we study the existence of bounded solutions to equation (1). The cases when the domains are \({\mathbb {N}}_{0}\) and \({\mathbb {Z}}\setminus {\mathbb {N}}\) are treated separately, while the results in the case when the domain is \({\mathbb {Z}}\) are obtained as some consequences of the considerations on the domains \({\mathbb {N}}_{0}\) and \({\mathbb {Z}}\setminus {\mathbb {N}}\).
Our first result is, among others, an extension of the result in Problem 1, so it could be folklore.
Theorem 1
Assume that
$$\begin{aligned} \limsup_{n\to+\infty} \vert q_{n} \vert :=q< 1 \end{aligned}$$
(4)
and that
\((f_{n})_{n\in {\mathbb {N}}_{0}}\)
is a bounded sequence. Then the following statements are true.

(a)
Every solution to equation (1) is bounded.

(b)
If
\(\lim_{n\to+\infty}f_{n}=0\), then every solution to equation (1) converges to zero.
Proof
(a) Since (4) holds, we have that there is \(n_{1}\in {\mathbb {N}}\) such that
$$\begin{aligned} \vert q_{n} \vert \le\frac{1+q}{2}\quad\mbox{for }n\ge n_{1}. \end{aligned}$$
(5)
Let
$$\begin{aligned} M_{1}:=\max\Bigl\{ \sqrt{2},\max_{j=\overline {0,n_{1}1}} \vert q_{j} \vert \Bigr\} . \end{aligned}$$
(6)
Using (5) and (6) in (2), as well as some standard estimates and sums, we have
$$\begin{aligned} \vert x_{n} \vert \le{}& \vert x_{0} \vert \prod _{j=0}^{n1} \vert q_{j} \vert + \sum_{i=0}^{n1} \vert f_{i} \vert \prod_{j=i+1}^{n1} \vert q_{j} \vert \end{aligned}$$
(7)
$$\begin{aligned} \le{}& \vert x_{0} \vert \prod_{j=0}^{n1} \vert q_{j} \vert + \Vert f \Vert _{\infty}\Biggl(\sum _{i=0}^{n_{1}2}\prod _{j=i+1}^{n1} \vert q_{j} \vert +\sum _{i=n_{1}1}^{n1}\prod_{j=i+1}^{n1} \vert q_{j} \vert \Biggr) \\ \le{}& \vert x_{0} \vert \biggl(\frac{1+q}{2} \biggr)^{nn_{1}}M_{1}^{n_{1}}+ \Vert f \Vert _{\infty}\biggl(\frac{1+q}{2} \biggr)^{nn_{1}}\sum _{i=0}^{n_{1}2}M_{1}^{n_{1}i1} \\ &{}+ \Vert f \Vert _{\infty}\sum_{i=n_{1}1}^{n1} \biggl(\frac{1+q}{2} \biggr)^{ni1} \\ \le{}& \vert x_{0} \vert \biggl(\frac{2M_{1}}{1+q} \biggr)^{n_{1}}+M_{1} \Vert f \Vert _{\infty}\biggl( \frac{2}{1+q} \biggr)^{n_{1}}\frac{M_{1}^{n_{1}1}1}{M_{1}1} +\frac{2 \Vert f \Vert _{\infty}}{1q}< \infty \end{aligned}$$
(8)
for \(n\ge n_{1}\), from which the boundedness of \((x_{n})_{n\in {\mathbb {N}}_{0}}\) follows.
(b) Since \(f_{n}\) converges to zero, we have that for each \(\varepsilon >0\), there is \(n_{2}\in {\mathbb {N}}\) such that
$$\begin{aligned} \vert f_{n} \vert < \varepsilon \quad\mbox{for }n\ge n_{2}. \end{aligned}$$
