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Research article

The asymptotic behavior of the reciprocal sum of generalized Fibonacci numbers

  • Received: 15 October 2024 Revised: 31 December 2024 Accepted: 15 January 2025 Published: 24 January 2025
  • Let (un)n0 be the special Lucas u-sequence defined by

    un+2=Aun+1Bun,u0=0,u1=1,

    where n0, B=±1, and A is an integer such that A24B>0. Let

    ak=1usmk,1umk+umk+l,1li=0umk+i,1umkumk+2l,1umkumk+2l1,1umk+C,

    where m,l are positive integers, s=1,2,3,4, and C is any constant. The aim of this paper is to find a form gn such that

    limn((k=nak)1gn)=0.

    For example, we show that

    limn((k=n1umk)1(umnum(n1)))=0.

    Citation: Hongjian Li, Kaili Yang, Pingzhi Yuan. The asymptotic behavior of the reciprocal sum of generalized Fibonacci numbers[J]. Electronic Research Archive, 2025, 33(1): 409-432. doi: 10.3934/era.2025020

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  • Let (un)n0 be the special Lucas u-sequence defined by

    un+2=Aun+1Bun,u0=0,u1=1,

    where n0, B=±1, and A is an integer such that A24B>0. Let

    ak=1usmk,1umk+umk+l,1li=0umk+i,1umkumk+2l,1umkumk+2l1,1umk+C,

    where m,l are positive integers, s=1,2,3,4, and C is any constant. The aim of this paper is to find a form gn such that

    limn((k=nak)1gn)=0.

    For example, we show that

    limn((k=n1umk)1(umnum(n1)))=0.



    In the past years, many mathematicians were interested in finding the formula for the integer part of the reciprocal tails of the convergent series. That is, find the explicit value of (k=nak)1 when k=1ak converges. The motivation of such research comes from the reciprocal sum of Fibonacci numbers. Let us recall that the Fibonacci sequence (Fn)n0 is defined by

    Fn+2=Fn+1+Fn,F0=0,F1=1,

    where n0. In [1], Ohtsuka and Nakamura proved

    (k=n1Fk)1={Fn2,if n2 is even;Fn21,if n1 is odd,

    and

    (k=n1F2k)1={Fn1Fn1,if n2 is even;Fn1Fn,if n1 is odd,

    where x denotes the greatest integer x. In [2], Wang proved

    (k=n1F3k)1={FnF2n1+Fn2F2n+111(14Fn25Fn),if n2 is even;FnF2n1+Fn2F2n+111(5Fn14Fn2),if n1 is odd,

    where F1=F1=1. In [3], Hwang et al. provided the relevant formula for the reciprocal sum of the fourth power of the Fibonacci numbers, that is,

    (k=n1F4k)1=F4nF4n1+2(1)n5F2n1{n+25},

    where {x}=xx. In addition, some mathematicians studied the reciprocal sums of Fibonacci, Lucas, and Pell sequences, such as [4,5,6,7], while some mathematicians studied the reciprocal sums of other types of sequences, such as [8,9,10].

    Many mathematicians also considered the asymptotic behavior of these sequences. That is, find a suitable function gn such that

    (k=nak)1gn

    when k=1ak converges. Here the notation AnBn means that

    limn(AnBn)=0.

    In [11], Lee et al. proved that

    (k=n1Fmkl)1FmnlFm(n1)l

    for any positive integer m and 0lm1. In [12], Marques et al. proved that for any positive integer m there exists a positive constant Cm such that

    (k=n1F2mk)1F2mnF2m(n1)+(1)mnCm.

    Moreover, they gave an explicit form for Cm as follows:

    Cm={2(L2m2)25F2m5,ifmis even,2(L2m+2)5L2m,ifmis odd,

    where Ln is the nth Lucas number. In [3], Hwang et al. studied the asymptotic behavior of the reciprocal sum of the fourth power of the Fibonacci numbers, and they proved that

    (k=n1F4k)1F4nF4n1+2(1)n5F2n1+2575.

    In [13], Lee and Park studied the asymptotic behavior of the reciprocal sum of FkFk+m, and they proved that

    (k=n1FkFk+2l)1Fn+l1Fn+l(F2l+(1)l)(1)n3

    and

    (k=n1FkFk+2l1)1F2n+l1(Fl1Fl+(1)l)(1)n3,

    where l is a positive integer. In addition, some mathematicians studied other types of asymptotic behavior, such as [14,15]. Inspired by the above results, we use the method of Yuan et al. [16] to study the asymptotic behavior of the sequences that are more general than the Fibonacci sequence. Let (un)n0 be the special Lucas u-sequence defined by

    un+2=Aun+1Bun,u0=0,u1=1, (1.1)

    where n0, B=±1, and A is an integer such that A24B>0. The relevant properties of the Lucas u-sequence can be found in Sun's book [17]. We know that the Binet formula is related to the sequence (un)n0 has the form

    un=αnβnαβ,n0, (1.2)

    where

    α,β=A±A24B2.

    Let

    ak=1usmk,1umk+umk+l,1li=0umk+i,1umkumk+2l,1umkumk+2l1,1umk+C,

    where m and l are positive integers, s=1,2,3,4, and C is any constant. The aim of this paper is to find a form gn such that

    (k=nak)1gn.

    The rest of this paper is organized as follows: in Section 2, we give our main results. In Section 3, we give the proof of our main results.

    Let un be defined by the second-order linear recurrence sequence (1.1). In this paper, we shall prove the following eight theorems.

