Asymptotics of large eigenvalues for some discrete unbounded Jacobi matrices

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Linear Algebra and its Applications

j o u r n a l h o m e p a g e : w w w . e l s e v i e r . c o m / l o c a t e / l a a

Asymptotics of large eigenvalues for some discrete

unbounded Jacobi matrices

Maria Malejki

AGH University of Science and Technology, Faculty of Applied Mathematics, al. Mickiewicza 30, 30-059 Kraków, Poland

A R T I C L E I N F O A B S T R A C T Article history:

Received 8 July 2007 Accepted 17 June 2009 Available online 19 July 2009

Submitted by V. Mehrmann AMS classiﬁcation: 47B25 47B36 15A18 Keywords: Tridiagonal matrix Jacobi matrix Self-adjoint operator Asymptotics Eigenvalue Point spectrum

The aim of this paper is to find asymptotic formulas for eigen-values of self-adjoint discrete operators inl2_(N)_{given by some} infinite symmetric Jacobi matrices. The approach used to calculate an asymptotic behaviour of eigenvalues is based on method of diag-onalization, Janas and Naboko’s lemma [J. Janas, S. Naboko, Infinite Jacobi matrices with unbounded entries: asymptotics of eigenval-ues and the transformation operator approach, SIAM J. Math. Anal. 36 (2) (2004) 643–658] and the Rozenbljum theorem [G.V. Rozen-bljum, Near-similarity of operators and the spectral asymptotic behaviour of pseudodifferential operators on the circle, (Russian) Trudy Maskov. Mat. Obshch. 36 (1978) 59–84]. The asymptotic for-mulas are given with use of eigenvalues and determinants of finite tridiagonal matrices.

1. Introduction

Finite and inﬁnite tridiagonal matrices appear in different sections of mathematics such as theory of orthogonal polynomials, numerical analysis, spectral theory of differential operators and theory of second order differential and difference equations (see [5,7,8,21,24], etc.). There are interesting examples of applications of tridiagonal matrices in mathematical physics [1,13,22,18]. Inﬁnite sym-metric tridiagonal matrices called Jacobi matrices have essential meaning. Recently, many authors

E-mail address:[email protected]

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studied different problems concerning spectral properties of linear operators associated with Ja-cobi matrices (see, e.g., [6,7,12,14,15,18,21]). We can also find articles that investigate and solve the problem when a linear operator defined by a Jacobi matrix has discrete spectrum i.e. it is compact or has a compact resolvent and its spectrum consists of eigenvalues of finite multiplicity (see, e.g., [4,10,13]).

In this article we calculate the asymptotic behaviour of eigenvalues of discrete operators, so we distinguish the papers [2,3,5,9,11,13,20,23,26], which concern similar problems. Sometimes it is pos-sible to calculate exact formulas for eigenvalues of Jacobi matrices (see, e.g., [8,17,20,11]), but it is not possible in general. So, asymptotic and approximate approaches to localize the point spectrum have essential meaning (see, e.g., [2,3,13,16,19,20,24,26]). In the ﬁrst instance, there are applied methods based on the idea that if discrete operators are near-similar in the sense of Rozenbljum then their point spectra are asymptotically close (see [19,13]). Therefore, methods of diagonalization are applied, i.e., there is sought a diagonal matrix which is near-similar to an operator associated with a Jacobi matrix. In [13] Janas and Naboko and in [2,3] Boutet de Monvel et al. introduced and developed the successive diagonalization method. The method we use in this article is closely related to the ideas from [2,3,13].

The aim of this paper is to ﬁnd some asymptotic formulas for eigenvalues of self-adjoint discrete operators inl2

(

N

)

with the canonical basis given by symmetric Jacobi matrices. We continue the research included in [11,16]. The class of Jacobi matrices, for which the method works, is described by conditions

(

C1

)

−

(

C3

)

formulated in Section 2. Although these conditions look restrictive, there is a large the class of Jacobi matrices, which satisfy these properties. At the end of the paper, we give some examples of Jacobi matrices that belong to this class and appear in applications. In Section 2 we formulate the main result of this paper. We put explicit formulas for an asymptotic behaviour of the point spectrum, which use eigenvalues and determinants of ﬁnite matrices. Section 3 includes every auxiliary construction and lemmas. In Section 4 we execute the diagonalization method to complete the proof of the main result. At the end, in Section 5, we give some examples and comments in order to explain how the obtained formulas work.

2. The main result

Let us consider an operatorJin the complex Hilbert spacel2

=

l2

(

N

)

with the canonical basis given by the tridiagonal symmetric Jacobi matrix

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ d1 c1 0

· · · ·

c1 d2 c2 0 . .. 0 c2 d3 c3 . ..

...

. ._. . ._. . ._. . ._. ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ (1)

andJacts on the maximum domain

D

(

J

)

=

{

fn

}

∞n=1

∈

l 2

_{: {}

cn−1fn−1

+

dnfn

+

cnfn+1

}

∞n=1

∈

l 2

(we here takec0

=

f0

=

0).

We assume the real sequences

{

dn

}

∞n=1and

{

cn

}

∞n=1satisfy the following conditions. There exist

c,C,N0

>

0,

α >

1 and

β

0 such that

(

C1

)

dn+1

−

dncnα−1, nN0,

(

C2

)

|

cn

|

Cnβ, n1,

(3)

Notice that condition

(

C3

)

implies that there existsp

∈

Nsuch that

(

C4

) α >

p

+

1

p

β

+

1

.

If

(

C1

)

−

(

C3

)

are satisﬁed then limn→∞

|

dn

| = +∞

and

lim inf n→∞ d2_n c2 n

+

c 2 n−1

= +∞

>

2

;

therefore, the operatorJis self-adjoint, has a compact resolvent and its spectrum is discrete by Janas and Naboko’s criterion (see [14]). The spectrum ofJconsists of the eigenvalues only, so we may denote

σ(

J

)

= {

λ

n

(

J

)

:

n

=

1, 2,

. . .

