Let (un)n≥0 be the special Lucas u-sequence defined by
un+2=Aun+1−Bun,u0=0,u1=1,
where n≥0, B=±1, and A is an integer such that A2−4B>0. Let
ak=1usmk,1umk+umk+l,1∑li=0umk+i,1umkumk+2l,1umkumk+2l−1,1umk+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)−1−gn)=0.
For example, we show that
limn→∞((∞∑k=n1umk)−1−(umn−um(n−1)))=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)n≥0 be the special Lucas u-sequence defined by
un+2=Aun+1−Bun,u0=0,u1=1,
where n≥0, B=±1, and A is an integer such that A2−4B>0. Let
ak=1usmk,1umk+umk+l,1∑li=0umk+i,1umkumk+2l,1umkumk+2l−1,1umk+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)−1−gn)=0.
For example, we show that
limn→∞((∞∑k=n1umk)−1−(umn−um(n−1)))=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)n≥0 is defined by
Fn+2=Fn+1+Fn,F0=0,F1=1, |
where n≥0. In [1], Ohtsuka and Nakamura proved
⌊(∞∑k=n1Fk)−1⌋={Fn−2,if n≥2 is even;Fn−2−1,if n≥1 is odd, |
and
⌊(∞∑k=n1F2k)−1⌋={Fn−1Fn−1,if n≥2 is even;Fn−1Fn,if n≥1 is odd, |
where ⌊x⌋ denotes the greatest integer ≤x. In [2], Wang proved
⌊(∞∑k=n1F3k)−1⌋={FnF2n−1+Fn−2F2n+⌊111(14Fn−2−5Fn)⌋,if n≥2 is even;FnF2n−1+Fn−2F2n+⌊111(5Fn−14Fn−2)⌋,if n≥1 is odd, |
where F−1=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⌋=F4n−F4n−1+2(−1)n5F2n−1−{n+25}, |
where {x}=x−⌊x⌋. 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)−1∼gn |
when ∑∞k=1ak converges. Here the notation An∼Bn means that
limn→∞(An−Bn)=0. |
In [11], Lee et al. proved that
(∞∑k=n1Fmk−l)−1∼Fmn−l−Fm(n−1)−l |
for any positive integer m and 0≤l≤m−1. In [12], Marques et al. proved that for any positive integer m there exists a positive constant Cm such that
(∞∑k=n1F2mk)−1∼F2mn−F2m(n−1)+(−1)mnCm. |
Moreover, they gave an explicit form for Cm as follows:
Cm={−2(L2m−2)25F2m√5,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)−1∼F4n−F4n−1+2(−1)n5F2n−1+2√575. |
In [13], Lee and Park studied the asymptotic behavior of the reciprocal sum of FkFk+m, and they proved that
(∞∑k=n1FkFk+2l)−1∼Fn+l−1Fn+l−(F2l+(−1)l)(−1)n3 |
and
(∞∑k=n1FkFk+2l−1)−1∼F2n+l−1−(Fl−1Fl+(−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)n≥0 be the special Lucas u-sequence defined by
un+2=Aun+1−Bun,u0=0,u1=1, | (1.1) |
where n≥0, B=±1, and A is an integer such that A2−4B>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)n≥0 has the form
un=αn−βnα−β,n≥0, | (1.2) |
where
α,β=A±√A2−4B2. |
Let
ak=1usmk,1umk+umk+l,1∑li=0umk+i,1umkumk+2l,1umkumk+2l−1,1umk+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)−1∼gn. |
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)−1∼umn−um(n−1). |
Theorem 2.2. For any positive integer m, we have
(∞∑k=n1u2mk)−1∼u2mn−u2m(n−1)+BmnCm, |
where Cm=2(1−Bm)(α−β)2−2(α2m−1)2(α−β)2(α4m−Bm).
Theorem 2.3. For any positive integer m, we have
(∞∑k=n1u3mk)−1∼u3mn−u3m(n−1)+3BmnQm(um(n+2)−um(n−3)), |
where Qm=u2m(1−(Bα)5m)(1−(Bβ)5m).
Theorem 2.4. For any positive integer m, we have
(∞∑k=n1u4mk)−1∼u4mn−u4m(n−1)+4BmnUm(u2m(n+1)−Bmu2m(n−2))+Vm, |
where Um=u2m(1−Bmα6m)(1−Bmβ6m) and Vm=(α4m−1)2(α−β)4(16(α4m−1)(α6m−Bm)2−10α8m−1).
Theorem 2.5. For all positive integers m and l, we have
(∞∑k=n1umk+umk+l)−1∼umn+l−um(n−1)+l+umn−um(n−1). |
Theorem 2.6. For all positive integers m and l, we have
(∞∑k=n1∑li=0umk+i)−1∼1α−1(umn+l+1−um(n−1)+l+1−umn+um(n−1)). |
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)−1∼u2mn+l−u2m(n−1)+l−Bmn(α2m−1)2α4m−BmCm,l, |
where Cm,l=u2l+2BlA2−4B(1−(1−Bm)(α4m−Bm)(α2m−1)2).
(ⅱ)
(∞∑k=n1umkumk+2l−1)−1∼α(1−β2m)u2mn+l−1−Bmn(α2m−1)2α4m−BmC′m,l, |
where C′m,l=ulul−1+Bl−1A2−4B(A−2(α−αβ2m)(α4m−Bm)(α2m−1)2).