(9)
Let \(n_{3}=\max\{n_{1},n_{2}\}\), where \(n_{1}\) is from (5), and
$$\begin{aligned} M_{2}:=\max\Bigl\{ \sqrt{2},\max_{j=\overline {0,n_{3}1}} \vert q_{j} \vert \Bigr\} . \end{aligned}$$
(10)
Using (5), (9) and (10) in (2), as well as some standard estimates and sums, we have
$$\begin{aligned} \vert x_{n} \vert \le{}& \vert x_{0} \vert \biggl( \frac{1+q}{2} \biggr)^{nn_{3}}M_{2}^{n_{3}}+ \Vert f \Vert _{\infty}\sum_{i=0}^{n_{3}1}\prod _{j=i+1}^{n1} \vert q_{j} \vert + \varepsilon \sum_{i=n_{3}}^{n1}\prod _{j=i+1}^{n1} \vert q_{j} \vert \\ \le{}& \vert x_{0} \vert \biggl(\frac{1+q}{2} \biggr)^{nn_{3}} M_{2}^{n_{3}}+ \Vert f \Vert _{\infty}\biggl(\frac{1+q}{2} \biggr)^{nn_{3}} \sum _{i=0}^{n_{3}1} M_{2}^{n_{3}i1} +\varepsilon \sum _{i=n_{3}}^{n1} \biggl(\frac{1+q}{2} \biggr)^{ni1} \\ \le{}& \vert x_{0} \vert \biggl(\frac{2M_{2}}{1+q} \biggr)^{n_{3}} \biggl(\frac{1+q}{2} \biggr)^{n}+ \Vert f \Vert _{\infty}\biggl(\frac{2}{1+q} \biggr)^{n_{3}} \frac {M_{2}^{n_{3}}1}{M_{2}1} \biggl(\frac{1+q}{2} \biggr)^{n}+ \frac{2\varepsilon }{1q} \end{aligned}$$
(11)
for \(n\ge n_{3}\).
Letting \(n\to+\infty\) in (11), we get
$$\limsup_{n\to+\infty} \vert x_{n} \vert \le \frac{2\varepsilon }{1q}. $$
From this, and since ε is an arbitrary positive number, the result follows. □
Remark 1
Theorem 1 is optimal in the sense that condition (4) cannot be replaced by the following one: there is \(n_{4}\in {\mathbb {N}}_{0}\) such that
$$\begin{aligned} \vert q_{n} \vert < 1\quad\mbox{for }n\ge n_{4}. \end{aligned}$$
(12)
Indeed, assume that \(q_{n}\in(0,1)\), \(n\in {\mathbb {N}}_{0}\), is an increasing sequence such that the sequence
$$Q_{n}:=\prod_{j=0}^{n1}q_{j},\quad n\in {\mathbb {N}}_{0}, $$
converges to \(Q\in(0,1)\) as \(n\to+\infty\), and that there are some numbers l and L such that
$$\begin{aligned} 0< l\le f_{n}\le L< \infty,\quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(13)
Then, by using (2) and the fact \(Q\le Q_{n}\le1\), \(n\in {\mathbb {N}}_{0}\), we have
$$\begin{aligned} \vert x_{n} \vert =Q_{n} \Biggl\vert x_{0}+ \sum_{j=0}^{n1}\frac {f_{j}}{Q_{j+1}} \Biggr\vert \ge Q \bigl\vert  \vert x_{0} \vert +nl \bigr\vert \to+ \infty \end{aligned}$$
(14)
as \(n\to+\infty\), from which it follows that not only there are unbounded solutions to equation (1) in this case, but that even none of the solutions to the equation in the case is bounded.
For example, let