}

,

where the eigenvalues are arranged increasingly.

We need some auxiliary deﬁnitions and notations to formulate the main result. If 1kldenote

Jk_l

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ dk ck ck dk+1 . .. . ._. _c l−1 cl−1 dl ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ and Dk_l

(λ)

=

detJ_lk

−

λ

.

Additionally, assumeDk_k₋₁

(λ)

=

1.

Let choose the parameterp1 that satisﬁes condition

(

C4

)

. For givennp,J_nn−₊_pp+₋₁1is a symmetric

(

2p

−

1

)

×

(

2p

−

1

)

-matrix. Notice that

J_nn−₊_pp+₋₁1

−

dn e∗_p

|

cn−1

|

2

+ |

cn

|

22Cnβ, wheree∗_p

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 .. . 1₍_p₎ 0 .. . ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

∈

R

2p−1_and

_∗

_{is the Euclidean norm in}_R2p−1_{; therefore, the Wilkinson}

the-orem (see [25] or [24]) implies that there exists an eigenvalue

λ

n

∈

RofJ n−p+1

n+p−1 such that

|

λ

n

−

dn

|

2Cnβ.

Let us ﬁx a sequence

{

λ

n

}

∞n=1

⊂

Rsuch that

Dn_n−₊p_p+₋1₁

(λ

n

)

=

0 and

|

λ

n

−

dn

|

2Cnβ, n1

.

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(Forn

∈ {

1,

. . .

,p

−

1

}

, we formally takeJ_nn₊−_pp+₋₁1

=

J_n1₊_p₋₁andDn_n₊−p_p₋+1₁

(λ)

=

D1_n₊_p₋₁

(λ)

.) In general,

λ

nis not an eigenvalue ofJ. However, we are going to prove that the sequence

{

λ

n

}

determines the

asymptotics of the point spectrum ofJ. Ifp2, from the Laplace formula we derive

Dn_n₊−p_p₋+1₁

(λ)

=

(

dn

−

λ)

D n−p+1 n−1

(λ)

D n+1 n+p−1

(λ)

+

−

c2_n₋₁D_nn−₋p₂+1

(λ)

Dn_n+₊1_p₋₁

(λ)

−

c_n2D_nn−₋₁p+1

(λ)

Dn_n+₊2_p₋₁

(λ).

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Notice that if nis large enough and

λ

∈

dn

−

2Cnβ,dn

+

2Cnβ

then the main diagonals of the matrices

λ

−

J_nn₋−₁p+1andJ_nn+₊1_p₋₁

−

λ

dominate, soDn_n−₋p₁+1

(λ) /

=

0 andDn_n+₊_p1₋₁

(λ) /

=

0. Therefore,

λ

nsatisﬁes

λ

n

=

dn

−

cn2−1 Dn_n−₋p₂+1

(λ

n

)

Dn_n−₋p₁+1

(λ

n

)

−

c_n2D n+2 n+p−1

(λ

n

)

Dn_n+₊1_p₋₁

(λ

n

)

.

(3) Forn1, denote wp,n

(λ)

=

dn

−

c2n−1 Dn_n−₋p₂+1

(λ)

Dn_n−₋p₁+1

(λ)

−

c 2 n Dn_n+₊2_p₋₁

(λ)

Dn_n+₊1_p₋₁

(λ)

, (4) then deﬁne

λ

(1) n

=

dn, (5)

λ

(i) n

=

wp,n

λ

(i−1) n , i2

.

(6)

(Forp

=

1 we simply take

λ

n

=

dnandw1,n

(λ)

=

dn.)

The main results of this paper are formulated as the following theorems.

Theorem 2.1.Let J be an operator in l2given by Jacobi matrix

(

1

)

satisfying conditions

(

C1

)

−

(

C3

)

and let p

∈

Nsatisfy

(

C4

)

then

1.if

{

λ

n

}

satisﬁes(2)then

λ

n

(

J

)

=

λ

n

+

O n−p α−p₊1 p β−1 , n

→ ∞;

2.

λ

n

(

J

)

=

λ

(nq)

+

O n−p α−p₊1 p β−1 , n

→ ∞

, where q

=

min i

∈

N

:

ip+₂1 and

λ

(q)

n is given according to

(

5

)

and

(

6

)

;

3.moreover,

λ

n

(

J

)

=

λ

(ni)

+

O nβ−(2i−1)(α−β−1) , n

→ ∞

, for1ip+₂1

.

Proof.The proof of this theorem we derive from Theorem4.2and Lemma3.4.

(

1

)

(

C1

)

−

(

C3

)

and let p

∈

Nsatisfy

(

C4

).

Assume that there is a sequence

{˜

λ

n

}

∞n=1satisfying

λ

n

(

J

)

= ˜

λ

n

+

O n−p α−p+1 p β−1 , n

→ ∞

, and let

_˜

λ

n

=

wp,n

(

λ

˜

n

)

, n1, where p

=

p

+

1or p

=

p

+

2

.

Then

λ

n

(

J

)

=

λ

˜

n

+

O n−p _α−p+1 p β−1 , n

→ ∞

.

Theorem2.2we derive from Theorem2.1but we need additional estimations to complete the proof. So, this proof is run at the end of Section 3.

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3. Auxiliary construction and lemmas

Let choose a parameterp

∈

Nthat satisﬁes condition

(

C4

)

. We look for sequences of vectorsf(n)

∈

l2

(

f(n)

1

)

and real numbers

λ

n, for which the values

(

J

−

λ

n

)

f(n)are small for largen. This

idea is motivated by the method of diagonalization [2,3,13] together with the Wilkinson theorem [25]. The sequence

{

λ

n

}

∞n=1 is determined by (2), so we are going to construct a system of vectors

f(n)

= {

f_kn

(λ

n

)

}

∞k=1

∈

l2forn1.