    Theorem 2.1. For any positive integer m, we have

    (k=n1umk)1umnum(n1).

    Theorem 2.2. For any positive integer m, we have

    (k=n1u2mk)1u2mnu2m(n1)+BmnCm,

    where Cm=2(1Bm)(αβ)22(α2m1)2(αβ)2(α4mBm).

    Theorem 2.3. For any positive integer m, we have

    (k=n1u3mk)1u3mnu3m(n1)+3BmnQm(um(n+2)um(n3)),

    where Qm=u2m(1(Bα)5m)(1(Bβ)5m).

    Theorem 2.4. For any positive integer m, we have

    (k=n1u4mk)1u4mnu4m(n1)+4BmnUm(u2m(n+1)Bmu2m(n2))+Vm,

    where Um=u2m(1Bmα6m)(1Bmβ6m) and Vm=(α4m1)2(αβ)4(16(α4m1)(α6mBm)210α8m1).

    Theorem 2.5. For all positive integers m and l, we have

    (k=n1umk+umk+l)1umn+lum(n1)+l+umnum(n1).

    Theorem 2.6. For all positive integers m and l, we have

    (k=n1li=0umk+i)11α1(umn+l+1um(n1)+l+1umn+um(n1)).

    Remark 2.1. Note that when l=1, the two main terms of Theorems 2.5 and 2.6 are different. However, there is no contradiction since they are equivalent.

    Theorem 2.7. For all positive integers m and l, we have

    (ⅰ)

    (k=n1umkumk+2l)1u2mn+lu2m(n1)+lBmn(α2m1)2α4mBmCm,l,

    where Cm,l=u2l+2BlA24B(1(1Bm)(α4mBm)(α2m1)2).

    (ⅱ)

    (k=n1umkumk+2l1)1α(1β2m)u2mn+l1Bmn(α2m1)2α4mBmCm,l,

    where Cm,l=ulul1+Bl1A24B(A2(ααβ2m)(α4mBm)(α2m1)2).

    Theorem 2.8. For any positive integer m and constant C, we have

    (k=n1umk+C)1umnum(n1)+Cαm1αm+1.

    Let A=1 and B=1 in (1.1). Then un becomes the nth Fibonacci number. So we can obtain some known results when we take some special values for m,A,B.

    If we take A=1 and B=1 in Theorem 2.2, then

    Cm=2(1Bm)(αβ)22(α2m1)2(αβ)2(α4mBm)=2(1(1)m)52(α2m1)25(α4m(1)m). (2.1)

    If m is even, then it follows from (2.1) that

    Cm=2(α2m1)25(α4m(1)m)=2(α2mαmβm)25(α4mα2mβ2m)=2(αmβm)25(α2mβ2m)=2(α2m+β2m2)(αβ)5(α2mβ2m)(αβ)=2(L2m2)55u2m=2(L2m2)25u2m5,

    where Ln is the nth Lucas number with L0=2,L1=1. If m is odd, then it follows from (2.1) that

    Cm=452(α2m1)25(α4m+1)=452(α2m+αmβm)25(α4m+α2mβ2m)=452(αm+βm)25(α2m+β2m)=452(α2m+β2m2)5(α2m+β2m)=4L2m5L2m2(L2m2)5L2m=2(L2m+2)5L2m.

    So we have the following corollary, which is given by [12].

    Corollary 2.1. Let Fn be the nth Fibonacci number with F0=0,F1=1 and let Ln be the nth Lucas number with L0=2,L1=1. Then for any positive integer m, we have

    (k=n1F2mk)1F2mnF2m(n1)+(1)mnCm,

    where

    Cm={2(L2m2)25F2m5,ifmis even,2(L2m+2)5L2m,ifmis odd.

    If we take m=A=1 and B=1 in Theorem 2.4, then

    Um=u2m(1Bmα6m)(1Bmβ6m)=1(1+α6)(1+β6)=12+α6+β6=120

    and

    Vm=(α4m1)2(αβ)4(16(α4m1)(α6mBm)210α8m1)=(α41)2(αβ)4(16(α41)(α6+1)210α81)=125(16(α41)3(α6+1)210(α41)α4+1)=125(451053)=2575,

    which imply that

    u4mnu4m(n1)+4BmnUm(u2m(n+1)Bmu2m(n2))+Vm=u4nu4n1+4(1)n120(u2n+1+u2n2)+2575=u4nu4n1+(1)n5(u2n+1+u2n2)+2575=u4nu4n1+(1)n5(u2n+1u2n1+u2n1+u2n2)+2575=u4nu4n1+(1)n5(u2n+u2n3)+2575=u4nu4n1+2(1)n5u2n1+2575.

    So we have the following corollary, which is given by [3, Corollary 4.3].

    Corollary 2.2. Let Fn be the nth Fibonacci number with F0=0,F1=1. Then we have

    (k=n1F4k)1F4nF4n1+2(1)n5F2n1+2575.