$$q_{n}=1\frac{1}{(n+2)^{2}}, \quad n\in {\mathbb {N}}_{0} $$
and
$$f_{n}=2+\sin n,\quad n\in {\mathbb {N}}_{0}. $$
Then
$$Q_{n}=\prod_{j=0}^{n1} \biggl(1 \frac{1}{(j+2)^{2}} \biggr)=\prod_{j=0}^{n1} \frac{(j+1)(j+3)}{(j+2)^{2}}=\frac {n!(n+2)!}{2(n+1)!^{2}}=\frac{n+2}{2(n+1)} $$
for \(n\in {\mathbb {N}}_{0}\), from which it follows that \(\lim_{n\to+\infty}Q_{n}=1/2\),
$$\begin{aligned} \frac{1}{2}\le Q_{n}\le1,\quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(15)
and along with (2) that
$$\begin{aligned} x_{n}=\frac{n+2}{2(n+1)} \Biggl(x_{0}+\sum _{j=0}^{n1}(2+\sin j)\frac{2(j+2)}{j+3} \Biggr). \end{aligned}$$
(16)
Using (15), the following obvious estimate
$$\begin{aligned} 1\le f_{n}\le3, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(17)
and the triangle inequality in (16), we get
$$\vert x_{n} \vert \ge\frac{1}{2} \biggl(\frac{4}{3} n \vert x_{0} \vert \biggr),\quad n\in {\mathbb {N}}_{0}, $$
from which it follows that each solution to equation (1) in this case is unbounded indeed.
If we assume that \(f_{n}\) is a positive sequence such that \(\lim_{n\to +\infty}f_{n}=0\) and \(\sum_{j=0}^{\infty}f_{j}=+\infty\) (for example, \(f_{n}=1/(n+1)\), \(n\in {\mathbb {N}}_{0}\)), and that \(q_{n}\) is chosen as above, then from (2) and the fact \(Q\le Q_{n}\le1\), \(n\in {\mathbb {N}}_{0}\), we have
$$\vert x_{n} \vert =Q_{n} \Biggl\vert x_{0}+ \sum_{j=0}^{n1}\frac{f_{j}}{Q_{j+1}} \Biggr\vert \ge Q \Biggl\vert  \vert x_{0} \vert +\sum _{j=0}^{n1}f_{j} \Biggr\vert \to+\infty $$
as \(n\to+\infty\), from which it follows that none of the solutions to equation (1) in this case converges to zero. Specially, every solution to the difference equation
$$x_{n+1}=\frac{(n+1)(n+3)}{(n+2)^{2}}x_{n}+\frac{1}{n+1},\quad n\in {\mathbb {N}}_{0}, $$
is unbounded.
Now we consider the case \(\liminf_{n\to+\infty} \vert q_{n} \vert >1\). In this case, we may assume that \(q_{n}\ne0\), \(n\in {\mathbb {N}}_{0}\), otherwise we can consider (1) for sufficiently large n for which, due to the condition \(\liminf_{n\to+\infty} \vert q_{n} \vert >1\), will hold \(q_{n}\ne 0\).
Theorem 2
Assume that
\((q_{n})_{n\in {\mathbb {N}}_{0}}\subset {\mathbb {C}}\setminus\{0\}\)
is a sequence satisfying the following condition:
$$\begin{aligned} \liminf_{n\to+\infty} \vert q_{n} \vert :=\hat{q}>1, \end{aligned}$$
(18)
and that
\((f_{n})_{n\in {\mathbb {N}}_{0}}\)
is a bounded sequence of complex numbers. Then the following statements are true.

