On the solutions of second order generalized difference equations

Abstract

In this article, the authors discuss ${\ell }_{2(\ell )}$ and $c_{0(\ell )}$ solutions of the second order generalized difference equation

$${\mathrm{\Delta }}_{\ell }^{2}u(k)+f(k,u(k))=0,k\in [a,\mathrm{\infty }),a>0$$

and we prove the condition for non existence of non-trivial solution where ${\mathrm{\Delta }}_{\ell }u(k)=u(k+\ell )-u(k)$ for $\ell >0$. Further we present some formulae and examples to find the values of finite and infinite series in number theory as application of ${\mathrm{\Delta }}_{\ell }$.

MSC:39A12, 39A70, 47B39, 39B60.

1 Introduction

Difference equations usually describe the evolution of some certain phenomena over time and are also important in describing dynamics for fundamentally discrete system, see [1]. For example, in the numerical integration, the standard approach is to use the difference equations. Similarly, the population dynamics have discrete generations; the size of the $(k+1)$st generation $u(k+1)$ is a function of the k th generation $u(k)$. This can be expressed as difference equation of the form

$$u(k+1)=f(u(k)),$$

see for example [2]. Further, the concept of difference equations with many examples in applications such as asymptotic behavior of solutions of difference equations were studied extensively by Elaydi [3] where the analytic and geometric approaches were also combined in order to studying difference equations. Further, in [3], both classical and modern treatment of the difference equations were presented in excellent form. For related results on difference equations, see [4–8]. In the present article, we study ${\ell }_{2(\ell )}$ and $c_{0(\ell )}$ solutions of the following second order generalized difference equation

$${\mathrm{\Delta }}_{\ell }^{2}u(k)+f(k,u(k))=0,k\in [a,\mathrm{\infty }),a>0,$$

(1)

where ${\mathrm{\Delta }}_{\ell }u(k)=u(k+\ell )-u(k)$ for $\ell >0$. We provide some related definitions and development for the present article.

The basic theory of difference equations is based on the operator Δ defined as

$$\mathrm{\Delta }u(k)=u(k+1)-u(k),k\in \mathbb{N},$$

(2)

where $\mathbb{N}=\{0,1,2,3,\dots ,\}$. Even though many authors [1–4] have suggested the definition of Δ as

$$\mathrm{\Delta }u(k)=u(k+\ell )-u(k),k\in \mathbb{N},\ell \in \mathbb{R}-\{0\}$$

(3)

and there are several research took place on this line. By defining ${\mathrm{\Delta }}_{\ell }$ and its inverse ${\mathrm{\Delta }}_{\ell }^{-1}$, many interesting results and applications in number theory as well as in fluid dynamics can be obtained. By extending the study for sequences of complex numbers and ℓ to be real, some new qualitative properties like rotatory, expanding, shrinking, spiral and weblike structures were studied for the solutions of difference equations involving ${\mathrm{\Delta }}_{\ell }$. For similar results, we refer to [9–13].

In particular, the ${\ell }_2$ and $c_0$ solutions of second order difference equations of (1) when $\ell =1$, were discussed in [8]. In this article, we discuss ${\ell }_{2(\ell )}$ and $c_{0(\ell )}$ solutions for the second order generalized difference Equation (1) and present some applications of ${\mathrm{\Delta }}_{\ell }$ in the finite and infinite series of number theory. Throughout this article, we use the following notation:

• (i) $[k]$ denotes the integer part of k,

• (ii) $\mathbb{N}=\{0,1,2,3,\dots \}$, $\mathbb{N}(a)=\{a,a+1,a+2,\dots \}$,

• (iii) ${\mathbb{N}}_{\ell }(j)=\{j,j+\ell ,j+2\ell ,\dots \}$ and $\mathbb{R}$ is the set of all real numbers.

2 Preliminaries

In this section, we present some of the preliminary definitions and related results which will be useful for future discussion. The following three definitions held in [9].

Definition 2.1 Let $u:[0,\mathrm{\infty })\to \mathbb{C}$ and $\ell \in (0,\mathrm{\infty })$ then, the generalized difference operator ${\mathrm{\Delta }}_{\ell }$ is defined as

$${\mathrm{\Delta }}_{\ell }u(k)=u(k+\ell )-u(k).$$

(4)

Similarly, the generalized difference operator of the r th kind is defined as

$${\mathrm{\Delta }}_{\ell }^r={\mathrm{\Delta }}_{\ell }\left({\mathrm{\Delta }}_{\ell }^{r-1}\right)\text{if}r\ge 2.$$

(5)

Definition 2.2 For arbitrary $x,y\in \mathbb{R}$ the h-factorial function is defined by

$$x_h^{(y)}=h^y\frac{\mathrm{\Gamma }(\frac{x}{h}+1)}{\mathrm{\Gamma }(\frac{x}{h}+1-y)},$$

(6)

where Γ is the Euler gamma function. Note that when $x=k$, $h=\ell$, $y=n\in \mathbb{N}(1)$ Definition 2.2 coincides with Definition 2.1.

Definition 2.3 Let $u(k)$, $k\in [0,\mathrm{\infty })$ be a real or complex valued function and $\ell \in (0,\mathrm{\infty })$. Then, the inverse of ${\mathrm{\Delta }}_{\ell }$ denoted by ${\mathrm{\Delta }}_{\ell }^{-1}$ and defined as follows.

$$\text{If}{\mathrm{\Delta }}_{\ell }v(k)=u(k),\text{then}v(k)={\mathrm{\Delta }}_{\ell }^{-1}u(k)+c_j,$$

(7)

where $c_j$ is a constant for all $k\in {\mathbb{N}}_{\ell }(j)$, $j=k-\lceil \frac{k}{\ell }\rceil \ell$.

Definition 2.4 The generalized polynomial factorial for $\ell >0$ is defined as

$$k_{\ell }^{(n)}=k(k-\ell )(k-2\ell )\cdots (k-(n-1)\ell ).$$

(8)

Lemma 2.5 If$\ell >0$and$n\in {\mathbb{N}}_{\ell }(1)$then,

$${\mathrm{\Delta }}_{\ell }^{-1}{k}_{\ell }^{(n)}=\frac{1}{(n+1)\ell }{(k-\ell )}_{\ell }^{(n+1)}+c_j$$

(9)

for all$k\in {\mathbb{N}}_{\ell }(j)$, $j=k-\lceil \frac{k}{\ell }\rceil \ell$and$c_j$is constant.