    If we take m=A=1 and B=1 in Theorem 2.7, then

    C1,l=u2l+2BlA24B(1(1Bm)(α4mBm)(α2m1)2)=u2l+2(1)l5(12(α4+1)(α21)2)=u2l+2(1)l5(12(α2+β2)(α+β)2)=u2l2(1)l (2.2)

    and

    C1,l=ulul1+Bl1A24B(A2(ααβ2m)(α4mBm)(α2m1)2)=ulul1+(1)l15(12(ααβ2)(α4+1)(α21)2)=ulul1+(1)l15(12(α4+1)(α21)2)=ulul1+(1)l. (2.3)

    Note that

    Bmn(α2m1)2α4mBm=(1)n(α21)2α4+1=(1)n(α+β)2α2+β2=(1)n3. (2.4)

    Then it follows from (2.2)–(2.4) that

    u2mn+lu2m(n1)+lBmn(α2m1)2α4mBmCm,l=u2n+lu2n1+l(1)n3C1,l=un+lun+l1+(1)n+l1(1)n3(u2l2(1)l)=un+lun+l1(1)n3(u2l2(1)l+3(1)l)=un+lun+l1(1)n3(u2l+(1)l)

    and

    α(1β2m)u2mn+l1Bmn(α2m1)2α4mBmCm,l=α(1β2)u2n+l1(1)n3C1,l=u2n+l1(1)n3(ulul1+(1)l).

    So we have the following corollary, which is given by [13, Sections 3 and 4].

    Corollary 2.3. Let Fn be the nth Fibonacci number with F0=0,F1=1. Then for any positive integer l, we have

    (k=n1FkFk+2l)1Fn+l1Fn+l(F2l+(1)l)(1)n3

    and

    (k=n1FkFk+2l1)1F2n+l1(Fl1Fl+(1)l)(1)n3.

    In this section, we give the proofs of the above eight theorems. We first give some identities that will be used in the proofs of our main results. Let m and k be positive integers. Then it follows from (1.2) that

    1umk=αβαmkβmk=αβαmk(1Bmkα2mk)1. (3.1)

    Moreover, we have the following identities:

    (1+x)1=1x+x2x31+x, (3.2a)
    (1x)1=1+x+x2+x31x, (3.2b)
    (1x)2=1+2x+3x2+x3(43x)(x1)2, (3.2c)
    (1x)3=1+3x+6x2+x3(1015x+6x2)(1x)3, (3.2d)
    (1x)4=1+4x+10x2+x3(2045x+36x210x3)(x1)4. (3.2e)

    To prove the above eight theorems, we may split the proofs into eight subsections as follows:

    In this subsection, we will provide a proof of Theorem 2.1.

    Proof. By (3.1) and (3.2b), we have

    1umk=αβαmk(1Bmkα2mk)1=αβαmk(1+Bmkα2mk+1α4mk+Bmkα4mk(α2mkBmk))=(αβ)(1αmk+Bmkα3mk+1α5mk+Bmkα5mk(α2mkBmk)). (3.3)

    Let n be a positive integer. Then it follows from (3.3) that

    k=n1umk=(αβ)(k=n1αmk+k=nBmkα3mk+k=n1α5mk+k=nBmkα5mk(α2mkBmk))=(αβ)(αmαmn(αm1)+Bmnα3mα3mn(α3mBm)+α5mα5mn(α5m1))+(αβ)k=nBmkα5mk(α2mkBmk) =αm(αβ)αmn(αm1)(1+Bmnα2m(αm1)α2mn(α3mBm)+α4m(αm1)α4mn(α5m1)+O(1α6mn)). (3.4)

    Here the notation f(x)=O(g(x)) means that there is a constant C such that |f(x)|Cg(x) for all large enough real numbers x. By (3.2a) and (3.4), we have

    (k=n1umk)1=αmn(αm1)αm(αβ)(1+Bmnα2m(αm1)α2mn(α3mBm)+α4m(αm1)α4mn(α5m1)+O(1α6mn))1=αmn(αm1)αm(αβ)(1Bmnα2m(αm1)α2mn(α3mBm)+O(1α4mn))=αmn(αm1)αm(αβ)Bmnαm(αm1)2αmn(αβ)(α3mBm)+O(1α3mn)=αmnαβαm(n1)αβ+O(1αmn)=αmnβmn+βmnαβαm(n1)βm(n1)+βm(n1)αβ+O(1αmn)=umnum(n1)+βmnβm(n1)αβ+O(1αmn).

    Then

    limn((k=n1umk)1(umnum(n1)))=limn(βmnβm(n1)αβ+O(1αmn))=0.

    So we have

    (k=n1umk)1umnum(n1).

    In this subsection, we will provide a proof of Theorem 2.2.

    Proof. By (3.1) and (3.2c), we have

    1u2mk=(αβ)2α2mk(1Bmkα2mk)2=(αβ)2α2mk(1+2Bmkα2mk+3α4mk+4Bmkα2mk3α4mk(Bmkα2mk1)2)=(αβ)2(1α2mk+2Bmkα4mk+3α6mk+4Bmkα2mk3α6mk(Bmkα2mk1)2)=(αβ)2(1α2mk+2Bmkα4mk+3α6mk+Rk), (3.5)

    where

    Rk=4Bmkα2mk3α6mk(Bmkα2mk1)2.

    Let n be a positive integer. Then it follows from (3.5) that

    k=n1u2mk=(αβ)2(k=n1α2mk+k=n2Bmkα4mk+k=n3α6mk+k=nRk)=(αβ)2(α2mα2mn(α2m1)+2Bmnα4mα4mn(α4mBm)+3α6mα6mn(α6m1)+k=nRk)=(αβ)2α2mα2mn(α2m1)(1+2Bmnα2m(α2m1)α2mn(α4mBm)+3α4m(α2m1)α4mn(α6m1))+(αβ)2α2mα2mn(α2m1)α2mn(α2m1)α2mk=nRk=(αβ)2α2mα2mn(α2m1)(1+ω), (3.6)

    where

    ω=2Bmnα2mnα2m(α2m1)(α4mBm)+3α4mnα4m(α4m+α2m+1)+α2mn(α2m1)α2mk=nRk.