(a)
There is a unique bounded solution to equation (1).

(b)
If
\(f_{n}\to0\)
as
\(n\to+\infty\), then the bounded solution also converges to zero as
\(n\to+\infty\).
Proof
(a) From (18), we have
$$\begin{aligned} \lim_{n\to+\infty} \Biggl\vert \prod_{j=0}^{n1}q_{j} \Biggr\vert =+\infty . \end{aligned}$$
(19)
Thus, from (2) and (19) we see that for a bounded solution to (1) it must be
$$\begin{aligned} x_{0}=\sum_{i=0}^{\infty}\frac{f_{i}}{\prod_{j=0}^{i}q_{j}}, \end{aligned}$$
(20)
and that the sum on the righthand side of (20) is finite (see [18]).
Using (20) in (2), it follows that
$$\begin{aligned} x_{n}=\prod_{j=0}^{n1}q_{j} \sum_{i=n}^{\infty}\frac {f_{i}}{\prod_{j=0}^{i}q_{j}}=\sum _{i=n}^{\infty}\frac{f_{i}}{\prod_{j=n}^{i}q_{j}},\quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(21)
Condition (18) implies that there is \(n_{5}\in {\mathbb {N}}\) such that
$$\begin{aligned} \vert q_{n} \vert \ge\frac{1+\hat{q}}{2}>1\quad\mbox{for }n\ge n_{5}. \end{aligned}$$
(22)
Let
$$\begin{aligned} M_{3}:=\min\Bigl\{ 1/\sqrt{2}, \min_{j=\overline {0,n_{5}1}} \vert q_{j} \vert \Bigr\} . \end{aligned}$$
(23)
Note that \(M_{3}>0\) due to the assumption \(q_{n}\ne0\), \(n\in {\mathbb {N}}_{0}\).
Hence, by using (22) and (23), we have
$$\begin{aligned} \Biggl\vert \sum_{i=0}^{\infty}\frac{f_{i}}{\prod_{j=0}^{i}q_{j}} \Biggr\vert \le{}& \Vert f \Vert _{\infty}\sum _{i=0}^{\infty}\frac{1}{\prod_{j=0}^{i} \vert q_{j} \vert } \\ \le{}& \Vert f \Vert _{\infty}\Biggl(\sum_{i=0}^{n_{5}1} \frac{1}{\prod_{j=0}^{i} \vert q_{j} \vert }+\frac{1}{M_{3}^{n_{5}}}\sum_{i=n_{5}}^{\infty}\biggl(\frac{2}{1+\hat{q}} \biggr)^{in_{5}+1} \Biggr) \\ \le{}& \Vert f \Vert _{\infty}\biggl(\frac{1}{M_{3}^{n_{5}}(1M_{3})}+ \frac {2}{M_{3}^{n_{5}}(\hat{q}1)} \biggr)< \infty, \end{aligned}$$
(24)
from which it follows that \(x_{0}\) defined in (20) is finite.
From (21) and (22), we have
$$\vert x_{n} \vert \le\sum_{i=n}^{\infty}\frac{ \vert f_{i} \vert }{\prod_{j=n}^{i} \vert q_{j} \vert }\le \Vert f \Vert _{\infty}\sum _{i=n}^{\infty}\biggl(\frac{2}{1+\hat{q}} \biggr)^{i+1n}=\frac {2 \Vert f \Vert _{\infty}}{\hat{q}1}, $$
for \(n\ge n_{5}\), from which the boundedness of the sequence in (21) follows. It is directly verified that the sequence satisfies equation (1), from which along with the unique choice of \(x_{0}\) in (20) it follows that it is a unique bounded solution to the equation.
(b) Since \(f_{n}\) tends to zero as \(n\to+\infty\), it follows that (9) holds for, say, \(n\ge n_{6}\). Using (9) and (22) in (21), we have
$$\begin{aligned} \vert x_{n} \vert \le& \sum_{i=n}^{\infty}\frac{ \vert f_{i} \vert }{\prod_{j=n}^{i} \vert q_{j} \vert }< \varepsilon \sum_{i=n}^{\infty}\biggl( \frac{2}{1+\hat{q}} \biggr)^{i+1n}=\frac{2\varepsilon }{\hat{q}1} \end{aligned}$$
(25)
for \(n\ge\max\{n_{5},n_{6}\}\).
Letting \(n\to+\infty\) in (25) the following is obtained:
$$\limsup_{n\to+\infty} \vert x_{n} \vert \le \frac{2\varepsilon }{\hat{q}1}. $$
From this and by using the fact that ε is an arbitrary positive number, we obtain that \(x_{n}\to0\) as \(n\to+\infty\), as desired. □
Remark 2
Theorem 2 is optimal in the sense that condition (18) cannot be replaced by the following one: there is \(n_{7}\in {\mathbb {N}}_{0}\) such that
$$\begin{aligned} \vert q_{n} \vert >1\quad\mbox{for }n\ge n_{7}. \end{aligned}$$
(26)
Indeed, assume that \(q_{n}>1\), \(n\in {\mathbb {N}}_{0}\), is an decreasing sequence such that the sequence
$$Q_{n}:=\prod_{j=0}^{n1}q_{j},\quad n\in {\mathbb {N}}_{0}, $$
converges to \(Q>1\) as \(n\to+\infty\), and that there are some numbers l and L such that (13) holds. Then, by using (2) and the fact \(Q\ge Q_{n}\ge1\), \(n\in {\mathbb {N}}_{0}\), we have