Lemma 2.6 ([13] Product formula)

Let$u(k)$and$v(k)$be any two functions. Then

$$\begin{array}{rcl}{\mathrm{\Delta }}_{\ell }\{u(k)v(k)\}& =& u(k+\ell ){\mathrm{\Delta }}_{\ell }v(k)+v(k){\mathrm{\Delta }}_{\ell }u(k)\\ =& v(k+\ell ){\mathrm{\Delta }}_{\ell }u(k)+u(k){\mathrm{\Delta }}_{\ell }v(k),\mathrm{\forall }k\in {\mathbb{N}}_{\ell }(a).\end{array}$$

(10)

Lemma 2.7 ([12])

Let$\ell >0$, $n\in \mathbb{N}(2)$, $k\in (\ell ,\mathrm{\infty })$and$k_{\ell }^{(n)}\ne 0$. Then,

$${\mathrm{\Delta }}_{\ell }^{-1}\frac{1}{k_{\ell }^{(n)}}=\frac{-1}{(n-1)\ell {(k-\ell )}_{\ell }^{(n-1)}}+c_j.$$

(11)

Definition 2.8 A function $u(k)$, $k\in [a,\mathrm{\infty })$ is said to be in the space ${\ell }_{2(\ell )}$, if

$$\sum _{\gamma =0}^{\mathrm{\infty }}|u(a+j+\gamma \ell ){|}^2<\mathrm{\infty }\text{for all}j\in [0,\ell ).$$

(12)

If ${lim}_{r\to \mathrm{\infty }}|u(a+j+r\ell )|=0$, for all $0\le j<\ell$ then $u(k)$ is said to be in the space $c_{0(\ell )}$.

Lemma 2.9 ([9] Summation formula of finite series)

If real valued function$u(k)$is defined for all$k\in [0,\mathrm{\infty })$, then

$${\mathrm{\Delta }}_{\ell }^{-1}u(k)=\sum _{r=1}^{\lceil \frac{k}{\ell }\rceil }u(k-r\ell )+c_j,$$

(13)

where$c_j$is a constant for all$k\in {\mathbb{N}}_{\ell }(j)$, $j=k-\lceil \frac{k}{\ell }\rceil \ell$. Since$[0,\mathrm{\infty })={\bigcup }_{0\le j<\ell }{N}_{\ell }(j)$, each complex number$c_j$, ($0\le j<\ell$) is called an initial value of$k\in N_{\ell }(j)$. Usually, each initial value$c_j$is taken from any one of the values$u(j)$, $u(j+\ell )$, $u(j+2\ell )$, etc.

Lemma 2.10 (Summation formula of infinite series)

If${lim}_{k\to \mathrm{\infty }}u(k)=0$and$\ell >0$, then

$${\mathrm{\Delta }}_{\ell }^{-1}u(k)=-\sum _{r=0}^{\mathrm{\infty }}u(k+r\ell ).$$

(14)

Proof Assume $z(k)={\sum }_{r=0}^{\mathrm{\infty }}u(k+r\ell )$. Then,

$${\mathrm{\Delta }}_{\ell }z(k)=\sum _{r=0}^{\mathrm{\infty }}u(k+\ell +r\ell )-\sum _{r=0}^{\mathrm{\infty }}u(k+r\ell )=-u(k).$$

Now, the proof follows from ${lim}_{k\to \mathrm{\infty }}u(k)=0$ and Definition 2.3. □

Theorem 2.11 If${lim}_{k\to \mathrm{\infty }}u(k)=0$and$\ell >0$, then

$${\mathrm{\Delta }}_{\ell }^{-2}u(k)=\sum _{r_1=0}^{\mathrm{\infty }}\sum _{r_2=0}^{\mathrm{\infty }}u(k+r_1\ell +r_2\ell ).$$

(15)

Proof The proof follows by taking ${\mathrm{\Delta }}_{\ell }^{-1}$ on (14). □

Corollary 2.12 Let$k\in [\ell ,\mathrm{\infty })$and$\ell \in (0,\mathrm{\infty })$. Then

$${\mathrm{\Delta }}_{\ell }^{-1}\frac{1}{k(k-\ell )}=-\frac{1}{\ell (k-\ell )}$$

and hence

$$\sum _{r=0}^{\mathrm{\infty }}\frac{1}{(k+r\ell )(k+r\ell -\ell )}=\frac{1}{\ell (k-\ell )}.$$

(16)

Proof The proof follows from Equation (14) and $c_j=0$ as $k\to \mathrm{\infty }$. □

The following example illustrates Corollary 2.12.

Example 2.13 Taking $\ell =0.8$, $k=1$ in (16), we obtain

$$\frac{1}{1\times 0.2}+\frac{1}{1.8\times 1}+\frac{1}{2.6\times 1.8}+\cdots =\frac{1}{0.8\times 0.2}.$$

The following example shows that $\frac{1}{k_{\ell }^{(n)}}\in c_{0(\ell )}$ and ${\ell }_{2(\ell )}$.

Example 2.14 Assume $n\in \mathbb{N}(2)$ and $k\in [n\ell ,\mathrm{\infty })$. Let $u(k)=\frac{1}{k_{\ell }^{(n)}}$. By Lemmas 2.7 and 2.10, we obtain

$$\frac{1}{(n-1)\ell k_{\ell }^{(n-1)}}=\sum _{r=0}^{\mathrm{\infty }}\frac{1}{{(k+r\ell )}_{\ell }^{(n)}}.$$

Since $c_j=0$ as $k\to \mathrm{\infty }$. Replacing k by $a+j$, we get

$$\sum _{r=0}^{\mathrm{\infty }}\frac{1}{{(a+j+r\ell )}_{\ell }^{(n)}}=\frac{1}{(n-1)\ell {(a+j)}_{\ell }^{(n-1)}},\text{for}a\ge n\ell .$$

(17)