Letn1 be ﬁxed. Then denote by

ϕ

(n)

₌

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ f_n(₋n)_p₊₁

(λ

n

)

...

fn(n)

(λ

n

)

...

f_n(₊n)_p₋₁

(λ

n

)

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ (7)

an eigenvector ofJn_n₊−_pp₋+₁1associated with

λ

n. Then Lemma3.2assures thatfn(n)

(λ

n

) /

=

0 fornlarge

enough; therefore, we may assume

f_n(n)

(λ

n

)

=

1

.

(8) Further, put f_n(₋n)_p

(λ

n

)

=

−

cn−p dn−p

−

λ

n f_n(₋n)_p₊₁

(λ

n

)

, (9) f_n(₊n)_p

(λ

n

)

=

−

cn+p−1 dn+p

−

λ

n f_n(₊n)_p₋₁

(λ

n

)

(10) and f_n(_±n)_i

(λ

n

)

=

0, forip

+

1

.

(11)

Finally, forn1, we put

f(n)

= {

f_k(n)

(λ

n

)

}

∞k=1

.

(12)

Because of (11), it is clear thatf(n)

∈

l2.

We may write the Jacobi matrix down as a system of blocks

J

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ J_n1₋_p₋₁ 0

...

cn−p−1 0

. . .

cn−p−1 dn−p cn−p0

. . .

cn−p 0

...

J n−p+1 n+p−1 0

...

cn+p−1 0

. . .

cn+p−1 dn+p cn+p0

. . .

cn+p 0

...

J n+p+1 ∞ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

.

(6)

Using (7)–(12) and the form ofJabove we calculate

(

J

−

λ

n

)

f(n)

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ J1_n₋_p₋₁

−

λ

n 0

+

⎛ ⎜ ⎜ ⎝ 0

...

cn−p−1fn(−n)p

(λ

n

)

⎞ ⎟ ⎟ ⎠ cn−p−10

+

(

dn−p

−

λ

n

)

f( n) n−p

(λ

n

)

+

cn−pf( n) n−p+1

(λ

n

)

⎛ ⎜ ⎜ ⎝ cn−pf( n) n−p

(λ

n

)

0

...

⎞ ⎟ ⎟ ⎠

+

J_nn₊−_pp₋+₁1

−

λ

n

ϕ

(n)

₊

⎛ ⎜ ⎜ ⎝ 0

...

cn+p−1fn(+n)p

(λ

n

)

⎞ ⎟ ⎟ ⎠ cn+p−1f( n) n+p−1

(λ

n

)

+

(

dn+p

−

λ

n

)

f( n) n+p

(λ

n

)

+

cn+p0 ⎛ ⎜ ⎜ ⎝ cn+pf( n) n+p

(λ

n

)

0

...

⎞ ⎟ ⎟ ⎠

+

J_∞n+p+1

−

λ

n 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

;

therefore, for large enoughn1, we obtain the following equality

(

J

−

λ

n

)

f(n)

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

. . .

0 cn−p−1fn(−n)p

(λ

n

)

0 cn−pf( n) n−p

(λ

n

)

0

...

0 cn+p−1fn(+n)p

(λ

n

)

0 cn+pf( n) n+p

(λ

n

)

0

. . .

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

.

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Clearly,

λ

nis not an eigenvalue ofJandf(n)is not an eigenvector but we are going to prove that the

sequence

{

λ

n

}

determines the asymptotics of the point spectrum ofJ.

Let us formulate the following simple lemma.

Lemma 3.1.Let A

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ a1 b1 b1 a2 . .. . ._. . ._. _b 2p−2 b2p−2 a2p−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ,

ϕ

=

⎛ ⎜ ⎜ ⎝

ϕ

1

...

ϕ

2p−1 ⎞ ⎟ ⎟ ⎠

(7)

and

λ

∈

Rbe such that

ϕ /

=

0and

(

A

−

λ)ϕ

=

0

.

Assume that

λ

is not an eigenvalue for any matrix Ai_j

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎝ a_i b_i bi ai+1 . .. . ._. . ._. bj−1 b_j₋₁ a_j ⎞ ⎟ ⎟ ⎟ ⎟ ⎠,where1ijp

−

1or p

+

1ij2p

−

1

.

Then 1.

ϕ

p

=

/

0 2.

ϕ

p−i

= −

bp−i detA1 p₋i₋1−λ detA1_p₋_i−λ

ϕ

p−i+1 and

ϕ

p+i

= −

bp+i−1 detAp₂+_p₋i+₁1

−

λ

det Ap₂+_p₋i₁

−

λ

ϕ

p+i−1 for i

∈ {

1,

. . .

,p

−

1

}

.

Proof.DenoteΦ_ji

=

⎛ ⎜ ⎝ ϕi .. . ϕj ⎞ ⎟

⎠, for 1ij2p

−

1. Notice that

(

A

−

λ)ϕ

=

0

⇔

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ A1_l₋₁

−

λ

Φ1 l−1

+

⎛ ⎜ ⎜ ⎝ 0

...

bl−1

ϕ

l ⎞ ⎟ ⎟ ⎠

=

0 b_⎛l−1

ϕ

l−1

+

(

al

−

λ)ϕ

l

+

bl

ϕ

l+1

=

0 ⎜ ⎜ ⎝ bl

ϕ

l

...

0 ⎞ ⎟ ⎟ ⎠

+

Al₂+_p₋1₁

−

λ

Φ₂l+_p₋1₁

=

0 (14)

forl

∈ {

1,

. . .

, 2p

−

1

}

. (We assume thatbk

=

0, ifk0 ork2p.)

Ad. 1.Consider (14) forl

=

p. If

ϕ

p

=

0 then

A1_p₋₁

−

λ

Φ_p1₋₁

=

0 andA₂p_p+₋1₁

−

λ

Φ₂p+_p₋1₁

=

0. But

λ

is not an eigenvalue for neither ofA1_p₋₁ andAp₂+_p₋1₁, soΦ_p1₋₁

=

0 andΦ₂p+_p₋1₁

=

0. This yields

ϕ

=

0, which contradicts the assumption since

ϕ /

=

0.