$$\begin{aligned} \vert x_{n} \vert =Q_{n} \Biggl\vert x_{0}+ \sum_{j=0}^{n1}\frac {f_{j}}{Q_{j+1}} \Biggr\vert \ge \bigl\vert  \vert x_{0} \vert +nl/Q \bigr\vert \to+ \infty \end{aligned}$$
(27)
as \(n\to+\infty\), from which it follows that all the solutions to equation (1) are unbounded in the case, so, it does not have bounded solutions.
For example, let
$$q_{n}=1+\frac{1}{n^{2}+4n+3},\quad n\in {\mathbb {N}}_{0}, $$
and
$$f_{n}=5+2\cos n, \quad n\in {\mathbb {N}}_{0}. $$
Then
$$Q_{n}=\prod_{j=0}^{n1} \biggl(1+ \frac{1}{j^{2}+4j+3} \biggr)=\prod_{j=0}^{n1} \frac{(j+2)^{2}}{(j+1)(j+3)}=\frac{2(n+1)}{n+2} $$
for \(n\in {\mathbb {N}}_{0}\), from which it follows that \(\lim_{n\to+\infty}Q_{n}=2\),
$$\begin{aligned} 1\le Q_{n}\le2,\quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(28)
and along with (2) that
$$\begin{aligned} x_{n}=\frac{2(n+1)}{n+2} \Biggl(x_{0}+\sum _{j=0}^{n1}(5+2\cos j)\frac{j+3}{2(j+2)} \Biggr). \end{aligned}$$
(29)
Using (28), the following obvious estimate
$$\begin{aligned} 3\le f_{n}\le7, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(30)
and the triangle inequality in (29), we get
$$\vert x_{n} \vert \ge\frac{3}{2} n \vert x_{0} \vert ,\quad n\in {\mathbb {N}}_{0}, $$
from which it follows that each solution to equation (1) in this case is unbounded.
If we assume that \(f_{n}\) is a positive sequence such that \(\lim_{n\to +\infty}f_{n}=0\) and \(\sum_{j=0}^{\infty}f_{j}=+\infty\) (for example, \(f_{n}=1/\ln(n+2)\), \(n\in {\mathbb {N}}_{0}\)), and that \(q_{n}\) is chosen as above, then from (2) and the fact \(1\le Q_{n}\le Q\), \(n\in {\mathbb {N}}_{0}\), we have
$$\vert x_{n} \vert =Q_{n} \Biggl\vert x_{0}+ \sum_{j=0}^{n1}\frac{f_{j}}{Q_{j+1}} \Biggr\vert \ge \Biggl\vert  \vert x_{0} \vert +\frac{1}{Q} \sum_{j=0}^{n1}f_{j} \Biggr\vert \to+\infty, $$
as \(n\to+\infty\), from which it follows that none of the solutions to equation (1) in this case converges to zero. Specially, every solution to the difference equation
$$x_{n+1}=\frac{(n+2)^{2}}{(n+1)(n+3)}x_{n}+\frac{1}{\ln(n+2)},\quad n\in {\mathbb {N}}_{0}, $$
is unbounded.
Now we consider the case when \(n\in {\mathbb {Z}}\setminus {\mathbb {N}}\). If in (1) is \(q_{n}\ne0\) for every \(n\in {\mathbb {Z}}\setminus {\mathbb {N}}_{0}\), then the sequence \(x_{n}\) is not only welldefined on the set \({\mathbb {N}}_{0}\), but also for every \(n\in {\mathbb {Z}}\). Indeed, if \(n\le0\), then from (1) we have
$$\begin{aligned} x_{n}=\frac{x_{(n1)}}{q_{n}}\frac {f_{n}}{q_{n}},\quad n\in {\mathbb {N}}. \end{aligned}$$
(31)
Using one of the methods for solving equation (1), from (31) the following is obtained:
$$\begin{aligned} x_{n}=\frac{x_{0}\sum_{j=1}^{n}f_{j}\prod_{l=1}^{j1}q_{l}}{\prod_{j=1}^{n}q_{j}} \end{aligned}$$
(32)
for \(n\in {\mathbb {N}}\).
Closed form formulas (2) and (32) together present the general solution to equation (1) on \({\mathbb {Z}}\), when \(q_{n}\ne0\), for \(n\in {\mathbb {Z}}\setminus {\mathbb {N}}_{0}\).
Now we formulate and prove the corresponding results to Theorems 1 and 2 concerning bounded solutions to equation (31). The results are dual to Theorems 1 and 2 and are essentially obtained from them by using the change of variables \(y_{n}=x_{n}\). However, there are some different details which are used later in the text. Because of this and for the completeness, we will sketch their proofs.
First, we consider equation (31) for the case
$$\begin{aligned} \liminf_{n\to\infty} \vert q_{n} \vert =:\tilde{q}>1. \end{aligned}$$
(33)
Theorem 3
Assume that
\((q_{n})_{n\in {\mathbb {N}}}\subset {\mathbb {C}}\setminus\{0\}\)
is a sequence satisfying condition (33), and that
\((f_{n})_{n\in {\mathbb {N}}}\)
is a bounded sequence of complex numbers. Then the following statements are true.