Since

$$|\frac{1}{{(a+j+r\ell )}_{\ell }^{(n)}}{|}^2<\frac{1}{{(a+j+r\ell )}_{\ell }^{(n)}},$$

for $a\ge n\ell$ thus Equation (17) yields

$$\sum _{r=0}^{\mathrm{\infty }}|u(a+j+r\ell ){|}^2<\sum _{r=0}^{\mathrm{\infty }}\frac{1}{{(a+j+r\ell )}_{\ell }^{(n)}}=\frac{1}{(n-1)\ell {(a+j)}^{(n-1)}}<\mathrm{\infty }.$$

By Definition 2.8, the function $\frac{1}{k_{\ell }^{(n)}}\in {\ell }_{2(\ell )}$. Since

$$\underset{r\to \mathrm{\infty }}{lim}\frac{1}{{(a+j+r\ell )}_{\ell }^{(n)}}=0,\frac{1}{k_{\ell }^{(n)}}\in c_{0(\ell )}.$$

Now taking $a=n\ell$ then $u(k)$ is an ${\ell }_{2(\ell )}$ space function.

3 Main results

In this section, we present the condition for non existence of non-trivial solution of (1).

Lemma 3.1 Let$a\ge 2\ell$and$k\in [a,\mathrm{\infty })$. Then

$$\frac{1}{k}<\frac{4}{(\sqrt{k+\ell }+\sqrt{k})(\sqrt{k}+\sqrt{k-\ell })}.$$

Proof We have

$$\begin{array}{c}\frac{4}{(\sqrt{k+\ell }+\sqrt{k})(\sqrt{k}+\sqrt{k-\ell })}\hfill \\ =\frac{4(\sqrt{k+\ell }-\sqrt{k})(\sqrt{k}-\sqrt{k-\ell })}{{\ell }^2}\hfill \\ =\frac{4}{{\ell }^2}\sqrt{k}\sqrt{k}[{(1+\frac{\ell }{k})}^{\frac{1}{2}}-1][1-{(1-\frac{\ell }{k})}^{\frac{1}{2}}]\hfill \\ =\frac{4k}{{\ell }^2}[1+\frac{1}{2}\frac{\ell }{k}-\frac{1}{2!}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{3!}\frac{1}{4}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3-\frac{1}{4!}\frac{1}{4}\frac{3}{2}\frac{5}{2}{\left(\frac{\ell }{k}\right)}^4+\cdots ]\hfill \\ \times [1-(1-\frac{1}{2}\frac{\ell }{k}-\frac{1}{2!}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^2-\frac{1}{3!}\frac{1}{4}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3-\frac{1}{4!}\frac{1}{4}\frac{3}{2}\frac{5}{2}{\left(\frac{\ell }{k}\right)}^4-\cdots )].\hfill \end{array}$$

Since each positive term is greater than the consecutive negative term in the first expression, we find

$$\begin{array}{c}\frac{4k}{{\ell }^2}[\frac{1}{2}\frac{\ell }{k}-\frac{1}{2!}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^2]\times [\frac{1}{2}\frac{\ell }{k}+\frac{1}{2!}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{3!}\frac{1}{4}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3+\frac{1}{4!}\frac{1}{4}\frac{3}{2}\frac{5}{2}{\left(\frac{\ell }{k}\right)}^4+\cdots ]\hfill \\ =\frac{4}{{\ell }^2}[\frac{\ell }{2}-\frac{\ell }{2}\frac{1}{4}\frac{\ell }{k}][\frac{1}{2}\frac{\ell }{k}+\frac{1}{2!}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{3!}\frac{1}{4}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3+\frac{1}{4!}\frac{1}{4}\frac{3}{2}\frac{5}{2}{\left(\frac{\ell }{k}\right)}^4+\cdots ]\hfill \\ =\frac{4}{{\ell }^2}\frac{\ell }{2}[\frac{1}{2}\frac{\ell }{k}+\frac{1}{2!}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{3!}\frac{1}{4}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3+\frac{1}{4!}\frac{1}{4}\frac{3}{2}\frac{5}{2}{\left(\frac{\ell }{k}\right)}^4+\cdots ]\hfill \\ -\frac{4}{{\ell }^2}\frac{\ell }{2}\frac{1}{4}\frac{\ell }{k}[\frac{1}{2}\frac{\ell }{k}+\frac{1}{2!}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{3!}\frac{1}{4}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3+\frac{1}{4!}\frac{1}{4}\frac{3}{2}\frac{5}{2}{\left(\frac{\ell }{k}\right)}^4+\cdots ]\hfill \\ =\frac{1}{k}+\frac{2}{\ell }[\frac{1}{2!}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{3!}\frac{1}{4}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3+\frac{1}{4!}\frac{1}{4}\frac{3}{2}\frac{5}{2}{\left(\frac{\ell }{k}\right)}^4+\cdots ]\hfill \\ -\frac{2}{\ell }[\frac{1}{2!}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{2!}\frac{1}{4}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^3+\frac{1}{3!}\frac{1}{4}\frac{1}{4}\frac{1}{4}{\left(\frac{\ell }{k}\right)}^4+\cdots ]\hfill \\ =\frac{1}{k}+\frac{2}{4\ell }[\frac{1}{3!}(\frac{3}{2}-\frac{3}{4}){\left(\frac{\ell }{k}\right)}^3+\frac{1}{4!}\frac{3}{2}(\frac{5}{2}-\frac{4}{4}){\left(\frac{\ell }{k}\right)}^4\hfill \\ +\frac{1}{5!}\frac{3}{2}\frac{5}{2}(\frac{7}{2}-\frac{5}{4}){\left(\frac{\ell }{k}\right)}^5+\frac{1}{6!}\frac{3}{2}\frac{5}{2}\frac{7}{2}(\frac{9}{2}-\frac{6}{4}){\left(\frac{\ell }{k}\right)}^6+\cdots ]>\frac{1}{k},\hfill \end{array}$$

since the second term is positive. □

Lemma 3.2 Let$a\ge 2\ell$and$k\in [a,\mathrm{\infty })$. Then

$$\frac{\sqrt{k+\ell }}{\sqrt{k}}-\frac{\sqrt{k}}{\sqrt{k+\ell }+\sqrt{k-\ell }}<1.$$