Theorem 2.8. For any positive integer m and constant C, we have
(∞∑k=n1umk+C)−1∼umn−um(n−1)+Cαm−1αm+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(1−Bm)(α−β)2−2(α2m−1)2(α−β)2(α4m−Bm)=2(1−(−1)m)5−2(α2m−1)25(α4m−(−1)m). | (2.1) |
If m is even, then it follows from (2.1) that
Cm=−2(α2m−1)25(α4m−(−1)m)=−2(α2m−αmβm)25(α4m−α2mβ2m)=−2(αm−βm)25(α2m−β2m)=−2(α2m+β2m−2)(α−β)5(α2m−β2m)(α−β)=−2(L2m−2)5√5u2m=−2(L2m−2)25u2m√5, |
where Ln is the nth Lucas number with L0=2,L1=1. If m is odd, then it follows from (2.1) that
Cm=45−2(α2m−1)25(α4m+1)=45−2(α2m+αmβm)25(α4m+α2mβ2m)=45−2(αm+βm)25(α2m+β2m)=45−2(α2m+β2m−2)5(α2m+β2m)=4L2m5L2m−2(L2m−2)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)−1∼F2mn−F2m(n−1)+(−1)mnCm, |
where
Cm={−2(L2m−2)25F2m√5,ifmis even,2(L2m+2)5L2m,ifmis odd. |
If we take m=A=1 and B=−1 in Theorem 2.4, then
Um=u2m(1−Bmα6m)(1−Bmβ6m)=1(1+α6)(1+β6)=12+α6+β6=120 |
and
Vm=(α4m−1)2(α−β)4(16(α4m−1)(α6m−Bm)2−10α8m−1)=(α4−1)2(α−β)4(16(α4−1)(α6+1)2−10α8−1)=125(16(α4−1)3(α6+1)2−10(α4−1)α4+1)=125(4√5−10√53)=2√575, |
which imply that
u4mn−u4m(n−1)+4BmnUm(u2m(n+1)−Bmu2m(n−2))+Vm=u4n−u4n−1+4(−1)n⋅120⋅(u2n+1+u2n−2)+2√575=u4n−u4n−1+(−1)n5⋅(u2n+1+u2n−2)+2√575=u4n−u4n−1+(−1)n5⋅(u2n+1−u2n−1+u2n−1+u2n−2)+2√575=u4n−u4n−1+(−1)n5⋅(u2n+u2n−3)+2√575=u4n−u4n−1+2(−1)n5u2n−1+2√575. |
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)−1∼F4n−F4n−1+2(−1)n5F2n−1+2√575. |
If we take m=A=1 and B=−1 in Theorem 2.7, then
C1,l=u2l+2BlA2−4B(1−(1−Bm)(α4m−Bm)(α2m−1)2)=u2l+2(−1)l5(1−2(α4+1)(α2−1)2)=u2l+2(−1)l5(1−2(α2+β2)(α+β)2)=u2l−2(−1)l | (2.2) |
and
C′1,l=ulul−1+Bl−1A2−4B(A−2(α−αβ2m)(α4m−Bm)(α2m−1)2)=ulul−1+(−1)l−15(1−2(α−αβ2)(α4+1)(α2−1)2)=ulul−1+(−1)l−15(1−2(α4+1)(α2−1)2)=ulul−1+(−1)l. | (2.3) |
Note that
Bmn(α2m−1)2α4m−Bm=(−1)n(α2−1)2α4+1=(−1)n(α+β)2α2+β2=(−1)n3. | (2.4) |
Then it follows from (2.2)–(2.4) that
u2mn+l−u2m(n−1)+l−Bmn(α2m−1)2α4m−BmCm,l=u2n+l−u2n−1+l−(−1)n3C1,l=un+lun+l−1+(−1)n+l−1−(−1)n3(u2l−2(−1)l)=un+lun+l−1−(−1)n3(u2l−2(−1)l+3(−1)l)=un+lun+l−1−(−1)n3(u2l+(−1)l) |
and
α(1−β2m)u2mn+l−1−Bmn(α2m−1)2α4m−BmC′m,l=α(1−β2)u2n+l−1−(−1)n3C′1,l=u2n+l−1−(−1)n3(ulul−1+(−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)−1∼Fn+l−1Fn+l−(F2l+(−1)l)(−1)n3 |
and
(∞∑k=n1FkFk+2l−1)−1∼F2n+l−1−(Fl−1Fl+(−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=α−βαmk−βmk=α−βαmk(1−Bmkα2mk)−1. | (3.1) |
Moreover, we have the following identities:
(1+x)−1=1−x+x2−x31+x, | (3.2a) |
(1−x)−1=1+x+x2+x31−x, | (3.2b) |
(1−x)−2=1+2x+3x2+x3(4−3x)(x−1)2, | (3.2c) |
(1−x)−3=1+3x+6x2+x3(10−15x+6x2)(1−x)3, | (3.2d) |