    Note that

    ω=2Bmnα2mnα2m(α2m1)(α4mBm)+O(1α4mn).

    Then we have

    ω2ω31+ω=O(1α4mn). (3.7)

    From (3.2a), (3.6), and (3.7), it follows that

    (k=n1u2mk)1=α2mn(α2m1)(αβ)2α2m(1+ω)1=α2mn(α2m1)(αβ)2α2m(1ω+ω2ω31+ω)=α2mn(α2m1)(αβ)2α2m(1ω+O(1α4mn))=α2mn(α2m1)(αβ)2α2m(12Bmnα2mnα2m(α2m1)(α4mBm)+O(1α4mn))=α2mn(α2m1)(αβ)2α2m2Bmn(αβ)2(α2m1)2(α4mBm)+O(1α2mn)=α2mn(αβ)2α2m(n1)(αβ)22Bmn(αβ)2(α2m1)2(α4mBm)+O(1α2mn)=(αmnβmn+βmnαβ)2(αm(n1)βm(n1)+βm(n1)αβ)22Bmn(αβ)2(α2m1)2(α4mBm)+O(1α2mn)=u2mnu2m(n1)+2Bmn(1Bm)(αβ)22Bmn(αβ)2(α2m1)2(α4mBm)+β2mn(α2m1)(αβ)2+O(1α2mn)=u2mnu2m(n1)+BmnCm+O(1α2mn),

    where

    Cm=2(1Bm)(αβ)22(α2m1)2(αβ)2(α4mBm).

    Then

    limn((k=n1u2mk)1(u2mnu2m(n1)+BmnCm))=0.

    So we have

    (k=n1u2mk)1u2mnu2m(n1)+BmnCm,

    where Cm=2(1Bm)(αβ)22(α2m1)2(αβ)2(α4mBm).

    In this subsection, we will provide a proof of Theorem 2.3.

    Proof. By (3.1) and (3.2d), we have

    1u3mk=(αβ)3α3mk(1Bmkα2mk)3=(αβ)3α3mk(1+3Bmkα2mk+6α4mk+10α4mk15Bmkα2mk+6α4mk(Bmkα2mk1)3)=(αβ)3(1α3mk+3Bmkα5mk+6α7mk+10α4mk15Bmkα2mk+6α7mk(Bmkα2mk1)3)=(αβ)3(1α3mk+3Bmkα5mk+6α7mk+Rk), (3.8)

    where

    Rk=10α4mk15Bmkα2mk+6α7mk(Bmkα2mk1)3.

    Let n be a positive integer. Then it follows from (3.8) that

    k=n1u3mk=(αβ)3(k=n1α3mk+k=n3Bmkα5mk+k=n6α7mk+k=nRk)=(αβ)3(α3mα3mn(α3m1)+3Bmnα5mα5mn(α5mBm)+6α7mα7mn(α7m1)+k=nRk)=(αβ)3α3mα3mn(α3m1)(1+3Bmnα2m(α3m1)α2mn(α5mBm)+6α4m(α3m1)α4mn(α7m1))+(αβ)3α3mα3mn(α3m1)α3mn(α3m1)α3mk=nRk=(αβ)3α3mα3mn(α3m1)(1+ω), (3.9)

    where

    ω=3Bmnα2mnα2m(α3m1)(α5mBm)+6α4mnα4m(α3m1)(α7m1)+α3mn(α3m1)α3mk=nRk.

    Note that

    ω=3Bmnα2mnα2m(α3m1)(α5mBm)+O(1α4mn).

    Then we have

    ω2ω31+ω=O(1α4mn). (3.10)

    From (3.2a), (3.9), and (3.10), it follows that

    (k=n1u3mk)1=((αβ)3α3mα3mn(α3m1))1(1+ω)1=α3mn(α3m1)(αβ)3α3m(1ω+ω2ω31+ω)=α3mn(α3m1)(αβ)3α3m(1ω+O(1α4mn))=α3mn(α3m1)(αβ)3α3m(13Bmnα2mnα2m(α3m1)(α5mBm)+O(1α4mn))=α3mn(α3m1)(αβ)3α3m3(Bα)mn(αβ)3(α3m1)2αm(α5mBm)+O(1αmn)=α3mn(αβ)3α3m(n1)(αβ)33(Bα)mn(αβ)3(α3m1)2αm(α5mBm)+O(1αmn)=(αmnβmn+βmnαβ)3(αm(n1)βm(n1)+βm(n1)αβ)33(Bα)mn(αβ)3(α3m1)2αm(α5mBm)+O(1αmn)=u3mnu3m(n1)+δ+O(1αmn),

    where

    δ=3(Bα)mn(1βm)+3(Bβ)mn(αm1)+β3mn(1(Bα)3m)(αβ)33(Bα)mn(αβ)3(α3m1)2αm(α5mBm)=3(Bα)mn(1βm)(αβ)33(Bα)mn(αβ)3(α3m1)2αm(α5mBm)+O(1αmn)=3(Bα)mn(αβ)3((α3m1)2+(1βm)αm(α5mBm)αm(α5mBm))+O(1αmn)=3(Bα)mn(αβ)3(α4m2Bmα2m+B2m(Bα)5m1)+O(1αmn)=3(Bα)mn(αβ)3α2m(αmβm)21Bmα5m+O(1αmn)=(αmβmαβ)23(Bα)m(n+2)αβ11(Bα)5m+O(1αmn).