(18)

Proof From the Binomial theorem for rational index, we find

$$\begin{array}{rcl}\frac{\sqrt{k+\ell }}{\sqrt{k}}-\frac{\sqrt{k}}{\sqrt{k+\ell }+\sqrt{k-\ell }}& =& {(1+\frac{\ell }{k})}^{\frac{1}{2}}-\frac{\sqrt{k}}{2\ell }[{(k+\ell )}^{\frac{1}{2}}-{(k-\ell )}^{\frac{1}{2}}]\\ =& 1+\frac{1}{2}\frac{\ell }{k}-\frac{1}{2!}\frac{1}{2}\frac{1}{2}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{3!}\frac{1}{2}\frac{1}{2}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3-\cdots \\ -\frac{k}{2\ell }[1+\frac{1}{2}\frac{\ell }{k}-\frac{1}{2!}\frac{1}{2}\frac{1}{2}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{3!}\frac{1}{2}\frac{1}{2}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3-\cdots \\ -(1-\frac{1}{2}\frac{\ell }{k}-\frac{1}{2!}\frac{1}{2}\frac{1}{2}{\left(\frac{\ell }{k}\right)}^2-\frac{1}{3!}\frac{1}{2}\frac{1}{2}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3-\cdots )]\\ =& 1+\frac{1}{2}\frac{\ell }{k}-\frac{1}{2!}\frac{1}{2}\frac{1}{2}{\left(\frac{\ell }{k}\right)}^2+\frac{1}{3!}\frac{1}{2}\frac{1}{2}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3-\cdots \\ -\frac{k}{2\ell }[\frac{\ell }{k}+\frac{1}{3!}\frac{1}{2}\frac{1}{2}\frac{3}{2}{\left(\frac{\ell }{k}\right)}^3+\cdots ].\end{array}$$

Since each negative terms is greater than the next consecutive positive term and $k\ge 2\ell$, we get

$$\frac{\sqrt{k+\ell }}{\sqrt{k}}-\frac{\sqrt{k}}{\sqrt{k+\ell }+\sqrt{k-\ell }}=1+\frac{1}{2}\frac{\ell }{k}-\frac{1}{2}=\frac{1}{2}+\frac{1}{2}\frac{\ell }{k}<1.$$

 □

Lemma 3.3 Let$a\ge 2\ell$. If

$${\mathrm{\Delta }}_{\ell }z(k)\le \alpha (k)+\beta (k)z(k)$$

(19)

and $\frac{-\ell }{k}<\beta <\frac{-{\ell }^2}{k^2}$ for all $k\in [a,\mathrm{\infty })$ then

$${\mathrm{\Delta }}_{\ell }(z(k)\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil -1}{(1+\beta (j+a+r\ell ))}^{-1})\le \alpha (k)\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil }{(1+\beta (j+a+r\ell ))}^{-1},$$

(20)

where$j=k-a-\lceil \frac{k-a}{\ell }\rceil \ell$.

Proof From the inequality (19) and $1+\beta (k)>0$ for all $k\in [a,\ell )$, we find,

$$\frac{z(k+\ell )}{1+\beta (k)}-z(k)\le \frac{\alpha (k)}{1+\beta (k)}$$

which yields,

$$\begin{array}{c}\frac{z(k+\ell )}{1+\beta (k)}\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil -1}{(1+\beta (j+a+r\ell ))}^{-1}-z(k)\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil -1}{(1+\beta (j+a+r\ell ))}^{-1}\hfill \\ \le \frac{\alpha (k)}{1+\beta (k)}\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil -1}{(1+\beta (j+a+r\ell ))}^{-1}.\hfill \end{array}$$

Now (20) follows by taking $r=\lceil \frac{k-a}{\ell }\rceil$ and $j+a+\lceil \frac{k-a}{\ell }\rceil \ell =k$. □

The following theorem shows the nonexistence of solutions of (3).

Theorem 3.4 For all$(k,u)\in [a,\mathrm{\infty })\times \mathbb{R}$, let the function$f(k,u)$be defined and

$$|f(k,u)|\le \frac{{\ell }^2}{2}{k}^{-2}|u|.$$

(21)

Then, if$u(k)\in {\ell }_{2(\ell )}$is a solution of (3), there exists a real$k_1\ge a$ ($a\ge 2\ell$) such that$u(k)=0$for all$k\in [k_1,\mathrm{\infty })$.

Proof Since $u(k)$ is a solution of (3) and belong to ${\ell }_{2(\ell )}$, we have ${\sum }_{r=0}^{\mathrm{\infty }}|u(a+j+r\ell ){|}^2<\mathrm{\infty }$ which yields ${lim}_{k\to \mathrm{\infty }}u(k)=0$ and hence

$$\underset{k\to \mathrm{\infty }}{lim}{\mathrm{\Delta }}_{\ell }u(k)=\underset{k\to \mathrm{\infty }}{lim}{\mathrm{\Delta }}_{\ell }^{2}u(k)=0.$$

(22)

By using Equations (3) and (22), and applying ${\mathrm{\Delta }}_{\ell }^{-1}$ on Equation (3) with Lemma 2.10, we obtain

$${\mathrm{\Delta }}_{\ell }u(k)=\sum _{r=0}^{\mathrm{\infty }}f(k+r\ell ,u(k+r\ell )).$$

(23)

Now by applying again ${\mathrm{\Delta }}_{\ell }^{-1}$ on both sides, and by Theorem 2.10, we get

$$u(k)=-\sum _{r=0}^{\mathrm{\infty }}\sum _{s=0}^{\mathrm{\infty }}f(k+r\ell +s\ell ,u(k+r\ell +s\ell ))$$

(24)

which yields

$$u(k)=-\sum _{r=0}^{\mathrm{\infty }}(r+1)f(k+r\ell ,u(k+r\ell )),k\in [a,\mathrm{\infty }).$$

(25)

Therefore, from (21), we obtain

$$|u(k)|\le \frac{{\ell }^2}{2}v(k),$$

(26)

where

$$v(k)=\sum _{r=0}^{\mathrm{\infty }}(r+1){(k+r\ell )}^{-2}|u(k+r\ell )|,\text{for all}k\in [a,\mathrm{\infty }).$$

(27)

Obviously $v(k)\ge 0$ for all $k\in [a,\mathrm{\infty })$ and ${lim}_{k\to \mathrm{\infty }}v(k)=0$.