(1−x)−4=1+4x+10x2+x3(20−45x+36x2−10x3)(x−1)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=α−βαmk(1−Bmkα2mk)−1=α−βαmk(1+Bmkα2mk+1α4mk+Bmkα4mk(α2mk−Bmk))=(α−β)(1αmk+Bmkα3mk+1α5mk+Bmkα5mk(α2mk−Bmk)). | (3.3) |
Let n be a positive integer. Then it follows from (3.3) that
∞∑k=n1umk=(α−β)(∞∑k=n1αmk+∞∑k=nBmkα3mk+∞∑k=n1α5mk+∞∑k=nBmkα5mk(α2mk−Bmk))=(α−β)(αmαmn(αm−1)+Bmnα3mα3mn(α3m−Bm)+α5mα5mn(α5m−1))+(α−β)∞∑k=nBmkα5mk(α2mk−Bmk) =αm(α−β)αmn(αm−1)(1+Bmnα2m(αm−1)α2mn(α3m−Bm)+α4m(αm−1)α4mn(α5m−1)+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(αm−1)αm(α−β)(1+Bmnα2m(αm−1)α2mn(α3m−Bm)+α4m(αm−1)α4mn(α5m−1)+O(1α6mn))−1=αmn(αm−1)αm(α−β)(1−Bmnα2m(αm−1)α2mn(α3m−Bm)+O(1α4mn))=αmn(αm−1)αm(α−β)−Bmnαm(αm−1)2αmn(α−β)(α3m−Bm)+O(1α3mn)=αmnα−β−αm(n−1)α−β+O(1αmn)=αmn−βmn+βmnα−β−αm(n−1)−βm(n−1)+βm(n−1)α−β+O(1αmn)=umn−um(n−1)+βmn−βm(n−1)α−β+O(1αmn). |
Then
limn→∞((∞∑k=n1umk)−1−(umn−um(n−1)))=limn→∞(βmn−βm(n−1)α−β+O(1αmn))=0. |
So we have
(∞∑k=n1umk)−1∼umn−um(n−1). |
In this subsection, we will provide a proof of Theorem 2.2.
Proof. By (3.1) and (3.2c), we have
1u2mk=(α−β)2α2mk(1−Bmkα2mk)−2=(α−β)2α2mk(1+2Bmkα2mk+3α4mk+4Bmkα2mk−3α4mk(Bmkα2mk−1)2)=(α−β)2(1α2mk+2Bmkα4mk+3α6mk+4Bmkα2mk−3α6mk(Bmkα2mk−1)2)=(α−β)2(1α2mk+2Bmkα4mk+3α6mk+Rk), | (3.5) |
where
Rk=4Bmkα2mk−3α6mk(Bmkα2mk−1)2. |
Let n be a positive integer. Then it follows from (3.5) that
∞∑k=n1u2mk=(α−β)2(∞∑k=n1α2mk+∞∑k=n2Bmkα4mk+∞∑k=n3α6mk+∞∑k=nRk)=(α−β)2(α2mα2mn(α2m−1)+2Bmnα4mα4mn(α4m−Bm)+3α6mα6mn(α6m−1)+∞∑k=nRk)=(α−β)2α2mα2mn(α2m−1)(1+2Bmnα2m(α2m−1)α2mn(α4m−Bm)+3α4m(α2m−1)α4mn(α6m−1))+(α−β)2α2mα2mn(α2m−1)⋅α2mn(α2m−1)α2m∞∑k=nRk=(α−β)2α2mα2mn(α2m−1)(1+ω), | (3.6) |
where
ω=2Bmnα2mn⋅α2m(α2m−1)(α4m−Bm)+3α4mn⋅α4m(α4m+α2m+1)+α2mn(α2m−1)α2m∞∑k=nRk. |
Note that
ω=2Bmnα2mn⋅α2m(α2m−1)(α4m−Bm)+O(1α4mn). |
Then we have
ω2−ω31+ω=O(1α4mn). | (3.7) |
From (3.2a), (3.6), and (3.7), it follows that
(∞∑k=n1u2mk)−1=α2mn(α2m−1)(α−β)2α2m(1+ω)−1=α2mn(α2m−1)(α−β)2α2m(1−ω+ω2−ω31+ω)=α2mn(α2m−1)(α−β)2α2m(1−ω+O(1α4mn))=α2mn(α2m−1)(α−β)2α2m(1−2Bmnα2mn⋅α2m(α2m−1)(α4m−Bm)+O(1α4mn))=α2mn(α2m−1)(α−β)2α2m−2Bmn(α−β)2⋅(α2m−1)2(α4m−Bm)+O(1α2mn)=α2mn(α−β)2−α2m(n−1)(α−β)2−2Bmn(α−β)2⋅(α2m−1)2(α4m−Bm)+O(1α2mn)=(αmn−βmn+βmnα−β)2−(αm(n−1)−βm(n−1)+βm(n−1)α−β)2−2Bmn(α−β)2⋅(α2m−1)2(α4m−Bm)+O(1α2mn)=u2mn−u2m(n−1)+2Bmn(1−Bm)(α−β)2−2Bmn(α−β)2⋅(α2m−1)2(α4m−Bm)+β2mn(α2m−1)(α−β)2+O(1α2mn)=u2mn−u2m(n−1)+BmnCm+O(1α2mn), |
where
Cm=2(1−Bm)(α−β)2−2(α2m−1)2(α−β)2(α4m−Bm). |
Then
limn→∞((∞∑k=n1u2mk)−1−(u2mn−u2m(n−1)+BmnCm))=0. |
So we have
(∞∑k=n1u2mk)−1∼u2mn−u2m(n−1)+BmnCm, |
where Cm=2(1−Bm)(α−β)2−2(α2m−1)2(α−β)2(α4m−Bm).