(a)
Every solution to (31) is bounded.

(b)
If
\(\lim_{n\to+\infty}f_{n}=0\), then every solution to (31) converges to zero as
\(n\to+\infty\).
Proof
(a) From (33) it follows that there is \(n_{8}\in {\mathbb {N}}\) such that
$$\begin{aligned} \frac{1}{ \vert q_{n} \vert }\le\frac{2}{1+\tilde{q}}\quad\mbox{for }n\ge n_{8}. \end{aligned}$$
(34)
Let
$$\begin{aligned} M_{5}:=\min\Bigl\{ 1/\sqrt{2},\min_{j=\overline {1,n_{8}1}} \vert q_{j} \vert \Bigr\} . \end{aligned}$$
(35)
Using (34) and (35) in (32), as well as some standard estimates and sums, we have
$$\begin{aligned} \vert x_{n} \vert \le{}&\frac{ \vert x_{0} \vert }{\prod_{j=1}^{n} \vert q_{j} \vert }+\sum _{j=1}^{n}\frac { \vert f_{j} \vert }{\prod_{l=j}^{n} \vert q_{l} \vert } \\ \le{}&\frac{ \vert x_{0} \vert }{M_{5}^{n_{8}1}}+ \frac{ \Vert f \Vert _{\infty}}{(1M_{5})M_{5}^{n_{8}1}}+\frac{2 \Vert f \Vert _{\infty}}{\tilde{q}1}< \infty \end{aligned}$$
(36)
for \(n\ge n_{8}\), from which the boundedness of \((x_{n})_{n\in {\mathbb {N}}_{0}}\) follows.
(b) Since \(f_{n}\) converges to zero as \(n\to+\infty\), we have that for every \(\varepsilon >0\), there is \(n_{9}\in {\mathbb {N}}\) such that
$$\begin{aligned} \vert f_{n} \vert < \varepsilon \quad\mbox{for }n\ge n_{9}. \end{aligned}$$
(37)
Let \(n_{10}=\max\{n_{8},n_{9}\}\), where \(n_{8}\) is from (34), and
$$\begin{aligned} M_{6}:=\min\Bigl\{ 1/\sqrt{2},\min_{j=\overline {1,n_{10}1}} \vert q_{j} \vert \Bigr\} . \end{aligned}$$
(38)
Using (34), (37) and (38) in (36), as well as some standard estimates and sums, we have
$$\begin{aligned} \vert x_{n} \vert \le \vert x_{0} \vert \biggl(\frac{1+\tilde{q}}{2M_{6}} \biggr)^{n_{10}1} \biggl(\frac{2}{1+\tilde{q}} \biggr)^{n}+ \Vert f \Vert _{\infty}\biggl(\frac{1+\tilde{q}}{2M_{6}} \biggr)^{n_{10}1} \biggl( \frac{2}{1+\tilde{q}} \biggr)^{n}\frac{1}{1M_{6}}+\frac{2\varepsilon }{\tilde{q}1} \end{aligned}$$
(39)
for \(n\ge n_{10}\).
Letting \(n\to+\infty\) in (39) and using the fact that ε is an arbitrary positive number, the result follows. □
Now we consider the case \(\limsup_{n\to+\infty} \vert q_{n} \vert <1\). If so, then a bounded solution \((x_{n})_{n\in {\mathbb {N}}}\) to (31) is obtained only if
$$\begin{aligned} x_{0}=\sum_{j=1}^{\infty}f_{j}\prod_{l=1}^{j1}q_{l}, \end{aligned}$$
(40)
and for such chosen \(x_{0}\), it is obtained
$$\begin{aligned} x_{n}=\frac{\sum_{j=n+1}^{\infty}f_{j}\prod_{l=1}^{j1}q_{l}}{\prod_{l=1}^{n}q_{l}}=\sum_{j=n+1}^{\infty}f_{j}\prod_{l=n+1}^{j1}q_{l} \end{aligned}$$
(41)
for \(n\in {\mathbb {N}}\).
Theorem 4
Assume that
\((q_{n})_{n\in {\mathbb {N}}}\subset {\mathbb {C}}\setminus\{0\}\)
is a sequence such that
$$\begin{aligned} \limsup_{n\to\infty} \vert q_{n} \vert :=q< 1, \end{aligned}$$
(42)
and that
\((f_{n})_{n\in {\mathbb {N}}}\)
is a bounded sequence of complex numbers. Then the following statements are true.