Ad. 2.Forl

∈ {

2,

. . .

,p

}

,A1_l₋₁

−

λ

is invertible. If we denote

A1_l₋₁

−

λ

₋1

=

(

mi,j

)

il,−j=11thenml−1,l−1

=

det

(

A1_l₋2−λ

)

det

(

A1 l₋1−λ

)

. From (14) we deriveΦ_l1₋₁

= −

A1_l₋₁

−

λ

−1 ⎛ ⎝ 0 .. . b_l₋₁ϕ_l ⎞ ⎠and thus

ϕ

l−1

= −

ml−1,l−1

bl−1

ϕ

l. So, if we substitutel

−

1

=

p

−

ithen we may write that

ϕ

p−i

= −

bp−i detA1 p₋i₋1−λ detA1 p−i−λ

ϕ

_p₋_i₊₁ fori

∈ {

1,

. . .

,p

−

1

}

. Letl

∈ {

p,

. . .

, 2p

−

2

}

, thenΦ₂l+_p₋1₁

= −

Al₂+_p₋1₁

−

λ

₋1 ⎛ ⎝ b_lϕ_l .. . 0 ⎞ ⎠and if we denote Al₂+_p₋1₁

−

λ

₋1

=

(

wi,j

)

i,j then

ϕ

l+1

= −

blw1,1

ϕ

l, where w1,1

=

detAl₂+_p2₋1−λ det Al2+p1₋1−λ

. But this means that

ϕ

p+i

=

−

bp+i−1

detAp₂+_p₋i+₁1−λ

(8)

Lemma 3.2. Let

(

C1

)

−

(

C4

)

hold

.

For n1we assume that

λ

nis an eigenvalue of J n−p+1

n+p−1such that

(

2

)

holds

.

Assume

(

7

)

,where

ϕ

(n)is an eigenvector of J_nn₊−p_p₋+₁1associated with

λ

n

.

Then there exists n0such that for nn0

1.fn(n)

(λ

n

) /

=

0, 2.f_n(₋n)_i

(λ

n

)

= −

cn−i Dn_n−₋p_i₋+₁1(λn) Dn_n−₋p_i+1(λn) f_n(₋n)_i₊₁

(λ

n

)

and f_n(₊n)_i

(λ

n

)

= −

cn+i−1 Dn_n+₊i_p+₋1₁

(λ

n

)

Dn_n+₊i_p₋₁

(λ

n

)

f_n(₊n)_i₋₁

(λ

n

)

, for i

∈ {

1,

. . .

,p

−

1

}

.

Proof.The eigenvalue

λ

nsatisﬁes

|

λ

n

−

dn

|

2Cnβ; therefore, ifnis large andn

−

p

+

1lkn

−

1

then Jl_k

−

λ

n

=

⎛ ⎜ ⎜ ⎝ dl

−

dn

+

O

(

nβ

)

cl cl . .. ck−1 ck−1 dk

−

dn

+

O

(

nβ

)

⎞ ⎟ ⎟ ⎠

is a non-singular matrix because of

(

C1

)

−

(

C3

)

this matrix has a dominating diagonal. Similarly, J_kl

−

λ

nis non-singular ifn

+

1lkn

+

p

−

1. Then we apply Lemma3.1toJ

n−p+1

n+p−1,

λ

nand

ϕ

(n)

to complete the proof.

Denote R(_in)

(λ)

=

D n−p+1 n−p+i−1

(λ)

Dn_n−₋p_p+₊1_i

(λ)

, i

∈ {

1,

. . .

,p

−

1

}

.

(15)

The Laplace formula, applied to the determinant ofJ_kn−p+1

−

λ

, implies the following equation Dn_k−p+1

(λ)

=

(

dk

−

λ)

Dkn−−1p+1

(λ)

−

c 2 k−1D n+p−1 k−2

(λ)

, kn

−

p

+

2

;

therefore, D_kn−₋₁p+1

(λ)

D_kn−p+1

(λ)

=

⎛ ⎝dk

−

λ

−

ck2−1 Dn_k−₋p₂+1

(λ)

Dn_k−₋p₁+1

(λ)

⎞ ⎠ −1

.

If we use (15) and putk

=

n

−

p

+

ito the formula above then we obtain R(_in)

(λ)

=

1

dn−p+i

−

λ

−

cn2−p+i−1R( n) i−1

(λ)

, fori

∈ {

2,

. . .

,p

−

1

};

(16)

additionally, directly from (15) we have

R(₁n)

(λ)

=

1 dn−p+1

−

λ

.

(17) Now, denote S_i(n)

(λ)

=

D n+p+1−i n+p−1

(λ)

Dn_n+₊p_p−₋i₁

(λ)

, i

∈ {

1,

. . .

,p

−

1

}

.

(18)

(9)

In this case, from the Laplace formula for the determinant ofJ_nk₊_p₋₁

−

λ

, we derive Dk_n₊_p₋₁

(λ)

=

(

dk

−

λ)

Dkn++1p−1

(λ)

−

c 2 kD k+2 n+p−1

(λ)

, kn

+

p

−

2 and Dk_n+₊1_p₋₁

(λ)

Dk_n₊_p₋₁

(λ)

=

⎛ ⎝dk

−

λ

−

c2k Dk_n+₊2_p₋₁

(λ)

Dk_n+₊1_p₋₁

(λ)

⎞ ⎠ −1 fork

∈ {

n

+

1,

. . .

,n

+

p

−

2

}

.

Again, if we putk

=

n

+

p

−

ito this formula and use (18) then we have

S_i(n)

(λ)

=

1 dn+p−i

−

λ

−

cn2+p−iS( n) i−1

(λ)

, fori

∈ {

2,

. . .

,p

−

1

}

.

(19) Moreover, (18) implies S₁(n)

(λ)

=

1 dn+p−1

−

λ

.