    Let Qm=u2m(1(Bα)5m)(1(Bβ)5m). Then we can obtain

    δ=3BmnQmαm(n+2)(1(Bβ)5m)αβ+O(1αmn)=3BmnQm(αm(n+2)αβαm(n3)αβ)+O(1αmn)=3BmnQm(αm(n+2)βm(n+2)+βm(n+2)αβαm(n3)βm(n3)+βm(n3)αβ)+O(1αmn)=3BmnQm(um(n+2)um(n3)+βm(n+2)βm(n3)αβ)+O(1αmn).

    Then

    limn((k=n1u3mk)1 (u3mnu3m(n1)+3BmnQm(um(n+2)um(n3))))=limn(3BmnQmβm(n+2)βm(n3)αβ+O(1αmn))=0.

    So we have

    (k=n1u3mk)1u3mnu3m(n1)+3BmnQm(um(n+2)um(n3)),

    where Qm=u2m(1(Bα)5m)(1(Bβ)5m).

    In this subsection, we will provide a proof of Theorem 2.4.

    Proof. By (3.1) and (3.2e), we have

    1u4mk=(αβ)4α4mk(1Bmkα2mk)4=(αβ)4α4mk(1+4Bmkα2mk+10α4mk+Rk)=(αβ)4(1α4mk+4Bmkα6mk+10α8mk+Rkα4mk), (3.11)

    where

    Rk=20Bmkα6mk45α4mk+36Bmkα2mk10α4mk(Bmkα2mk1)4.

    Let n be a positive integer. Then it follows from (3.11) that

    k=n1u4mk=(αβ)4(k=n1α4mk+k=n4Bmkα6mk+k=n10α8mk+k=nRkα4mk)=(αβ)4(α4mα4mn(α4m1)+4Bmnα6mα6mn(α6mBm)+10α8mα8mn(α8m1)+k=nRkα4mk)=(αβ)4α4mα4mn(α4m1)(1+4Bmnα2m(α4m1)α2mn(α6mBm)+10α4m(α4m1)α4mn(α8m1))+(αβ)4α4mα4mn(α4m1)α4mn(α4m1)α4mk=nRkα4mk=(αβ)4α4mα4mn(α4m1)(1+ω), (3.12)

    where

    ω=4Bmnα2m(α4m1)α2mn(α6mBm)+10α4m(α4m1)α4mn(α8m1)+α4mn(α4m1)α4mk=nRkα4mk.

    Note that

    ω=4Bmnα2m(α4m1)α2mn(α6mBm)+10α4m(α4m1)α4mn(α8m1)+O(1α6mn).

    Then we have

    ω2ω31+ω=16α4m(α4m1)2α4mn(α6mBm)2+O(1α6mn). (3.13)

    By (3.2a), (3.12), and (3.13), we have

    (k=n1u4mk)1=((αβ)4α4mα4mn(α4m1))1(1+ω)1=α4mn(α4m1)(αβ)4α4m(1ω+ω2ω31+ω)=α4mn(α4m1)(αβ)4α4m(1ω+16α4m(α4m1)2α4mn(α6mBm)2+O(1α6mn))=α4mn(α4m1)(αβ)4α4m(14Bmnα2m(α4m1)α2mn(α6mBm)+1α4mnCm+O(1α6mn))=α4mn(α4m1)(αβ)4α4m4Bmnα2mn(α4m1)2(αβ)4α2m(α6mBm)+(α4m1)(αβ)4α4mCm+O(1α2mn)=α4mn(αβ)4α4m(n1)(αβ)44Bmnα2mn(α4m1)2(αβ)4α2m(α6mBm)+(α4m1)(αβ)4α4mCm+O(1α2mn)=(αmnβmn+βmnαβ)4(αm(n1)βm(n1)+βm(n1)αβ)44Bmnα2mn(α4m1)2(αβ)4α2m(α6mBm)+(α4m1)(αβ)4α4mCm+O(1α2mn)=u4mnu4m(n1)+δ+Vm+O(1α2mn),

    where

    Cm=16α4m(α4m1)2(α6mBm)210α4m(α4m1)α8m1,
    Vm=(α4m1)(αβ)4α4mCm=(α4m1)(αβ)4α4m(16α4m(α4m1)2(α6mBm)210α4m(α4m1)α8m1)=(α4m1)2(αβ)4(16(α4m1)(α6mBm)210α8m1)

    and

    δ=4Bmnα2mn(αβ)4((1Bmβ2m)α2m(α6mBm)(α4m1)2α2m(α6mBm))=4Bmnα2mn(αβ)4(α4m2Bmα2m+B2m1Bmα6m)=4Bmnα2mn(αβ)4α2m(αmβm)21Bmα6m=(αmβmαβ)24Bmn(1Bmα6m)(1Bmβ6m)α2m(n+1)(1Bmβ6m)(αβ)2=4Bmnu2m(1Bmα6m)(1Bmβ6m)(α2m(n+1)(αβ)2Bmα2m(n2)(αβ)2).

    Let Um=u2m(1Bmα6m)(1Bmβ6m). Then we can obtain

    δ=4BmnUm((αm(n+1)βm(n+1)+βm(n+1)αβ)2Bm(αm(n2)βm(n2)+βm(n2)αβ)2)=4BmnUm(u2m(n+1)Bmu2m(n2)+Bmβ2m(n2)β2m(n2)(αβ)2)=4BmnUm(u2m(n+1)Bmu2m(n2))+O(1α2m(n2)).