If $v(k+j)=0$, $\mathrm{\forall }j\in [0,\ell )$, for some $k=k_1\ge a$, then $(r+1){(k+j+r\ell )}^{-2}u(k+j+r\ell )=0$ for all $r=0,1,2,\dots$ . Hence $u(k)=0$ for all $k\ge k_1$. In this case, the proof is complete.

Now, we suppose that $v(k)>0$ for all $k\in [a,\mathrm{\infty })$, from (27), we have

$${\mathrm{\Delta }}_{\ell }v(k)=-\sum _{r=0}^{\mathrm{\infty }}{(k+r\ell )}^{-2}|u(k+r\ell )|$$

and

$${\mathrm{\Delta }}_{\ell }^{2}v(k)=k^{-2}|u(k)|.$$

From (26), we have

$${\mathrm{\Delta }}_{\ell }^{2}v(k)\le \frac{{\ell }^2}{2}{k}^{-2}v(k)\text{for all}k\in [a,\mathrm{\infty }).$$

(28)

From (27), $a\ge 2\ell$, $\frac{r+1}{k+r\ell }\le \frac{1}{\ell }$, by Schwartz’s inequality, we obtain

$$v(k)\le {\ell }^{-1}\sum _{r=0}^{\mathrm{\infty }}{(k+r\ell )}^{-1}|u(k+r\ell )|\le {\ell }^{-1}{\left(\sum _{r=0}^{\mathrm{\infty }}{(k+r\ell )}^{-2}\right)}^{\frac{1}{2}}{\left(\sum _{r=0}^{\mathrm{\infty }}\right|u(k+r\ell ){|}^2)}^{\frac{1}{2}}.$$

By using Corollary 2.12, we get

$$v(k)\le {\ell }^{-\frac{3}{2}}\frac{1}{\sqrt{k-\ell }}{\left(\sum _{r=0}^{\mathrm{\infty }}\right|u(k+r\ell ){|}^2)}^{\frac{1}{2}}.$$

If $w(k)={\ell }^{\frac{3}{2}}\sqrt{k-\ell }v(k)$, then

$$w(k)\le {\left(\sum _{r=0}^{\mathrm{\infty }}{\parallel u(k+r\ell )\parallel }^2\right)}^{\frac{1}{2}},\text{for all}k\in [a,\mathrm{\infty }).$$

(29)

Hence we have

$$w(k)\to 0\text{and}w(k)>0,\mathrm{\forall }k\in [a,\mathrm{\infty }).$$

(30)

By applying Lemma 2.6 to Equation (29) twice, we obtain

$${\mathrm{\Delta }}_{\ell }^{2}w(k)={\ell }^{\frac{3}{2}}(\sqrt{k+\ell }{\mathrm{\Delta }}_{\ell }^{2}v(k)+2{\mathrm{\Delta }}_{\ell }v(k){\mathrm{\Delta }}_{\ell }\sqrt{k}+v(k){\mathrm{\Delta }}_{\ell }^2\sqrt{k-\ell }).$$

(31)

Again from Lemma 2.6 and Equation (29), we obtain

$${\mathrm{\Delta }}_{\ell }v(k)={\ell }^{-\frac{3}{2}}(\frac{1}{\sqrt{k}}{\mathrm{\Delta }}_{\ell }w(k)+w(k){\mathrm{\Delta }}_{\ell }\frac{1}{\sqrt{k-\ell }}).$$

(32)

From (31), (32) and by Lemma 2.6, we find that

$$\begin{array}{c}{\mathrm{\Delta }}_{\ell }(\frac{1}{k-\ell }{\mathrm{\Delta }}_{\ell }w(k))\hfill \\ =\frac{1}{k}{\mathrm{\Delta }}_{\ell }^{2}w(k)-\left(\frac{\ell }{k(k-\ell )}\right){\mathrm{\Delta }}_{\ell }w(k)\hfill \\ =\frac{{\ell }^{\frac{3}{2}}}{k}\{\sqrt{k+\ell }{\mathrm{\Delta }}_{\ell }^{2}v(k)+2{\mathrm{\Delta }}_{\ell }v(k){\mathrm{\Delta }}_{\ell }\sqrt{k}+v(k){\mathrm{\Delta }}_{\ell }^2\sqrt{k-\ell }\}\hfill \\ -\left(\frac{\ell }{k(k-\ell )}\right){\mathrm{\Delta }}_{\ell }w(k)\hfill \\ =\frac{{\ell }^{\frac{3}{2}}}{k}\{\sqrt{k+\ell }{\mathrm{\Delta }}_{\ell }^{2}v(k)+2{\ell }^{-\frac{3}{2}}[\frac{1}{\sqrt{k}}{\mathrm{\Delta }}_{\ell }w(k)+2\frac{w(k)}{k}{\mathrm{\Delta }}_{\ell }\frac{1}{\sqrt{k-\ell }}]{\mathrm{\Delta }}_{\ell }\sqrt{k}\hfill \\ +\frac{{\ell }^{\frac{3}{2}}}{k}v(k){\mathrm{\Delta }}_{\ell }^2\sqrt{k-\ell }\}-\left(\frac{\ell }{k(k-\ell )}\right){\mathrm{\Delta }}_{\ell }w(k)\hfill \\ ={\ell }^{\frac{3}{2}}\left(\frac{\sqrt{k+\ell }}{k}\right){\mathrm{\Delta }}_{\ell }^{2}v(k)+\frac{2}{k}{\ell }^{\frac{3}{2}}\sqrt{k-\ell }v(k){\mathrm{\Delta }}_{\ell }\frac{1}{\sqrt{k-\ell }}{\mathrm{\Delta }}_{\ell }\sqrt{k}\hfill \\ +\frac{{\ell }^{\frac{3}{2}}}{k}v(k){\mathrm{\Delta }}_{\ell }^2\sqrt{k-\ell }+\frac{2}{k\sqrt{k}}{\mathrm{\Delta }}_{\ell }w(k){\mathrm{\Delta }}_{\ell }\sqrt{k}-\frac{\ell }{k(k-\ell )}{\mathrm{\Delta }}_{\ell }w(k)\hfill \\ \le {\ell }^{\frac{3}{2}}\left(\frac{{\ell }^2\sqrt{k+\ell }}{2k^3}\right)v(k)+\frac{2{\ell }^{\frac{3}{2}}}{k}\sqrt{k-\ell }v(k){\mathrm{\Delta }}_{\ell }\sqrt{k}{\mathrm{\Delta }}_{\ell }\frac{1}{\sqrt{k-\ell }}\hfill \\ +\frac{{\ell }^{\frac{3}{2}}}{k}v(k){\mathrm{\Delta }}_{\ell }^2\sqrt{k-\ell }\hfill \\ +(\frac{2(k-\ell )}{k\sqrt{k}}{\mathrm{\Delta }}_{\ell }\sqrt{k}-\frac{\ell }{k})\frac{1}{k-\ell }{\mathrm{\Delta }}_{\ell }w(k)\hfill \end{array}$$