(20)

Lemma 3.3. Let R(_in)

(λ)

and S_i(n)

(λ)

for n1 and i

∈ {

1,

. . .

,p

−

1

}

be deﬁned by

(

15

)

and

(

18

)

, respectively

.

Assume conditions

(

C1

)

−

(

C4

)

are satisﬁed

.

Then the following facts hold

.

1.There exist M,n0

>

0such that for nn0

R(_in)

(λ)

M nα−1, S_i(n)

(λ)

M nα−1 for

λ

∈

dn

−

2Cnβ,dn

+

2Cnβ and i

∈ {

1,

. . .

,p

−

1

}

.

2.There exist M,n0

>

0such that for nn0

R(_in)

(

λ)

˜

−

R_i(n)

(

λ)

˜

M n2(α−1)

|˜

λ

−

_λ

_˜

_|

and S_i(n)

(

λ)

˜

−

S_i(n)

(

λ)

˜

M n2(α−1)

|˜

λ

−

_˜

λ

|

for

λ

˜

,

λ

˜

∈

dn

−

2Cnβ,dn

+

2Cnβ and i

∈ {

1,

. . .

,p

−

1

}

.

Proof.Ad.1. Choose

λ

=

dn

+

rn, wherern

∈

−

2Cnβ, 2Cnβ. From (17) and

(

C1

)

−

(

C3

)

, we obtain

R(₁n)

(λ)

=

1 dn−p+1

−

dn

−

rn 1 cnα−1

₋

₂_Cnβ C1 1 nα−1

fornn1, wheren1is large enough andC1

>

0 is independent onn. Using induction, we continue the

proof. Assume thati

∈ {

2,

. . .

,p

−

1

}

and

R(_i₋n)₁

(λ)

Ci−1

nα−1, nni−1,

then using (15) and

(

C1

)

−

(

C3

)

again, we estimate

R(_in)

(λ)

=

dn−p+i

−

λ

−

c12n−p+i−1R( n) i−1

(λ)

1

|

dn−p+i

−

dn

−

rn

| −

C2

(

n

−

p

+

i

−

1

)

2βCi−1n1−α 1 c

(

n

−

1

)

α−1

₋

₂_Cnβ

₋

_C_n2β_n1−α Ci 1 nα−1,

(10)

wherennini−1for somenilarge enough andCi

>

0 can be chosen independently onn. Finally,

letM

=

max

{

C1,

. . .

,Cp−1

}

andn0

=

np−1.

The proof, that the functionsS(_in)satisfy the same properties, is similar.

Ad.2.Let

λ

˜

=

dn

+ ˜

r,

λ

˜

=

dn

+ ˜˜

rand

˜

r,

˜˜

r

∈ [−

2Cnβ, 2Cnβ

]

. The proof of this part of lemma is also

obtained by induction. Because of (17) we have

R(₁n)

(

λ)

˜

−

R₁(n)

(

λ)

˜

=

_λ

_˜

_{− ˜}

_λ

(

dn−p+1

− ˜

λ)(

dn−p+1

−

λ)

˜

|

_λ

_˜

_{− ˜}

_λ

_|

|

(

dn−p+1

−

dn

− ˜

r

)(

dn−p+1

−

dn

− ˜˜

r

)

|

_˜

λ

− ˜

λ

|

(

cnα−1

−

₂_Cnβ

)

2 C1 n2(α−1)

|˜

λ

−

_˜

λ

|

,

for someC1

>

0 andnlarge enough, where the last inequality comes directly from

(

C1

)

,

(

C2

)

and

(

C3

)

. Then we continue the inductive step of the proof. Leti

∈ {

2,

. . .

,p

−

1

}

, from (15) we derive

R(_in)

(

λ)

˜

−

R(_in)

(

λ)

˜

=

_λ

_˜

₊

_c2 n−p+i−1R( n) i−1

(

λ)

˜

− ˜

λ

−

c2n−p+i−1R( n) i−1

(

λ)

˜

dn−p+i

− ˜

λ

−

cn2−p+i−1R( n) i−1

(

λ)

˜

dn−p+i

−

λ

˜

−

c2n−p+i−1R( n) i−1

(

λ)

˜

|

_˜

λ

− ˜

λ

| +

C2

(

n

−

p

+

i

−

1

)

2βR(_i₋n)₁

(

λ))

˜

−

R_i(₋n)₁

(

λ)

˜

|

dn−p+i

−

dn

| − |˜

r

| −

M1nβ

|

dn−p+i

−

dn

| − |˜˜

r

| −

M1nβ ,

notice that the ﬁrst point of this lemma implies that c2_n₋_p₊_i₋₁R_i(₋n)₁

(

λ)

˜

M1 n 2_β

nα−1 M1nβ and

c2_n₋_p₊_i₋₁R_i(₋n)₁

(

λ)

˜

M1nβ. Now, we apply the inductive assumption to the estimates above and obtain

R(_in)

(

λ)

˜

−

R_i(n)

(

λ)

˜

|

_˜

λ

− ˜

λ

| +

C2

(

n

−

p

+

i

−

1

)

2βCi−1n−2(α−1)

|

λ

˜

− ˜

λ

|

dn−p+i

−

dn

| −

M2nβ 2 Ci n2(α−1)

|

_λ

_˜

_{− ˜}

_λ

_|

,

for largen. Finally, letM

=

max

{

C1,

. . .

,Cp−1

}

. The inequalities forS( n)

i we obtain with use of the same

methods.

Lemma 3.4. Let

{

λ

n

}

∞n=1satisfy

(

2

)

and

λ

( i)

n be given by

(

5

)

and

(

6

)

for i,n1. Under conditions

(

C1

)

−

(

C4

)

,for i2

λ

n

=

λ

(ni)

+

O nβ n(2i−1)(α−β−1) , n

→ ∞

.