    Then

    limn((k=n1u4mk)1(u4mnu4m(n1)+4BmnUm(u2m(n+1)Bmu2m(n2))+Vm))=limn(O(1α2m(n2))+O(1α2mn))=0.

    So we have

    (k=n1u4mk)1u4mnu4m(n1)+4BmnUm(u2m(n+1)Bmu2m(n2))+Vm,

    where Um=u2m(1Bmα6m)(1Bmβ6m) and Vm=(α4m1)2(αβ)4(16(α4m1)(α6mBm)210α8m1).

    In this subsection, we will provide a proof of Theorem 2.5.

    Proof. By (1.2) and (3.2b), we have

    1umk+umk+l=(αmkβmkαβ+αmk+lβmk+lαβ)1=(αmk(1+αl)αβ)1(1Bmk(1+βl)α2mk(1+αl))1=αβαmk(1+αl)(1+Bmk(1+βl)α2mk(1+αl)+(1+βl)2α4mk(1+αl)2+Rk)=αβ1+αl(1αmk+Bmk(1+βl)α3mk(1+αl)+(1+βl)2α5mk(1+αl)2+Rkαmk), (3.14)

    where

    Rk=Bmk(1+βl)α4mk(1+αl)(α2mk(1+αl)Bmk(1+βl)).

    Let n be a positive integer. Then it follows from (3.14) that

    k=n1umk+umk+l=αβ1+αl(k=n1αmk+k=nBmk(1+βl)α3mk(1+αl)+k=n(1+βl)2α5mk(1+αl)2+k=nRkαmk)=αβ1+αl(αmαmn(αm1)+Cm)=αm(αβ)αmn(αm1)(1+αl)(1+αmn(αm1)αmCm), (3.15)

    where

    Cm=Bmnα3m(1+βl)α3mn(α3mBm)(1+αl)+α5m(1+βl)2α5mn(α5m1)(1+αl)2+k=nRkαmk.

    Then we have

    αmn(αm1)αmCm=O(1α2mn). (3.16)

    By (3.2a), (3.15), and (3.16), we have

    (k=n1umk+umk+l)1=(αm(αβ)αmn(1+αl)(αm1))1(1+O(1α2mn))1=αmn(1+αl)(αm1)αm(αβ)(1+O(1α2mn))=αmn+αmn+lαm(n1)αm(n1)+lαβ+O(1αmn)=αmn+lβmn+l+βmn+lαβαm(n1)+lβm(n1)+l+βm(n1)+lαβ+αmnβmn+βmnαβαm(n1)βm(n1)+βm(n1)αβ+O(1αmn)=umn+lum(n1)+l+umnum(n1)+βmn+βmn+lβm(n1)βm(n1)+lαβ+O(1αmn).

    Then

    limn((k=n1umk+umk+l)1(umn+lum(n1)+l+umnum(n1)))=limn(βmn+βmn+lβm(n1)βm(n1)+lαβ+O(1αmn))=0.

    So we have

    (k=n1umk+umk+l)1umn+lum(n1)+l+umnum(n1).

    In this subsection, we will provide a proof of Theorem 2.6.

    Proof. By (1.2), we have

    li=0umk+i=1αβ(αmkli=0αiβmkli=0βi)=1αβ(αmk(1αl+1)1αβmk(1βl+1)1β)=αmk(1αl+1)(αβ)(1α)(1Bmk(1α)(1βl+1)α2mk(1β)(1αl+1)). (3.17)

    From (3.2b) and (3.17), it follows that

    1li=0umk+i=(αmk(1αl+1)(αβ)(1α))1(1Bmk(1α)(1βl+1)α2mk(1β)(1αl+1))1=(αβ)(1α)αmk(1αl+1)(1+Bmk(1α)(1βl+1)α2mk(1β)(1αl+1)+(1α)2(1βl+1)2α4mk(1β)2(1αl+1)2+Rk)=(αβ)(1α)(1αl+1)(1αmk+Bmk(1α)(1βl+1)α3mk(1β)(1αl+1)+(1α)2(1βl+1)2α5mk(1β)2(1αl+1)2)+(αβ)(1α)(1αl+1)Rkαmk, (3.18)

    where

    Rk=(1βl+1)3(1α)3Bmk(4αmk(1β)(1αl+1)3Bmk(1α)(1βl+1))α4mk(1β)2(1αl+1)2(α2mk(1β)(1αl+1)Bmk(1α)(1βl+1)).

    Let n be a positive integer. Then it follows from (3.18) that

    k=n1li=0umk+i=(αβ)(1α)(1αl+1)(k=n1αmk+Cm)=(αβ)(1α)(1αl+1)(αmαmn(αm1)+Cm)=αm(αβ)(1α)αmn(αm1)(1αl+1)(1+αmn(αm1)αmCm), (3.19)

    where

    Cm=k=nBmk(1α)(1βl+1)α3mk(1β)(1αl+1)+k=n(1α)2(1βl+1)2α5mk(1β)2(1αl+1)2+k=nRkαmk=Bmnα3m(1α)(1βl+1))α3mn(α3mBm)(1β)(1αl+1)+α5m(1α)2(1βl+1)2α5mn(α5m1)(1β)2(1αl+1)2+k=nRkαmk=O(1α3mn).