which in view of (28), (30) gives

$${\mathrm{\Delta }}_{\ell }z(k)\le \alpha (k)+\beta (k)z(k),$$

(33)

where

$$z(k)=\frac{1}{k-\ell }{\mathrm{\Delta }}_{\ell }w(k),$$

(34)

$$\alpha (k)={\ell }^{\frac{3}{2}}(\frac{{\ell }^2\sqrt{k+\ell }}{2k^3}+\frac{2}{k}\sqrt{k-\ell }{\mathrm{\Delta }}_{\ell }\sqrt{k}{\mathrm{\Delta }}_{\ell }\frac{1}{\sqrt{k-\ell }}+\frac{1}{k}{\mathrm{\Delta }}_{\ell }^2\sqrt{k-\ell })v(k)$$

(35)

and

$$\beta (k)=\left(\frac{2(k-\ell )}{k\sqrt{k}}\right){\mathrm{\Delta }}_{\ell }\sqrt{k}-\frac{\ell }{k}.$$

(36)

Since $(\frac{2(k-\ell )}{k\sqrt{k}}){\mathrm{\Delta }}_{\ell }\sqrt{k}>0$, from ${(1+\frac{\ell }{k})}^{\frac{1}{2}}<1+\frac{1}{2}\frac{\ell }{k}$, we obtain

$$-\frac{\ell }{k}<\beta (k)<-\frac{{\ell }^2}{k^2},\text{where}k\in [a,\mathrm{\infty }).$$

(37)

Further, since

$$\begin{array}{rcl}\sqrt{k}\sqrt{k-\ell }{\mathrm{\Delta }}_{\ell }\sqrt{k}{\mathrm{\Delta }}_{\ell }\frac{1}{\sqrt{k-\ell }}& =& (\sqrt{k+\ell }-\sqrt{k})(\sqrt{k-\ell }-\sqrt{k})\\ =& -\frac{{\ell }^2}{(\sqrt{k+\ell }+\sqrt{k})(\sqrt{k-\ell }+\sqrt{k})}\end{array}$$

and

From Lemmas 3.1 and 3.2

$$\begin{array}{rcl}\gamma (k)& <& \frac{{\ell }^{\frac{3}{2}}}{k\sqrt{k}}(\frac{{\ell }^2\sqrt{k+\ell }}{2\sqrt{k}}\frac{4}{(\sqrt{k+\ell }+\sqrt{k})(\sqrt{k}+\sqrt{k-\ell })}\\ +\frac{-2{\ell }^2+\ell \sqrt{k}(\sqrt{k-\ell }-\sqrt{k+\ell })}{(\sqrt{k+\ell }+\sqrt{k})(\sqrt{k}+\sqrt{k-\ell })})v(k)\\ =& \frac{2{\ell }^{\frac{3}{2}}}{k\sqrt{k}(\sqrt{k+\ell }+\sqrt{k})(\sqrt{k}+\sqrt{k-\ell })}(\frac{{\ell }^2\sqrt{k+\ell }}{\sqrt{k}}-\frac{{\ell }^2\sqrt{k}}{\sqrt{k+\ell }+\sqrt{k-\ell }}-{\ell }^2)v(k)\\ =& \frac{2{\ell }^{\frac{7}{2}}}{k\sqrt{k}(\sqrt{k+\ell }+\sqrt{k})(\sqrt{k}+\sqrt{k-\ell })}(\frac{\sqrt{k+\ell }}{\sqrt{k}}-\frac{\sqrt{k}}{\sqrt{k+\ell }+\sqrt{k-\ell }}-1)v(k).\end{array}$$

(38)

By Lemma 3.2, we find $\gamma (k)<0$ for all $k\in [a,\mathrm{\infty })$. Thus from Lemma 3.3 and $\gamma (k)<0$, we find

$${\mathrm{\Delta }}_{\ell }(z(k)\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil -1}{(1+\beta (j+a+r\ell ))}^{-1})<0,\text{for all}k\in [a+\ell ,\mathrm{\infty }).$$

That is,

$$z(k)\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil -1}{(1+\beta (j+a+r\ell ))}^{-1}$$

is decreasing by ℓ steps.

If

$$z(k)\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil -1}{(1+\beta (j+a+r\ell ))}^{-1}>0$$

for all $k\in [a+\ell ,\mathrm{\infty })$, then $z(k)>0$. From (34) we find ${\mathrm{\Delta }}_{\ell }w(k)>0$ and hence $w(k)$ is increasing by ℓ steps, but this contradicts (30).