(a)
There is a unique bounded solution to (31).

(b)
If
\(\lim_{n\to+\infty}f_{n}=0\), then the bounded solution
\(x_{n}\)
also converges to zero as
\(n\to+\infty\).
Proof
(a) From (42) we have that there is \(n_{11}\in {\mathbb {N}}\) such that
$$\begin{aligned} \vert q_{n} \vert < \frac{1+q}{2}\quad\mbox{for }n\ge n_{11}. \end{aligned}$$
(43)
By using (43) and some simple estimates in (41), we have
$$\begin{aligned} \vert x_{n} \vert \le\sum_{j=n+1}^{\infty} \vert f_{j} \vert \prod_{l=n+1}^{j1} \vert q_{l} \vert < \Vert f \Vert _{\infty}\sum _{j=n+1}^{\infty}\biggl(\frac {1+q}{2} \biggr)^{jn1}=\frac{2 \Vert f \Vert _{\infty}}{1q} \end{aligned}$$
(44)
for \(n\ge n_{11}\), from which the boundedness of sequence (41) easily follows. A simple calculation shows that the sequence satisfies equation (31). Since \(x_{0}\) is uniquely determined by the convergent series in (40), it follows that the sequence is a unique bounded solution to (31).
(b) Since \(\lim_{n\to+\infty}f_{n}=0\), we have that for every \(\varepsilon >0\), there is \(n_{12}\in {\mathbb {N}}\) such that (37) holds for \(n\ge n_{12}\).
From this, (41) and (43), we have that
$$\begin{aligned} \vert x_{n} \vert \le\sum_{j=n+1}^{\infty} \vert f_{j} \vert \prod_{l=n+1}^{j1} \vert q_{l} \vert < \varepsilon \sum_{j=n+1}^{\infty}\biggl(\frac{1+q}{2} \biggr)^{jn1}=\frac{2\varepsilon }{1q} \end{aligned}$$
(45)
for \(n\ge\max\{n_{11},n_{12}\}\). Letting \(n\to+\infty\) in (45) and since ε is an arbitrary positive number, we obtain \(\lim_{n\to+\infty}x_{n}=0\), as desired. □
From Theorems 14 the following four interesting corollaries are obtained.
From Theorems 1 and 3 we obtain the following corollary.
Corollary 1
Consider equation (1) for
\(n\in {\mathbb {Z}}\). Assume that
\((q_{n})_{n\in {\mathbb {Z}}}\)
and
\((f_{n})_{n\in {\mathbb {Z}}}\)
are sequences of complex numbers such that
\(q_{n}\ne0\), \(n\in {\mathbb {N}}\), \(\limsup_{n\to+\infty} \vert q_{n} \vert <1\)
and
\(\liminf_{n\to+\infty} \vert q_{n} \vert >1\), and
$$\begin{aligned} \sup_{n\in {\mathbb {Z}}} \vert f_{n} \vert < \infty. \end{aligned}$$
(46)
Then the following statements are true.