In this subsection, we will provide a proof of Theorem 2.3.
Proof. By (3.1) and (3.2d), we have
1u3mk=(α−β)3α3mk(1−Bmkα2mk)−3=(α−β)3α3mk(1+3Bmkα2mk+6α4mk+10α4mk−15Bmkα2mk+6α4mk(Bmkα2mk−1)3)=(α−β)3(1α3mk+3Bmkα5mk+6α7mk+10α4mk−15Bmkα2mk+6α7mk(Bmkα2mk−1)3)=(α−β)3(1α3mk+3Bmkα5mk+6α7mk+Rk), | (3.8) |
where
Rk=10α4mk−15Bmkα2mk+6α7mk(Bmkα2mk−1)3. |
Let n be a positive integer. Then it follows from (3.8) that
∞∑k=n1u3mk=(α−β)3(∞∑k=n1α3mk+∞∑k=n3Bmkα5mk+∞∑k=n6α7mk+∞∑k=nRk)=(α−β)3(α3mα3mn(α3m−1)+3Bmnα5mα5mn(α5m−Bm)+6α7mα7mn(α7m−1)+∞∑k=nRk)=(α−β)3α3mα3mn(α3m−1)(1+3Bmnα2m(α3m−1)α2mn(α5m−Bm)+6α4m(α3m−1)α4mn(α7m−1))+(α−β)3α3mα3mn(α3m−1)⋅α3mn(α3m−1)α3m∞∑k=nRk=(α−β)3α3mα3mn(α3m−1)(1+ω), | (3.9) |
where
ω=3Bmnα2mn⋅α2m(α3m−1)(α5m−Bm)+6α4mn⋅α4m(α3m−1)(α7m−1)+α3mn(α3m−1)α3m∞∑k=nRk. |
Note that
ω=3Bmnα2mn⋅α2m(α3m−1)(α5m−Bm)+O(1α4mn). |
Then we have
ω2−ω31+ω=O(1α4mn). | (3.10) |
From (3.2a), (3.9), and (3.10), it follows that
(∞∑k=n1u3mk)−1=((α−β)3α3mα3mn(α3m−1))−1(1+ω)−1=α3mn(α3m−1)(α−β)3α3m(1−ω+ω2−ω31+ω)=α3mn(α3m−1)(α−β)3α3m(1−ω+O(1α4mn))=α3mn(α3m−1)(α−β)3α3m(1−3Bmnα2mn⋅α2m(α3m−1)(α5m−Bm)+O(1α4mn))=α3mn(α3m−1)(α−β)3α3m−3(Bα)mn(α−β)3⋅(α3m−1)2αm(α5m−Bm)+O(1αmn)=α3mn(α−β)3−α3m(n−1)(α−β)3−3(Bα)mn(α−β)3⋅(α3m−1)2αm(α5m−Bm)+O(1αmn)=(αmn−βmn+βmnα−β)3−(αm(n−1)−βm(n−1)+βm(n−1)α−β)3−3(Bα)mn(α−β)3⋅(α3m−1)2αm(α5m−Bm)+O(1αmn)=u3mn−u3m(n−1)+δ+O(1αmn), |
where
δ=3(Bα)mn(1−βm)+3(Bβ)mn(αm−1)+β3mn(1−(Bα)3m)(α−β)3−3(Bα)mn(α−β)3⋅(α3m−1)2αm(α5m−Bm)=3(Bα)mn(1−βm)(α−β)3−3(Bα)mn(α−β)3⋅(α3m−1)2αm(α5m−Bm)+O(1αmn)=3(Bα)mn(α−β)3(−(α3m−1)2+(1−βm)αm(α5m−Bm)αm(α5m−Bm))+O(1αmn)=−3(Bα)mn(α−β)3(α4m−2Bmα2m+B2m(Bα)5m−1)+O(1αmn)=3(Bα)mn(α−β)3⋅α2m(αm−βm)21−Bmα5m+O(1αmn)=(αm−βmα−β)2⋅3(Bα)m(n+2)α−β⋅11−(Bα)5m+O(1αmn). |
Let Qm=u2m(1−(Bα)5m)(1−(Bβ)5m). Then we can obtain
δ=3BmnQmαm(n+2)(1−(Bβ)5m)α−β+O(1αmn)=3BmnQm(αm(n+2)α−β−αm(n−3)α−β)+O(1αmn)=3BmnQm(αm(n+2)−βm(n+2)+βm(n+2)α−β−αm(n−3)−βm(n−3)+βm(n−3)α−β)+O(1αmn)=3BmnQm(um(n+2)−um(n−3)+βm(n+2)−βm(n−3)α−β)+O(1αmn). |
Then
limn→∞((∞∑k=n1u3mk)−1 −(u3mn−u3m(n−1)+3BmnQm(um(n+2)−um(n−3))))=limn→∞(3BmnQmβm(n+2)−βm(n−3)α−β+O(1αmn))=0. |
So we have
(∞∑k=n1u3mk)−1∼u3mn−u3m(n−1)+3BmnQm(um(n+2)−um(n−3)), |
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=(α−β)4α4mk(1−Bmkα2mk)−4=(α−β)4α4mk(1+4Bmkα2mk+10α4mk+Rk)=(α−β)4(1α4mk+4Bmkα6mk+10α8mk+Rkα4mk), | (3.11) |