    Then we have

    αmn(αm1)αmCm=O(1α2mn). (3.20)

    By (3.2a), (3.19), and (3.20), we have

    (k=n1li=0umk+i)1=(αm(α1)(αβ)αmn(αl+11)(αm1))1(1+O(1α2mn))1=αmn(αl+11)(αm1)αm(α1)(αβ)(1+O(1α2mn))=αmn+l+1αm(n1)+l+1αmn+αm(n1)(α1)(αβ)+O(1αmn)=αm(n1)βm(n1)+βm(n1)(α1)(αβ)αm(n1)+l+1βm(n1)+l+1+βm(n1)+l+1(α1)(αβ)αmnβmn+βmn(α1)(αβ)+αmn+l+1βmn+l+1+βmn+l+1(α1)(αβ)+O(1αmn)=1α1(umn+l+1um(n1)+l+1umn+um(n1))+βmn+l+1βmn+βm(n1)βm(n1)+l+1(α1)(αβ)+O(1αmn).

    Then

    limn((k=n1li=0umk+i)11α1(umn+l+1um(n1)+l+1umn+um(n1)))=limn(βmn+l+1βmn+βm(n1)βm(n1)+l+1(α1)(αβ)+O(1αmn))=0.

    So we have

    (k=n1li=0umk+i)11α1(umn+l+1um(n1)+l+1umn+um(n1)).

    In this subsection, we will provide a proof of Theorem 2.7.

    Proof. Let h be a positive integer. By (1.2), we have

    1umkumk+h=(αβ)2((αmkβmk)(αmk+hβmk+h))1=(αβ)2α2mk+h((1Bmkα2mk)(1Bmk+hα2mk+2h))1=(αβ)2α2mk+h(1Bmkα2mkBmk+hα2mk+2h+Bhα4mk+2h)1=(αβ)2α2mk+h(1η)1, (3.21)

    where

    η=Bmkα2mk+Bmk+hα2mk+2hBhα4mk+2h=Bmkα2mk+Bmk+hα2mk+2h+O(1α4mk).

    Then we have

    η2+η31η=O(1α4mk). (3.22)

    By (3.2b), (3.21), and (3.22), we have

    1umkumk+h=(αβ)2α2mk+h(1+η+η2+η31η)=(αβ)2α2mk+h(1+η+O(1α4mk))=(αβ)2α2mk+h(1+Bmkα2mk(1+Bhα2h)+O(1α4mk))=(αβ)2αh(1α2mk+Bmkα4mk(1+Bhα2h)+O(1α6mk)). (3.23)

    Let n be a positive integer. Then it follows from (3.23) that

    k=n1umkumk+h=(αβ)2αh(k=n1α2mk+(1+Bhα2h)k=nBmkα4mk)+O(1α6mn)=(αβ)2αh(α2mα2mn(α2m1)+(1+Bhα2h)Bmnα4mα4mn(α4mBm))+O(1α6mn)=(αβ)2α2mα2mn+h(α2m1)(1+(1+Bhα2h)Bmnα2m(α2m1)α2mn(α4mBm)+O(1α4mn))=(αβ)2α2mα2mn+h(α2m1)(1+ω), (3.24)

    where

    ω=(1+Bhα2h)Bmnα2m(α2m1)α2mn(α4mBm)+O(1α4mn).

    Then we have

    ω2ω31+ω=O(1α4mn). (3.25)

    By (3.2a), (3.24), and (3.25), we have

    (k=n1umkumk+h)1=((αβ)2α2mα2mn+h(α2m1))1(1+ω)1=α2mn+h(α2m1)(αβ)2α2m(1ω+ω2ω31+ω)=α2mn+h(α2m1)(αβ)2α2m(1ω+O(1α4mn))=α2mn+h(α2m1)(αβ)2α2m(1(1+Bhα2h)Bmnα2m(α2m1)α2mn(α4mBm)+O(1α4mn))=α2mn+hα2m(n1)+h(αβ)2(1+Bhα2h)Bmn(α2m1)2αh(αβ)2(α4mBm)+O(1α2mn). (3.26)

    (ⅰ) If we take h=2l, then it follows from (3.26) that

    (k=n1umkumk+2l)1=(αmn+lβmn+l+βmn+lαβ)2(αm(n1)+lβm(n1)+l+βm(n1)+lαβ)2(1+1α4l)Bmn(α2m1)2α2l(αβ)2(α4mBm)+O(1α2mn)=u2mn+lu2m(n1)+l+δ+O(1α2mn),

    where

    δ=2(Bmn+lBm(n1)+l)(αβ)2(1+1α4l)Bmn(α2m1)2α2l(αβ)2(α4mBm)=Bmn(α2m1)2α4mBm(α2l+β2l(αβ)22Bl(1Bm)(α4mBm)(αβ)2(α2m1)2)=Bmn(α2m1)2α4mBm(u2l+2Bl(αβ)22Bl(1Bm)(α4mBm)(αβ)2(α2m1)2)=Bmn(α2m1)2α4mBm(u2l+2Bl(αβ)2(1(1Bm)(α4mBm)(α2m1)2)).

    Let Cm,l=u2l+2Bl(αβ)2(1(1Bm)(α4mBm)(α2m1)2). Then

    limn((k=n1umkumk+2l)1(u2mn+lu2m(n1)+lBmn(α2m1)2α4mBmCm,l))=0.

    So we have

    (k=n1umkumk+2l)1u2mn+lu2m(n1)+lBmn(α2m1)2α4mBmCm,l,

    where Cm,l=u2l+2BlA24B(1(1Bm)(α4mBm)(α2m1)2).