(a)
Every solution to (1) is bounded on
\({\mathbb {Z}}\).

(b)
If
$$\begin{aligned} \lim_{n\to\pm\infty}f_{n}=0, \end{aligned}$$
(47)
then, for every solution
\((x_{n})_{n\in {\mathbb {Z}}}\)
to (1), we have
\(\lim_{n\to\pm\infty}x_{n}=0\).
From Theorems 1 and 4 we obtain the following corollary.
Corollary 2
Consider equation (1) for
\(n\in {\mathbb {Z}}\). Assume that
\((q_{n})_{n\in {\mathbb {Z}}}\)
and
\((f_{n})_{n\in {\mathbb {Z}}}\)
are sequences of complex numbers such that
\(q_{n}\ne0\), \(n\in {\mathbb {N}}\), \(\limsup_{n\to+\infty} \vert q_{n} \vert <1\)
and
\(\limsup_{n\to+\infty} \vert q_{n} \vert <1\), and that (46) holds. Then the following statements are true.

(a)
There is a unique bounded solution to (1) on
\({\mathbb {Z}}\).

(b)
If (47) holds, then, for the bounded solution
\((x_{n})_{n\in {\mathbb {Z}}}\), we have
\(\lim_{n\to\pm\infty}x_{n}=0\).
From Theorems 2 and 3 we obtain the following corollary.
Corollary 3
Consider equation (1) for
\(n\in {\mathbb {Z}}\). Assume that
\((q_{n})_{n\in {\mathbb {Z}}}\)
and
\((f_{n})_{n\in {\mathbb {Z}}}\)
are sequences of complex numbers such that
\(q_{n}\ne0\), \(n\in {\mathbb {N}}\), \(\liminf_{n\to+\infty} \vert q_{n} \vert >1\)
and
\(\liminf_{n\to+\infty} \vert q_{n} \vert >1\), and that (46) holds. Then the following statements are true.

(a)
There is a unique bounded solution to (1) on
\({\mathbb {Z}}\).

(b)
If (47) holds, then, for the bounded solution
\((x_{n})_{n\in {\mathbb {Z}}}\), we have
\(\lim_{n\to\pm\infty}x_{n}=0\).
From Theorems 2 and 4 we obtain the following corollary.
Corollary 4
Consider equation (1) for
\(n\in {\mathbb {Z}}\). Assume that
\((q_{n})_{n\in {\mathbb {Z}}}\)
and
\((f_{n})_{n\in {\mathbb {Z}}}\)
are sequences of complex numbers such that
\(q_{n}\ne0\), \(n\in {\mathbb {N}}\), \(\liminf_{n\to+\infty} \vert q_{n} \vert >1\)
and
\(\limsup_{n\to+\infty} \vert q_{n} \vert <1\), and that (46) holds. Then the following statements are true.

(a)
There is a unique bounded solution to (1) on
\({\mathbb {Z}}\)
if and only if
$$x_{0}=\sum_{j=1}^{\infty}f_{j}\prod_{l=1}^{j1}q_{l}= \sum_{i=0}^{\infty}\frac{f_{i}}{\prod_{j=0}^{i}q_{j}}. $$

(b)
If (47) holds, then, for the bounded solution
\((x_{n})_{n\in {\mathbb {Z}}}\), we have
\(\lim_{n\to\pm\infty}x_{n}=0\).