where
Rk=20Bmkα6mk−45α4mk+36Bmkα2mk−10α4mk(Bmkα2mk−1)4. |
Let n be a positive integer. Then it follows from (3.11) that
∞∑k=n1u4mk=(α−β)4(∞∑k=n1α4mk+∞∑k=n4Bmkα6mk+∞∑k=n10α8mk+∞∑k=nRkα4mk)=(α−β)4(α4mα4mn(α4m−1)+4Bmnα6mα6mn(α6m−Bm)+10α8mα8mn(α8m−1)+∞∑k=nRkα4mk)=(α−β)4α4mα4mn(α4m−1)(1+4Bmnα2m(α4m−1)α2mn(α6m−Bm)+10α4m(α4m−1)α4mn(α8m−1))+(α−β)4α4mα4mn(α4m−1)⋅α4mn(α4m−1)α4m∞∑k=nRkα4mk=(α−β)4α4mα4mn(α4m−1)(1+ω), | (3.12) |
where
ω=4Bmnα2m(α4m−1)α2mn(α6m−Bm)+10α4m(α4m−1)α4mn(α8m−1)+α4mn(α4m−1)α4m∞∑k=nRkα4mk. |
Note that
ω=4Bmnα2m(α4m−1)α2mn(α6m−Bm)+10α4m(α4m−1)α4mn(α8m−1)+O(1α6mn). |
Then we have
ω2−ω31+ω=16α4m(α4m−1)2α4mn(α6m−Bm)2+O(1α6mn). | (3.13) |
By (3.2a), (3.12), and (3.13), we have
(∞∑k=n1u4mk)−1=((α−β)4α4mα4mn(α4m−1))−1(1+ω)−1=α4mn(α4m−1)(α−β)4α4m(1−ω+ω2−ω31+ω)=α4mn(α4m−1)(α−β)4α4m(1−ω+16α4m(α4m−1)2α4mn(α6m−Bm)2+O(1α6mn))=α4mn(α4m−1)(α−β)4α4m(1−4Bmnα2m(α4m−1)α2mn(α6m−Bm)+1α4mnC′m+O(1α6mn))=α4mn(α4m−1)(α−β)4α4m−4Bmnα2mn(α4m−1)2(α−β)4α2m(α6m−Bm)+(α4m−1)(α−β)4α4mC′m+O(1α2mn)=α4mn(α−β)4−α4m(n−1)(α−β)4−4Bmnα2mn(α4m−1)2(α−β)4α2m(α6m−Bm)+(α4m−1)(α−β)4α4mC′m+O(1α2mn)=(αmn−βmn+βmnα−β)4−(αm(n−1)−βm(n−1)+βm(n−1)α−β)4−4Bmnα2mn(α4m−1)2(α−β)4α2m(α6m−Bm)+(α4m−1)(α−β)4α4mC′m+O(1α2mn)=u4mn−u4m(n−1)+δ+Vm+O(1α2mn), |
where
C′m=16α4m(α4m−1)2(α6m−Bm)2−10α4m(α4m−1)α8m−1, |
Vm=(α4m−1)(α−β)4α4mC′m=(α4m−1)(α−β)4α4m(16α4m(α4m−1)2(α6m−Bm)2−10α4m(α4m−1)α8m−1)=(α4m−1)2(α−β)4(16(α4m−1)(α6m−Bm)2−10α8m−1) |
and
δ=4Bmnα2mn(α−β)4((1−Bmβ2m)α2m(α6m−Bm)−(α4m−1)2α2m(α6m−Bm))=4Bmnα2mn(α−β)4(α4m−2Bmα2m+B2m1−Bmα6m)=4Bmnα2mn(α−β)4⋅α2m(αm−βm)21−Bmα6m=(αm−βmα−β)2⋅4Bmn(1−Bmα6m)(1−Bmβ6m)⋅α2m(n+1)(1−Bmβ6m)(α−β)2=4Bmnu2m(1−Bmα6m)(1−Bmβ6m)(α2m(n+1)(α−β)2−Bmα2m(n−2)(α−β)2). |
Let Um=u2m(1−Bmα6m)(1−Bmβ6m). Then we can obtain
δ=4BmnUm((αm(n+1)−βm(n+1)+βm(n+1)α−β)2−Bm(αm(n−2)−βm(n−2)+βm(n−2)α−β)2)=4BmnUm(u2m(n+1)−Bmu2m(n−2)+Bmβ2m(n−2)−β2m(n−2)(α−β)2)=4BmnUm(u2m(n+1)−Bmu2m(n−2))+O(1α2m(n−2)). |
Then
limn→∞((∞∑k=n1u4mk)−1−(u4mn−u4m(n−1)+4BmnUm(u2m(n+1)−Bmu2m(n−2))+Vm))=limn→∞(O(1α2m(n−2))+O(1α2mn))=0. |
So we have
(∞∑k=n1u4mk)−1∼u4mn−u4m(n−1)+4BmnUm(u2m(n+1)−Bmu2m(n−2))+Vm, |
where Um=u2m(1−Bmα6m)(1−Bmβ6m) and Vm=(α4m−1)2(α−β)4(16(α4m−1)(α6m−Bm)2−10α8m−1).