(

There exist Ci

>

0and ni

>

0such that

λ

n

−

λ

( i)

n Ci n β

(11)

Proof.Because of (15) and (18) and the deﬁnition of

λ

(ni)given by formula (6), we may write

λ

(i) n

=

dn

−

cn2−1R( n) p−1

λ

(i−1) n

−

c2_nS_p(n₋)₁

λ

(i−1) n fori2.

First, we prove that

λ

(ni)

∈

dn

−

2Cnβ,dn

+

2Cnβ

, if nni, for some large ni. This holds for

λ

(1)

n

=

dn, so we start to proceed by induction. Indeed, if we assume

λ

(ni)

∈

dn

−

2Cnβ,dn

+

2Cnβ

for nnithen Lemma3.3impliescn2−1R(

n) p−1

(λ

( i) n

)

,cn2S( n) p−1

λ

(i) n M n β nα−β−1

<

Cnβ fornni+1

>

ni, so

λ

(ni+1)

∈

dn

−

2Cnβ,dn

+

2Cnβ fornni+1. Notice that

λ

n

∈

dn

−

2Cnβ,dn

+

2Cnβ and

λ

n

=

dn

−

cn2−1R( n) p−1

(λ

n

)

−

c2nS( n) p−1

(λ

n

)

because of (2), (3), (15) and (18). Therefore; ifnis large enough, with use of Lemma3.3, we estimate

λ

n

−

λ

(n1)c 2 n−1R( n) p−1

(λ

n

)

+

cn2S( n) p−1

(λ

n

)

M n2β nα−1, nn1, and

λ

n

−

λ

( 2) n cn2−1R( n) p−1

(λ

n

)

−

R( n) p−1

λ

(1) n

+

c2nS( n) p−1

(λ

n

)

−

S( n) p−1

λ

(1) n Mn2β n2_(α₋1₎

λ

n

−

λ

( 1) n , nn2

.

If we assumei2 and

λ

n

−

λ

(ni) Mi n2(i−1)(α−β−1)

λ

n

−

λ

(n1), nni then

λ

n

−

λ

( i+1) n c2n−1R( n) p−1

(λ

n

)

−

R( n) p−1

λ

(i) n

+

c2nS( n) p−1

(λ

n

)

−

S( n) p−1

λ

(i) n Mn2β n2(α−1)

λ

n

−

λ

(i) n Mn2β n2_(α₋1₎ Mi n2₍i₋1_)(α₋_β₋1₎

λ

n

−

λ

( 1) n

=

_n2i_(αM₋i+_β1₋1₎

λ

n

−

λ

( 1) n ,

wherenni+1

>

ni, because of Lemma3.3. Finally, ifi1 then

λ

n

−

λ

(ni)

M_inβ n(2i−1)(α−β−1)

fornni, whereniis large enough.

Lemma 3.5. Assume

(

C1

)

−

(

C4

)

hold

.

Let

{

λ

n

}

∞n=1 satisfy

(

2

)

and f( n) k

(λ

n

)

, n,k1, be given by

(

7

)

–

(

11

).

Then f_n(_±n)_i

(λ

n

)

=

O ₁ ni(α−β−1) , n

→ ∞

, for i

∈ {

1, 2,

. . .

,p

}

.

Proof.Notice that due to (2) we can write

λ

n

=

dn

+

rn, where

|

rn

|

2Cnβ. Using (15) and Lemma

3.2we derive

f_n(₋n)₁

(λ

n

)

=

−

cn−1R( n) p−1

(λ

n

)

,

becausefn(n)

(λ

n

)

=

1. Then by Lemma3.3we may estimate

f_n(₋n)₁

(λ

n

)

C

(

n

−

1

)

β M nα−1 C nα−β−1

(12)

fornlarge enough. Then we proceed with induction. Assumef_n(₋₍n)_i₋₁₎

(λ

n

)

_n₍i₋1M_)(αi−₋1_β₋1₎. From Lemma 3.2and (15) we derive f_n(₋n)_i

(λ

n

)

= −

cn−iR( n) p−i

(λ

n

)

f( n) n−(i−1)

(λ

n

)

;

therefore, f_n(₋n)_i

(λ

n

)

Cnβ M nα−1 Mi−1 n(i−1)(α−β−1)

=

Mi ni(α−β−1),

forip(we takeR(₀n)

(λ)

=

_d 1

n₋p−λbecause of (9)).

Using Eqs. (18) and (10) and Lemma3.2, we apply similar methods for

f_n(₊n)_i

(λ

n

)

n

(

1ip

)

.

Proof of Theorem 2.2.Letp

=

p

+

1 orp

=

p

+

2. Ifpsatisﬁes

(

C4

)

thenpalso satisﬁes this in-equality. Let

{

α

n

}

be a sequence satisfying (2) with the parameterpi.e.

α

n

=

wp,n

(α

n

)

and

|

α

n

−

dn

|

2Cnα,n1. Theorem2.1implies that

α

n

−

λ

n

(

J

)

=

O

n−p _α−p+1

p β−1

. From this fact and the assumption of Theorem2.2we derive

α

n

− ˜

λ

n

=

(α

n

−

λ

n

(

J

))

+

(λ

n

(

J

)

− ˜

λ

n

)

=

O n−p _α−p+1 p β−1

+

O n−p α−p+1 p β−1

=

O n−p α−p₊1 p β−1

.

(21)

Applying notations (3), (15) and (18) and Lemma3.3for the parameterpinstead ofp, we observe that

|

α

n

−

λ

˜

n

| =

wp,n

(α

n

)

−

wp,n

(

λ

˜

n

)

c2_n₋₁R_p(n₋)₁

(α

n

)

−

R(_pn₋)1

(

λ

˜

n

)

+

c 2 nS( n) p−1

(α

n

)

−

S (n) p−1

(

λ

˜

n

)

Bn−2(α−β−1)

|

α

n

− ˜

λ

n

|

,

for a constantB

>

0 and largen. So, (21) implies

|

α

n

−

λ

˜

n

|

Bn−2(α−β−1)n− pα−p+_p1β−1

=

Bn−(p+2) α−p+3 p+2β−1

for largen. Finally, we obtain

λ

n

(

J

)

=

α

n

+

O n−p _α−p+1 p β−1

=

λ

˜

n

+

O n−p _α−p+1 p β−1 ,

asn

→ ∞

, so the proof is complete.