    (ⅱ) If we take h=2l1, then it follows from (3.26) that

    (k=n1umkumk+2l1)1=α2mn+2l1α2m(n1)+2l1(αβ)2(1+Bα4l2)Bmn(α2m1)2α2l1(αβ)2(α4mBm)+O(1α2mn)=(ααβ2m)α2mn+2l2(αβ)2(1+Bα4l2)Bmn(α2m1)2α2l1(αβ)2(α4mBm)+O(1α2mn)=(ααβ2m)(α2mn+2l2β2mn+2l2+β2mn+2l2(αβ)2)(1+Bα4l2)Bmn(α2m1)2α2l1(αβ)2(α4mBm)+O(1α2mn)=(ααβ2m)u2mn+l1+δ+O(1α2mn),

    where

    δ=2Bmn+l1(ααβ2m)(αβ)2(1+B2l1α4l2)Bmn(α2m1)2α2l1(αβ)2(α4mBm)=Bmn(α2m1)2α4mBm(αlαl1+β2l1(αβ)22Bl1(ααβ2m)(α4mBm)(αβ)2(α2m1)2)=Bmn(α2m1)2α4mBm(ulul1+Bl1(α+β)(αβ)22Bl1(ααβ2m)(α4mBm)(αβ)2(α2m1)2)=Bmn(α2m1)2α4mBm(ulul1+Bl1(αβ)2((α+β)2(ααβ2m)(α4mBm)(α2m1)2)).

    Let Cm,l=ulul1+Bl1(αβ)2((α+β)2(ααβ2m)(α4mBm)(α2m1)2). Then

    limn((k=n1umkumk+2l1)1((ααβ2m)u2mn+l1Bmn(α2m1)2α4mBmCm,l))=0.

    So we have

    (k=n1umkumk+2l1)1(ααβ2m)u2mn+l1Bmn(α2m1)2α4mBmCm,l,

    where Cm,l=ulul1+Bl1A24B(A2(ααβ2m)(α4mBm)(α2m1)2).

    In this subsection, we will provide a proof of Theorem 2.8.

    Proof. By (1.2), we have

    umk+C=αmkβmkαβ+C=αmkαβ(1+C(αβ)αmkBmkα2mk)=αmkαβ(1+η), (3.27)

    where

    η=C(αβ)αmkBmkα2mk.

    Then we have

    η2η31+η=O(1α2mk). (3.28)

    From (3.2a), (3.27), and (3.28), it follows that

    1umk+C=(αmkαβ)1(1+η)1=αβαmk(1η+η2η31+η)=αβαmk(1η+O(1α2mk))=αβαmk(1C(αβ)αmk+O(1α2mk))=(αβ)(1αmkC(αβ)α2mk+O(1α3mk)). (3.29)

    Let n be a positive integer. Then it follows from (3.29) that

    k=n1umk+C=k=nαβαmkk=nC(αβ)2α2mk+O(1α3mn)=αm(αβ)αmn(αm1)Cα2m(αβ)2α2mn(α2m1)+O(1α3mn)=(αβ)αmαmn(αm1)(1C(αβ)αmαmn(αm+1)+O(1α2mn))=(αβ)αmαmn(αm1)(1ω), (3.30)

    where

    ω=C(αβ)αmαmn(αm+1)+O(1α2mn).

    Then we have

    ω2+ω31ω=O(1α2mn). (3.31)

    By (3.2b), (3.30), and (3.31), we have

    (k=n1umk+C)1=((αβ)αmαmn(αm1))1(1ω)1=αmn(αm1)(αβ)αm(1+ω+ω2+ω31ω)=αmn(αm1)(αβ)αm(1+ω+O(1α2mn))=αmn(αm1)(αβ)αm(1+C(αβ)αmαmn(αm+1)+O(1α2mn))=αmnαm(n1)αβ+C(αm1)αm+1+O(1αmn)=αmnβmn+βmnαβαm(n1)βm(n1)+βm(n1)αβ+C(αm1)αm+1+O(1αmn)=umnum(n1)+βmnβm(n1)αβ+C(αm1)αm+1+O(1αmn).

    Then

    limn((k=n1umk+C)1(umnum(n1)+C(αm1)αm+1))=limn(βmnβm(n1)αβ+O(1αmn))=0.

    So we have

    (k=n1umk+C)1umnum(n1)+Cαm1αm+1.

    Let (un)n0 be the special Lucas u-sequence defined by un+2=Aun+1Bun,u0=0,u1=1, where n0, B=±1, and A is an integer such that A24B>0. In this paper, we study the asymptotic behavior of the sequences involving un. In Section 1, we give the definition of the asymptotic behavior and introduce the asymptotic behavior of some sequences. In Section 2, we give the asymptotic formulas for (k=nak)1, where

    ak=1usmk,1umk+umk+l,1li=0umk+i,1umkumk+2l,1umkumk+2l1,1umk+C,

    m,l are positive integers, s=1,2,3,4, and C is any constant. In Section 3, we give the proof of these results.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    Hongjian Li was supported by the Project of Guangdong University of Foreign Studies (Grant No. 2024RC063). Pingzhi Yuan was supported by the National Natural Science Foundation of China (Grant No. 12171163) and the Basic and Applied Basic Research Foundation of Guangdong Province (Grant No. 2024A1515010589).

    The authors declare there are no conflicts of interest.



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