In this subsection, we will provide a proof of Theorem 2.5.
Proof. By (1.2) and (3.2b), we have
1umk+umk+l=(αmk−βmkα−β+αmk+l−βmk+lα−β)−1=(αmk(1+αl)α−β)−1(1−Bmk(1+βl)α2mk(1+αl))−1=α−βαmk(1+αl)(1+Bmk(1+βl)α2mk(1+αl)+(1+βl)2α4mk(1+αl)2+Rk)=α−β1+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=α−β1+αl(∞∑k=n1αmk+∞∑k=nBmk(1+βl)α3mk(1+αl)+∞∑k=n(1+βl)2α5mk(1+αl)2+∞∑k=nRkαmk)=α−β1+αl(αmαmn(αm−1)+Cm)=αm(α−β)αmn(αm−1)(1+αl)(1+αmn(αm−1)αmCm), | (3.15) |
where
Cm=Bmnα3m(1+βl)α3mn(α3m−Bm)(1+αl)+α5m(1+βl)2α5mn(α5m−1)(1+αl)2+∞∑k=nRkαmk. |
Then we have
αmn(αm−1)αmCm=O(1α2mn). | (3.16) |
By (3.2a), (3.15), and (3.16), we have
(∞∑k=n1umk+umk+l)−1=(αm(α−β)αmn(1+αl)(αm−1))−1(1+O(1α2mn))−1=αmn(1+αl)(αm−1)αm(α−β)(1+O(1α2mn))=αmn+αmn+l−αm(n−1)−αm(n−1)+lα−β+O(1αmn)=αmn+l−βmn+l+βmn+lα−β−αm(n−1)+l−βm(n−1)+l+βm(n−1)+lα−β+αmn−βmn+βmnα−β−αm(n−1)−βm(n−1)+βm(n−1)α−β+O(1αmn)=umn+l−um(n−1)+l+umn−um(n−1)+βmn+βmn+l−βm(n−1)−βm(n−1)+lα−β+O(1αmn). |
Then
limn→∞((∞∑k=n1umk+umk+l)−1−(umn+l−um(n−1)+l+umn−um(n−1)))=limn→∞(βmn+βmn+l−βm(n−1)−βm(n−1)+lα−β+O(1αmn))=0. |
So we have
(∞∑k=n1umk+umk+l)−1∼umn+l−um(n−1)+l+umn−um(n−1). |
In this subsection, we will provide a proof of Theorem 2.6.
Proof. By (1.2), we have
l∑i=0umk+i=1α−β(αmkl∑i=0αi−βmkl∑i=0βi)=1α−β(αmk(1−αl+1)1−α−βmk(1−βl+1)1−β)=αmk(1−αl+1)(α−β)(1−α)(1−Bmk(1−α)(1−βl+1)α2mk(1−β)(1−αl+1)). | (3.17) |
From (3.2b) and (3.17), it follows that
1l∑i=0umk+i=(αmk(1−αl+1)(α−β)(1−α))−1(1−Bmk(1−α)(1−βl+1)α2mk(1−β)(1−αl+1))−1=(α−β)(1−α)αmk(1−αl+1)(1+Bmk(1−α)(1−βl+1)α2mk(1−β)(1−αl+1)+(1−α)2(1−βl+1)2α4mk(1−β)2(1−αl+1)2+Rk)=(α−β)(1−α)(1−αl+1)(1αmk+Bmk(1−α)(1−βl+1)α3mk(1−β)(1−αl+1)+(1−α)2(1−βl+1)2α5mk(1−β)2(1−αl+1)2)+(α−β)(1−α)(1−αl+1)⋅Rkαmk, | (3.18) |
where
Rk=(1−βl+1)3(1−α)3Bmk(4αmk(1−β)(1−αl+1)−3Bmk(1−α)(1−βl+1))α4mk(1−β)2(1−αl+1)2(α2mk(1−β)(1−αl+1)−Bmk(1−α)(1−βl+1)). |
Let n be a positive integer. Then it follows from (3.18) that
∞∑k=n1∑li=0umk+i=(α−β)(1−α)(1−αl+1)(∞∑k=n1αmk+Cm)=(α−β)(1−α)(1−αl+1)(αmαmn(αm−1)+Cm)=αm(α−β)(1−α)αmn(αm−1)(1−αl+1)(1+αmn(αm−1)αmCm), | (3.19) |
where
Cm=∞∑k=nBmk(1−α)(1−βl+1)α3mk(1−β)(1−αl+1)+∞∑k=n(1−α)2(1−βl+1)2α5mk(1−β)2(1−αl+1)2+∞∑k=nRkαmk=Bmnα3m(1−α)(1−βl+1))α3mn(α3m−Bm)(1−β)(1−αl+1)+α5m(1−α)2(1−βl+1)2α5mn(α5m−1)(1−β)2(1−αl+1)2+∞∑k=nRkαmk=O(1α3mn). |
Then we have
αmn(αm−1)αmCm=O(1α2mn). | (3.20) |
By (3.2a), (3.19), and (3.20), we have