If there exists a real $K\ge a+\ell$ such that

$$z(K)\prod _{r=0}^{\lceil \frac{K-a}{\ell }\rceil -1}{(1+\beta (j+a+r\ell ))}^{-1}=p_j<0$$

for all $0\le j<\ell$, then

$$z(k)\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil -1}{(1+\beta (j+a+r\ell ))}^{-1}<p_j$$

for all $k\in [K,\mathrm{\infty })$, that is,

$$z(k)<p_j\prod _{r=0}^{\lceil \frac{k-a}{\ell }\rceil -1}(1+\beta (j+a+r\ell )).$$

However from (37), since $1+\beta (k)>(k-\ell )/k>0$ and $j=k-\lceil \frac{k-a}{\ell }\rceil \ell$, it follows that $z(k)<p_j(j+a-\ell )/(k-\ell )$, and hence from (34), we find ${\mathrm{\Delta }}_{\ell }w(k)<p_j(j+a-\ell )$. Further, since

$$w(k)\to 0,k\ge K+2\ell \Rightarrow \frac{1}{\ell }(k-K-\ell )\ge 1$$

we get $w(k+\ell )<w(k)+p_j(j+a-\ell )$ which yields $w(k)<w(k-\ell )+p_j(j+a-\ell )$ and hence we get

$$w(k)<w(K+\ell )+\frac{1}{\ell }{p}_j(j+a-\ell )(k-K-\ell )$$

for all $k\in [K+2\ell ,\mathrm{\infty })$, since

$$k\ge K+2\ell \Rightarrow k-K\ge 2\ell ,\frac{1}{\ell }(k-K-\ell )\ge 1.$$

But this implies that $w(k)\to -\mathrm{\infty }$, and again we get a contradiction to (30).

Thus combining the above arguments, we conclude that our assumption $v(k)>0$ for all $k\in [a,\mathrm{\infty })$ is not correct, and this completes the proof. □

Theorem 3.5 For all$(k,u)\in [0,\mathrm{\infty })\times \mathbb{R}$, let the function$f(k,u)$be defined and

$$|f(k,u)|\le {\ell }^q{k}^{-q}|u|,q>\frac{5}{2}.$$

(39)

Then, if$u(k)$is a solution of (3) $\in c_{0(\ell )}$, there exists an integer$k_1\ge a$ ($a\ge 4\ell$) such that$u(k)=0$for all$k\in [k_1,\mathrm{\infty })$.

Proof Let $u(k)$ be a solution of (3) such that ${lim}_{k\to \mathrm{\infty }}|u(k)|=0$. Then,

$$\underset{k\to \mathrm{\infty }}{lim}{\mathrm{\Delta }}_{\ell }u(k)=\underset{k\to \mathrm{\infty }}{lim}{\mathrm{\Delta }}_{\ell }^{2}u(k)=0$$

for all $\ell >0$. Thus, for this solution also the relation (24) holds. Further, since there exists a constant $c_j>0$ such that $|u(k)|\le c_j$ for all $k\in [a,\mathrm{\infty })$, where $0\le j=k-\lceil \frac{k}{\ell }\rceil \ell <\ell$, we find that

$$\begin{array}{rcl}\sum _{r=0}^{\mathrm{\infty }}(r+1)\left|f((k+r\ell ),u(k+r\ell ))\right|& \le & \sum _{r=0}^{\mathrm{\infty }}(r+\frac{k}{\ell }{\ell }^q{(k+r\ell )}^{-q}|u(k+r\ell )|)\\ =& \sum _{r=0}^{\mathrm{\infty }}{(k+r\ell )}^{1-q}{\ell }^{q-1}|u(k+r\ell )|\\ \le & c_j{\ell }^{q-1}\sum _{r=0}^{\mathrm{\infty }}{(k+r\ell )}^{1-q}\text{where}j=k-\lceil \frac{k}{\ell }\rceil \ell \\ =& c_j{\ell }^{q-1}[k^{1-q}+\sum _{r=1}^{\mathrm{\infty }}{(k+r\ell )}^{1-q}]\\ =& c_j{\ell }^{q-1}[k^{1-q}+{\ell }^{1-q}\sum _{r=1}^{\mathrm{\infty }}{(\frac{k}{\ell }+r)}^{1-q}]\\ =& c_j{\ell }^{q-1}[k^{1-q}+{\ell }^{1-q}{[\frac{{(\frac{k}{\ell })}^{2-q}}{2-q}+r]}_{\frac{k}{\ell }}^{\mathrm{\infty }}]\\ =& c_j{\ell }^{q-1}[k^{1-q}+\left(\frac{k^{2-q}}{\ell (q-2)}\right)]<\mathrm{\infty },\end{array}$$

for all $k\in [k_1,\mathrm{\infty })$. Therefore, this solution also has the representation (24).

Now as in Theorem 3.4, we define

$$\overline{v}(k)=\sum _{r=0}^{\mathrm{\infty }}(r+1){(k+r\ell )}^{-q}|u(k+r\ell )|=\sum _{r=0}^{\mathrm{\infty }}{\ell }^{-q}(r+1){(\frac{k}{\ell }+r)}^{-q}|u(k+r\ell )|.$$

Since $q>\frac{5}{2}$, we find

$$\overline{v}(k)\le {\ell }^{-q}\sum _{r=0}^{\mathrm{\infty }}(r+1){(\frac{k}{\ell }+r)}^{-2}|u(k+r\ell )|={\ell }^{2-q}\sum _{r=0}^{\mathrm{\infty }}(r+1){(k+r)}^{-2}|u(k+r\ell )|$$

then it follows that

$$\overline{v}(k)\le {\ell }^{2-q}\left(\frac{{\ell }^{-\frac{3}{2}}}{\sqrt{k-\ell }}\right){\left\{\sum _{r=0}^{\mathrm{\infty }}\right|u(k+r\ell ){|}^2\}}^{\frac{1}{2}}.$$

Hence we define

$$\begin{array}{c}\overline{w}(k)={\ell }^{q-\frac{1}{2}}\sqrt{k-\ell }\overline{v}(k),\hfill \\ \overline{z}(k)=\frac{1}{k-\ell }{\mathrm{\Delta }}_{\ell }\overline{w}(k),\hfill \\ \overline{\gamma }(k)={\ell }^{q-\frac{1}{2}}({\ell }^q\frac{\sqrt{k+\ell }}{2k^{q+1}}+\frac{2}{k}\sqrt{k-\ell }{\mathrm{\Delta }}_{\ell }\sqrt{k}{\mathrm{\Delta }}_{\ell }\frac{1}{\sqrt{k-\ell }}+\frac{1}{k}{\mathrm{\Delta }}_{\ell }^2\sqrt{k-\ell })\overline{v}(k),\hfill \\ \overline{\beta }(k)=\left(\frac{2(k-\ell )}{k\sqrt{k}}\right){\mathrm{\Delta }}_{\ell }\sqrt{k}-\frac{\ell }{k},\hfill \end{array}$$

and applying similar arguments as in the previous theorem one can see that there exists a positive integer $k_1$ such that $u(k)=0$ for all $k\in [k_1,\mathrm{\infty })$. □

In the next we present some formulae and examples to find the values of finite and infinite series in number theory as application of ${\mathrm{\Delta }}_{\ell }$. First of all we need the following theorem.