4. Diagonalization and operator methods

Let

{

en

}

∞n=1be the standard orthonormal basis inl2. BySwe denote the shift operator onl2given on

the basis vectors as followsSen

=

en+1,n1. ThenS∗stands for the adjoint operator toS. Obviously,

(13)

S

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 0

· · ·

1 0 . .. 0 1 . .. . ._. . ._. ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , S∗

=

⎛ ⎜ ⎜ ⎜ ⎝ 0 1 0

· · ·

0 0 1 . .. . ._. . ._. . ._. ⎞ ⎟ ⎟ ⎟ ⎠

.

Denote diagQ

= {

qn

}

∞n=1, ifQ

=

⎛ ⎜ ⎜ ⎝ q₁ 0 0 q2 . .. . ._. . ._. ⎞ ⎟ ⎟

⎠is a diagonal operator inl2represented by a diagonal matrix. Let Λ

=

⎛ ⎜ ⎜ ⎝

λ

1 0 0

λ

2 . ._. ⎞ ⎟ ⎟ ⎠ (22)

be a diagonal operator with diagΛ

= {

λ

n

}

∞n=1satisfying (2).

Letf(n)

=

f_k(n)

(λ

n

)

∞

k=1be given as in (7)–(12). Put

F

=

f(1)

;

f(2)

;

. . .

, (23)

i.e.Fis an inﬁnite matrix, in which n-th column is given by sequencef(n). Notice that we may describe the structure ofFas follow:

F

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1 f₁(2)

(λ

2

)

. .. 0 f₂(1)

(λ

1

)

1 fn(−n)p

(λ

n

)

0

...

f₃(2)

(λ

2

)

...

fn(+n+1−1)p

(λ

n+1

)

f₁(₊1)_p

(λ

1

)

...

. .. f( n) n−1

(λ

n

)

...

. .. 0 f₂(₊2)_p

(λ

2

)

1 f( n+1) n

(λ

n+1

)

0 f_n(₊n)₁

(λ

n

)

1 . .. . ._.

...

_f(n+1) n+2

(λ

n+1

)

. .. fn(+n)p

(λ

n

)

...

. .. 0 f_n(₊n+₁₊1)_p

(λ

n+1

)

0 . .. ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

=

I

+

K, whereK

=

p_i₌₁SiWi

+

ViS∗i

andWi,Viare diagonal operators inl2with diagWi

=

f_n(₊n)_i

(λ

n

)

∞ n=1 and diagVi

=

f_n(₋n)_i

(λ

n

)

∞ n=i+1

=

fn(n+i)

(λ

n+i

)

∞

n=1. Lemma3.5yieldsKis a compact operator

(14)

Denote by E(n)

=

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

. . .

cn−p−1f( n) n−p

(λ

n

)

0 cn−pf( n) n−p

(λ

n

)

0

...

0 cn+p−1f( n) n+p

(λ

n

)

0 cn+pf( n) n+p

(λ

n

)

. . .

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ (n−p−1) (n−p+1) (n+p−1) (n+p+1)

the column in formula (13). Then we construct an inﬁnite matrix E

=

E(1)

;

E(2)

;

E(3)

;

. . .

, (24)

whereE(n)is treated as an n-th column ofE. Inl2with the standard basis, the matrixEgives an operator, that may be written in a 4-diagonal form

E

=

Sp−1A1

+

Sp+1A2

+

B1S∗p−1

+

B2S∗p+1,

whereA1,A2,B1,B2are diagonal operators with

diagA1

=

cn+p−1fn(+n)p

(λ

n

)

∞ n=1, diagA2

=

cn+pf( n) n+p

(λ

n

)

∞ n=1, diagB1

=

cn−pf( n) n−p

(λ

n

)

np

=

cn−1f( n−1+p) n−1

(λ

n−1+p

)

∞ n=1,

(

c0

=

0

)

, diagB2

=

cn−p−1f( n) n−p

(λ

n

)

np+2

=

cnf( n+p+1) n+1

(λ

n+p+1

)

∞ n=1

.

Notice thatEis a compact operator inl2. Indeed, from Lemma3.5we derive cn±pfn(±n)p

(λ

n

)

=

O nβ np(α−β−1)

=

O ⎛ ⎝ 1 np α−p₊1 p β−1 ⎞ ⎠ and cn±p−1f( n) n±p

(λ

n

)

=

O ⎛ ⎝ 1 np α−p₊1 p β−1 ⎞ ⎠, n

→ ∞

, so the sequences above converge to 0 because of

(

C4

)

. Moreover;

Een

=

E(n)

=

O ⎛ ⎝ 1 np α−p+1 p β−1 ⎞ ⎠, n

→ ∞

.

(25)

Finally, we can rewrite (13) as an operator equation

JF

−

FΛ

=

E

+

E, (26)

whereJis given by (1),Fby (23),Eby (24) andEis a ﬁnite dimensional operator, that is represented in the canonical basis by a matrix with a ﬁnite number of non-zero entries only. The operatorEappears

(15)

in (26) because the properties off_k(n)

(λ

n

)

, described in Lemma3.2, hold for largenand we may need

some corrections for smalln.