(∞∑k=n1l∑i=0umk+i)−1=(αm(α−1)(α−β)αmn(αl+1−1)(αm−1))−1(1+O(1α2mn))−1=αmn(αl+1−1)(αm−1)αm(α−1)(α−β)(1+O(1α2mn))=αmn+l+1−αm(n−1)+l+1−αmn+αm(n−1)(α−1)(α−β)+O(1αmn)=αm(n−1)−βm(n−1)+βm(n−1)(α−1)(α−β)−αm(n−1)+l+1−βm(n−1)+l+1+βm(n−1)+l+1(α−1)(α−β)−αmn−βmn+βmn(α−1)(α−β)+αmn+l+1−βmn+l+1+βmn+l+1(α−1)(α−β)+O(1αmn)=1α−1(umn+l+1−um(n−1)+l+1−umn+um(n−1))+βmn+l+1−βmn+βm(n−1)−βm(n−1)+l+1(α−1)(α−β)+O(1αmn). |
Then
limn→∞((∞∑k=n1∑li=0umk+i)−1−1α−1(umn+l+1−um(n−1)+l+1−umn+um(n−1)))=limn→∞(βmn+l+1−βmn+βm(n−1)−βm(n−1)+l+1(α−1)(α−β)+O(1αmn))=0. |
So we have
(∞∑k=n1∑li=0umk+i)−1∼1α−1(umn+l+1−um(n−1)+l+1−umn+um(n−1)). |
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=(α−β)2((αmk−βmk)(αmk+h−βmk+h))−1=(α−β)2α2mk+h((1−Bmkα2mk)(1−Bmk+hα2mk+2h))−1=(α−β)2α2mk+h(1−Bmkα2mk−Bmk+hα2mk+2h+Bhα4mk+2h)−1=(α−β)2α2mk+h(1−η)−1, | (3.21) |
where
η=Bmkα2mk+Bmk+hα2mk+2h−Bhα4mk+2h=Bmkα2mk+Bmk+hα2mk+2h+O(1α4mk). |
Then we have
η2+η31−η=O(1α4mk). | (3.22) |
By (3.2b), (3.21), and (3.22), we have
1umkumk+h=(α−β)2α2mk+h(1+η+η2+η31−η)=(α−β)2α2mk+h(1+η+O(1α4mk))=(α−β)2α2mk+h(1+Bmkα2mk(1+Bhα2h)+O(1α4mk))=(α−β)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=(α−β)2αh(∞∑k=n1α2mk+(1+Bhα2h)∞∑k=nBmkα4mk)+O(1α6mn)=(α−β)2αh(α2mα2mn(α2m−1)+(1+Bhα2h)Bmnα4mα4mn(α4m−Bm))+O(1α6mn)=(α−β)2α2mα2mn+h(α2m−1)(1+(1+Bhα2h)Bmnα2m(α2m−1)α2mn(α4m−Bm)+O(1α4mn))=(α−β)2α2mα2mn+h(α2m−1)(1+ω), | (3.24) |
where
ω=(1+Bhα2h)Bmnα2m(α2m−1)α2mn(α4m−Bm)+O(1α4mn). |
Then we have
ω2−ω31+ω=O(1α4mn). | (3.25) |
By (3.2a), (3.24), and (3.25), we have
(∞∑k=n1umkumk+h)−1=((α−β)2α2mα2mn+h(α2m−1))−1(1+ω)−1=α2mn+h(α2m−1)(α−β)2α2m(1−ω+ω2−ω31+ω)=α2mn+h(α2m−1)(α−β)2α2m(1−ω+O(1α4mn))=α2mn+h(α2m−1)(α−β)2α2m(1−(1+Bhα2h)Bmnα2m(α2m−1)α2mn(α4m−Bm)+O(1α4mn))=α2mn+h−α2m(n−1)+h(α−β)2−(1+Bhα2h)Bmn(α2m−1)2αh(α−β)2(α4m−Bm)+O(1α2mn). | (3.26) |
(ⅰ) If we take h=2l, then it follows from (3.26) that
(∞∑k=n1umkumk+2l)−1=(αmn+l−βmn+l+βmn+lα−β)2−(αm(n−1)+l−βm(n−1)+l+βm(n−1)+lα−β)2−(1+1α4l)Bmn(α2m−1)2α2l(α−β)2(α4m−Bm)+O(1α2mn)=u2mn+l−u2m(n−1)+l+δ+O(1α2mn), |
where
δ=2(Bmn+l−Bm(n−1)+l)(α−β)2−(1+1α4l)Bmn(α2m−1)2α2l(α−β)2(α4m−Bm)=−Bmn(α2m−1)2α4m−Bm(α2l+β2l(α−β)2−2Bl(1−Bm)(α4m−Bm)(α−β)2(α2m−1)2)=−Bmn(α2m−1)2α4m−Bm(u2l+2Bl(α−β)2−2Bl(1−Bm)(α4m−Bm)(α−β)2(α2m−1)2)=−Bmn(α2m−1)2α4m−Bm(u2l+2Bl(α−β)2(1−(1−Bm)(α4m−Bm)(α2m−1)2)). |
Let Cm,l=u2l+2Bl(α−β)2(1−(1−Bm)(α4m−Bm)(α2m−1)2). Then
limn→∞((∞∑k=n1umkumk+2l)−1−(u2mn+l−u2m(n−1)+l−Bmn(α2m−1)2α4m−BmCm,l))=0. |
So we have
(∞∑k=n1umkumk+2l)−1∼u2mn+l−u2m(n−1)+l−Bmn(α2m−1)2α4m−BmCm,l, |
where Cm,l=u2l+2BlA2−4B(1−(1−Bm)(α4m−Bm)(α2m−1)2).