Theorem 3.6 Let$k\in [\ell ,\mathrm{\infty })$and$\ell \in (0,\mathrm{\infty })$. Then

$$\sum _{r=1}^{[\frac{k}{\ell }]+s}\frac{{(k-r\ell +2\ell )}^2-3{\ell }^2}{{\ell }^r{(k-r\ell +4\ell )}_{\ell }^{(2)}{(k-r\ell +\ell )}_{\ell }^{(\lceil \frac{k-r\ell +\ell }{\ell }\rceil )}}=\frac{c_j}{{\ell }^{\lceil \frac{k}{\ell }\rceil }}-\frac{1}{(k+3\ell )k_{\ell }^{(\lceil \frac{k}{\ell }\rceil )}},$$

(40)

where$s=-1$for$k\in {\mathbb{N}}_{\ell }(\ell )$, $s=0$for$k\notin {\mathbb{N}}_{\ell }(\ell )$and each$c_j$is a constant for all$k\in {\mathbb{N}}_{\ell }(j)$, $j=k-[\frac{k}{\ell }]\ell$. In particular$c_j$is obtained from (40) by substituting$k=\ell +j$. Further

$$\sum _{r=0}^{\mathrm{\infty }}\frac{{(k+r\ell )}^3-{\ell }^3}{{\ell }^r{({(k+r\ell )}^2-2{\ell }^2)}_{\ell }^{(2)}{(k+r\ell +\ell )}_{\ell }^{(\lceil \frac{k+r\ell +\ell }{\ell }\rceil )}}=\frac{1}{({(k-\ell )}^2-2{\ell }^2)k_{\ell }^{(\lceil \frac{k}{\ell }\rceil )}}.$$

(41)

Proof By Definition 2.1, we find

$${\mathrm{\Delta }}_{\ell }^{-1}\frac{({(k+2\ell )}^2-3{\ell }^2){\ell }^{\lceil \frac{k}{\ell }\rceil }}{{(k+4\ell )}_{\ell }^{(2)}{(k+\ell )}_{\ell }^{(\lceil \frac{k+\ell }{\ell }\rceil )}}=c_j-\frac{{\ell }^{\lceil \frac{k}{\ell }\rceil }}{(k+3\ell )k_{\ell }^{(\lceil \frac{k}{\ell }\rceil )}}$$

and (40) follows by Lemma 2.9 and

$$\frac{{(k-([\frac{k}{\ell }]+s)\ell +2\ell )}^2-3{\ell }^2}{{(k-([\frac{k}{\ell }]+s)\ell +4\ell )}_{\ell }^{(2)}{(k-([\frac{k}{\ell }]+s)\ell +\ell )}_{\ell }^{(\lceil \frac{k-([\frac{k}{\ell }]+s)\ell +\ell }{\ell }\rceil )}}\ge 0.$$

 □

The following example illustrates Theorem 3.6.

Example 3.7 By taking $\ell =1.7$, $k=2$ and $j=0.3$ in (40), we get $c_j=\frac{85}{81}$ and hence (40) becomes

$$\begin{array}{c}\sum _{r=1}^{[\frac{k}{\ell }]}\frac{{(k-1.7r+2(1.7))}^2-3{(1.7)}^2}{{1.7}^r{(k-1.7r+4(1.7))}_{1.7}^{(2)}{(k-1.7r+1.7)}_{1.7}^{(\lceil \frac{k-1.7r+1.7}{1.7}\rceil )}}\hfill \\ =\frac{85}{81{(1.7)}^{\lceil \frac{k}{1.7}\rceil }}-\frac{1}{(k+3(1.7))k_{1.7}^{(\lceil \frac{k}{1.7}\rceil )}},k=2,3.7,5.4,\dots \hfill \end{array}$$

Example 3.8 Taking $\ell =3.5$ in (41), we obtain

$$\begin{array}{c}\sum _{r=0}^{\mathrm{\infty }}\frac{{(k+3.5r)}^3-{3.5}^3}{{3.5}^r{({(k+3.5r)}^2-2{(3.5)}^2)}_{\ell }^{(2)}{(k+3.5r+3.5)}_{3.5}^{(\lceil \frac{k+3.5r+3.5}{3.5}\rceil )}}\hfill \\ =\frac{1}{({(k-3.5)}^2-2{(3.5)}^2)k_{3.5}^{(\lceil \frac{k}{3.5}\rceil )}}.\hfill \end{array}$$

In particular, when $k=9$, above series becomes

$$\begin{array}{c}\frac{9^3-{3.5}^3}{{(9^2-2{(3.5)}^2)}_{3.5}^{(2)}{12.5}_{3.5}^{(4)}}+\frac{{12.5}^3-{3.5}^3}{3.5{({12.5}^2-2{(3.5)}^2)}_{3.5}^{(2)}{16}_{3.5}^{(5)}}\hfill \\ +\frac{{16}^3-{3.5}^3}{{3.5}^2{({16}^2-2{(3.5)}^2)}_{3.5}^{(2)}{19.5}_{3.5}^{(6)}}+\cdots =\frac{1}{({5.5}^2-2{(3.5)}^2)9_{3.5}^{(3)}}.\hfill \end{array}$$

4 Concluding remarks

In the difference equations there are several interesting development, see for example, [4–6], and [8–16]. Recently, in [7], the fractional h-difference equations was studied. In the present work we study the ${\ell }_{2(\ell )}$ and $c_{0(\ell )}$ solutions of the second order generalized difference equation

$${\mathrm{\Delta }}_{\ell }^{2}u(k)+f(k,u(k))=0,k\in [a,\mathrm{\infty }),a>0$$

and we prove the condition for non existence of non-trivial solution.