In [13] there is proved that there exists a bounded invertible inl2operatorFsuch that the matrix representation ofF

−

Fhas a ﬁnite number of non-zero entries only. Therefore, (26) implies

JF

=

FΛ

+

E

+

M, (27)

whereM

=

E

+

J

(

F

−

F

)

−

(

F

−

F

)

Λ. The choice ofEandFand the tridiagonal formula ofJ guar-antee that the entries of the matrixM, except of a ﬁnite number, are equal to zero. BecauseJandΛare self-adjoint, from (27) we derive

F∗J

=

ΛF∗

+

E∗

+

M∗,

whereF∗,E∗,M∗mean adjoint toF,EandM, respectively.F∗is invertible, so we notice J

=

(

F∗

)

−1Λ

+

(

E∗

+

M∗

)(

F∗

)

−1F∗,

i.e.,Jis similar toΛ

+

R, whereR

=

(

E∗

+

M∗

)(

F∗

)

−1. Moreover;Ris a compact operator because E∗andM∗are compact and

(

F∗

)

−1is bounded. The condition of similarity preserves the structure of spectra, so

λ

n

(

J

)

=

λ

n

(

Λ

+

R

)

, n1

.

(28)

The following lemma is proved in [13].

Lemma 4.1 (Janas, Naboko).Let D be a self-adjoint diagonal operator in a Hilbert space H given by

Den

=

μ

nen,where

{

en

}

is an orthonormal basis in H and simple eigenvalues

μ

n

→ ∞

are ordered

by

|

μ

k

|

μ

k+1

|

.

Assume that

|

μ

i

−

μ

k

|

>

0if i

=

/

k

.

If R is a compact

(

not necessary self-adjoint

)

operator in H then the operator T

=

D

+

R has discrete spectrum,which consists of complex eigenvalues

λ

n

(

T

)

and

λ

n

(

T

)

=

μ

n

+

O

(

R∗en

)

,n

→ ∞

.

If we apply the lemma above to the self-adjoint diagonal operatorΛand the compact operatorR then

λ

n

(

Λ

+

R

)

=

λ

n

+

OR∗en, n

→ ∞

.

(29)

Finally, from (28), (29) and (25) we derive

λ

n

(

J

)

=

λ

n

+

OF−1

(

E

+

M

)

en

=

λ

n

+

O

(

Een

)

=

λ

n

+

O n−p α−p+1 p β−1 , n

→ ∞

.

The considerations above yield the following theorem

(

1

)

(

C1

)

−

(

C3

)

and let p

∈

Nsatisfy

(

C4

).

Then

λ

n

(

J

)

=

λ

n

+

0 n−p α−p+1 p β−1 , n

→ ∞

, where

{

λ

n

}

∞n=1satisﬁes

(

2

).

Remark 4.1.Instead of Lemma4.1we can use the Rozenbljum theorem ([19]) to obtain the result

above.

5. Examples and applications

In this section we give examples of applications of Theorems2.1and2.2to Jacobi matrices related to problems that appear in literature.

(16)

Proposition 5.1.Let J be a Jacobi matrix with dn

=

n2

+

an

√

n

+

bn, cn

=

ρ

n

√

n

+

γ

n,

where

{

an

}

,

{

bn

}

,

{

ρ

n

}

and

{

γ

n

}

are bounded real sequences

.

Then the following asymptotic formulas are

true 1.

λ

n

(

J

)

=

dn

−

c2n₋1 dn−1−dn

−

c2 n dn+1−dn

+

O 1 √ n , n

→ ∞

.

2.

λ

n

(

J

)

=

n2

+

an

√

n

+

bn

+

ρ 2 n₋1−ρn2 2

+

ρ2 n₋1An−ρn2An+1 4n

+

Pn

+

O 1 n ,as n

→ ∞

,where An

=

an−1

√

n

−

1

−

an

√

n (30) and Pn

=

γ

n

ρ

n

√

n

−

γ

n−1

ρ

n−1

√

n

−

1 n

.

(31)

3.if we assume an

=

0and

γ

n

=

0then

λ

n

(

J

)

=

n2

+

bn

+

ρ

2 n−1

−

ρ

2 n 2

+

1 4n

ρ

2 n−1Wn

+

ρ

2nVn

+

O ₁ n2 , where

{

Wn

}

and

{

Vn

}

are bounded sequences given by

Wn

= −

1

+

bn−1

−

bn

−

ρ

2 n−1

−

ρ

n2 2

+

1 4

ρ

2 n−2, (32) Vn

=

1

+

bn+1

−

bn

−

ρ

2 n−1

−

ρ

n2 2

−

1 4

ρ

2 n+1

.

(33)

Proof.Ad.1. Obviously

{

dn

}

and

{

cn

}

satisfy

(

C1

)

−

(

C3

)

with

α

=

2 and

β

=

1₂. Thenp

=

2 is the

smallest positive integer that satisﬁes

(

C4

)

. From (4) we calculate w2,n

(λ)

=

dn

−

c2_n₋₁ dn−1

−

λ

−

c_n2 dn+1

−

λ

.

To apply Theorem2.1we ﬁndq

=

min

i

∈

N

:

ip+₂1

=

2, p

α

−

p+1 p

β

−

1

=

1 2 and

λ

(2) n

=

w2,n

(

dn

)

=

dn

−

c_n2₋₁ dn−1

−

dn

−

c2_n dn+1

−

dn

.

Therefore, with use of Theorem2.1, we obtain

λ

n

(

J

)

=

λ

(n2)

+

O 1

√

n

=

n2

+

an

√

n

+

bn

+

ρ

2 n−1

−

ρ

n2 2

+

O 1

√

n

.

Ad.2.If we want to obtain a more exact asymptotics then we choosep

=

3 that satisﬁes

(

C4

)

and we apply Theorem2.1again. We calculateq

=

2,p

α

−

p+_p1

β

−

1

=

1 and according to (4)–(6) we have w3,n

(λ)

=

dn

−

c_n2₋₁ dn−1

−

λ

−

cn2−2

/(

dn−2

−

λ)

−

cn2 dn+1

−

λ

−

cn2+1

/(

dn+2

−

λ)

and

λ

(n2)

=

w3,n

(

dn

)

. Finally, we obtain

λ

n

(

J

)

=

λ

( 2) n

+

O 1 n

=

n2

+

an

√

n

+

bn

+

ρ 2 n−1−ρ 2 n 2

+

ρ2 n−1An−ρn2An₊1 4n

+

Pn

+

O ₁ n ,