(ⅱ) If we take h=2l−1, then it follows from (3.26) that
(∞∑k=n1umkumk+2l−1)−1=α2mn+2l−1−α2m(n−1)+2l−1(α−β)2−(1+Bα4l−2)Bmn(α2m−1)2α2l−1(α−β)2(α4m−Bm)+O(1α2mn)=(α−αβ2m)α2mn+2l−2(α−β)2−(1+Bα4l−2)Bmn(α2m−1)2α2l−1(α−β)2(α4m−Bm)+O(1α2mn)=(α−αβ2m)(α2mn+2l−2−β2mn+2l−2+β2mn+2l−2(α−β)2)−(1+Bα4l−2)Bmn(α2m−1)2α2l−1(α−β)2(α4m−Bm)+O(1α2mn)=(α−αβ2m)u2mn+l−1+δ+O(1α2mn), |
where
δ=2Bmn+l−1(α−αβ2m)(α−β)2−(1+B2l−1α4l−2)Bmn(α2m−1)2α2l−1(α−β)2(α4m−Bm)=−Bmn(α2m−1)2α4m−Bm(αlαl−1+β2l−1(α−β)2−2Bl−1(α−αβ2m)(α4m−Bm)(α−β)2(α2m−1)2)=−Bmn(α2m−1)2α4m−Bm(ulul−1+Bl−1(α+β)(α−β)2−2Bl−1(α−αβ2m)(α4m−Bm)(α−β)2(α2m−1)2)=−Bmn(α2m−1)2α4m−Bm(ulul−1+Bl−1(α−β)2((α+β)−2(α−αβ2m)(α4m−Bm)(α2m−1)2)). |
Let C′m,l=ulul−1+Bl−1(α−β)2((α+β)−2(α−αβ2m)(α4m−Bm)(α2m−1)2). Then
limn→∞((∞∑k=n1umkumk+2l−1)−1−((α−αβ2m)u2mn+l−1−Bmn(α2m−1)2α4m−BmC′m,l))=0. |
So we have
(∞∑k=n1umkumk+2l−1)−1∼(α−αβ2m)u2mn+l−1−Bmn(α2m−1)2α4m−BmC′m,l, |
where C′m,l=ulul−1+Bl−1A2−4B(A−2(α−αβ2m)(α4m−Bm)(α2m−1)2).
In this subsection, we will provide a proof of Theorem 2.8.
Proof. By (1.2), we have
umk+C=αmk−βmkα−β+C=αmkα−β(1+C(α−β)αmk−Bmkα2mk)=αmkα−β(1+η), | (3.27) |
where
η=C(α−β)αmk−Bmkα2mk. |
Then we have
η2−η31+η=O(1α2mk). | (3.28) |
From (3.2a), (3.27), and (3.28), it follows that
1umk+C=(αmkα−β)−1(1+η)−1=α−βαmk(1−η+η2−η31+η)=α−βαmk(1−η+O(1α2mk))=α−βαmk(1−C(α−β)αmk+O(1α2mk))=(α−β)(1αmk−C(α−β)α2mk+O(1α3mk)). | (3.29) |
Let n be a positive integer. Then it follows from (3.29) that
∞∑k=n1umk+C=∞∑k=nα−βαmk−∞∑k=nC(α−β)2α2mk+O(1α3mn)=αm(α−β)αmn(αm−1)−Cα2m(α−β)2α2mn(α2m−1)+O(1α3mn)=(α−β)αmαmn(αm−1)(1−C(α−β)αmαmn(αm+1)+O(1α2mn))=(α−β)αmαmn(αm−1)(1−ω), | (3.30) |
where
ω=C(α−β)αmαmn(αm+1)+O(1α2mn). |
Then we have
ω2+ω31−ω=O(1α2mn). | (3.31) |
By (3.2b), (3.30), and (3.31), we have
(∞∑k=n1umk+C)−1=((α−β)αmαmn(αm−1))−1(1−ω)−1=αmn(αm−1)(α−β)αm(1+ω+ω2+ω31−ω)=αmn(αm−1)(α−β)αm(1+ω+O(1α2mn))=αmn(αm−1)(α−β)αm(1+C(α−β)αmαmn(αm+1)+O(1α2mn))=αmn−αm(n−1)α−β+C(αm−1)αm+1+O(1αmn)=αmn−βmn+βmnα−β−αm(n−1)−βm(n−1)+βm(n−1)α−β+C(αm−1)αm+1+O(1αmn)=umn−um(n−1)+βmn−βm(n−1)α−β+C(αm−1)αm+1+O(1αmn). |
Then
limn→∞((∞∑k=n1umk+C)−1−(umn−um(n−1)+C(αm−1)αm+1))=limn→∞(βmn−βm(n−1)α−β+O(1αmn))=0. |
So we have
(∞∑k=n1umk+C)−1∼umn−um(n−1)+Cαm−1αm+1. |
Let (un)n≥0 be the special Lucas u-sequence defined by un+2=Aun+1−Bun,u0=0,u1=1, where n≥0, B=±1, and A is an integer such that A2−4B>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,1∑li=0umk+i,1umkumk+2l,1umkumk+2l−1,1umk+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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