# On Reciprocal Series of Generalized Fibonacci Numbers with Subscripts in Arithmetic Progression

Authors: Neşe Ömür
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We investigate formulas for closely related series of the forms:∑n=0∞1/(Uan+b+c),
∑n=0∞(-1)nUan+b/(Uan+b+c)2,
∑n=0∞U2(an+b)/(Uan+b2+c)2 for certain values of a, b, and c.


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## Body

## 1. Introduction

Letp be a nonzero integer such that Δ=p2+4≠0. The generalized Fibonacci and Lucas sequences are defined by the following recurrences:(1.1)Un+1=pUn+Un-1,Vn+1=pVn+Vn-1,
where U0=0,U1=1 and V0=2,V1=p, respectively. When p=1,Un=Fn (nth Fibonacci number) and Vn=Ln (nth Lucas number).Ifα and β are the roots of equation x2-px-1=0, the Binet formulas of the sequences {Un} and {Vn} have the forms:(1.2)Un=αn-βnα-β,Vn=αn+βn,
respectively.In [1], Backstrom developed formulas for closely related series of the form:(1.3)∑n=0∞1Fan+b+c,
for certain values of a, b, and c. For example, he obtained the following series:(1.4)∑n=0∞1F2n+1+FK=K52LK,∑n=0∞1F(2n+1)K+2t+FK={(5-5Ft/Lt)2LK,teven,(5-Lt/Ft)2LK,todd,
where K represents an odd integer and t is an integer in the range -(K-1)/2 to (K-1)/2 inclusive. Also, he gave the similar results for Lucas numbers.In [2], Popov found in explicit form series of the form:(1.5)∑n=0∞1Fan+b±c,∑n=0∞1Fan+bFcn+d,∑n=0∞1Fan+b2±Fcn+d2,
for certain values of a, b, c, and d.In [3], Popov generalized some formulas of Backstrom [1] related to sums of reciprocal series of Fibonacci and Lucas numbers. For example,
(1.6)Δ∑n=0∞(-q)s+nrV(2n+1)r+2s-(-q)s+nrVr={βsUrUs,|βα|<1,αsUrUs,|αβ|<1,
where s and r are integers.In [4], Gauthier found the closed form expressions for the following sums:(1.7)∑k=1m(-1)knf(2k+1)nf(k+1)n2fkn2,m,n≥1,∑k=0m(-1)knf(2k+1)nl(k+1)n2lkn2,m,n≥0,
where for x≠0 an indeterminate, the generalized Fibonacci and Lucas polynomials {fn}n and {ln}n are given by the following recurrences:(1.8)fn+2=xfn+1+fn,f0=0,f1=1,n≥0,ln+2=xln+1+ln,l0=2,l1=x,n≥0,
respectively.In this paper, we investigate formulas for closely related series of the forms:(1.9)∑n=0∞1Uan+b+c,∑n=0∞(-1)nUan+b(Uan+b+c)2,∑n=0∞U2(an+b)(Uan+b2+c)2,
for certain values of a, b and c.


## 2. On Some Series of Reciprocals of Generalized Fibonacci Numbers

In this section, firstly, we will give the following lemmas for further use.Lemma 2.1.
Letn be an arbitrary nonzero integer. For integer  m≥1,
(2.1)∑k=1m(-1)knU(2k+1)nU(k+1)n2Ukn2=14Un(Vn2Un2-V(m+1)n2U(m+1)n2),
and for integer m≥0,
(2.2)∑k=0m(-1)knU(2k+1)nV(k+1)n2Vkn2=U(m+1)n24UnV(m+1)n2.Proof.
We give the proof of Lemma2.1 as the proofs of the sums in [4], using the following equalities:
(2.3)U(2k+1)nU(k+1)nUkn=12(VknUkn+V(k+1)nU(k+1)n),U(2k+1)nV(k+1)nVkn=12(U(k+1)nV(k+1)n+UknVkn),(-1)knUnU(k+1)nUkn=12(VknUkn-V(k+1)nU(k+1)n),(-1)knUnV(k+1)nVkn=12(U(k+1)nV(k+1)n-UknVkn).Lemma 2.2.
For arbitrary integersn and  t,
(2.4)V2n-(-1)n-tV2t=ΔUn-tUn+t,V2n+(-1)n-tV2t=Vn-tVn+t,(2.5)Un2-(-1)n-tUt2=Un-tUn+t,Vn2-(-1)n-tVt2=ΔUn-tUn+t.Proof.
From Binet formulas of sequences{Un} and {Vn}, the desired results are obtained.Theorem 2.3.
For an odd integert,
(2.6)∑n=1m1U(2n+1)t+Ut=12Vt(2-V2tU2t-2-V2(m+1)tU2(m+1)t),∑n=1m1U(2n+1)t-Ut=12Vt(2+V2tU2t-2+V2(m+1)tU2(m+1)t).Proof.
By replacingn with (2n+1)t in (2.5), we have
(2.7)U(2n+1)t2-Ut2=U2ntU2(n+1)t,
or
(2.8)1U(2n+1)t+Ut=U(2n+1)t-UtU2ntU2(n+1)t.
Taking r=(2n+1)t and s=t in the equality VsUr=Ur+s+(-1)sUr-s [5], the equality (2.8) is rewritten as follows:
(2.9)1U(2n+1)t+Ut=1Vt(1U2nt+(-1)tU2(n+1)t)-UtU2ntU2(n+1)t.
We have the sum
(2.10)∑n=1m1U(2n+1)t+Ut=1Vt∑n=1m(1U2nt+(-1)tU2(n+1)t)-Ut∑n=1m1U2ntU2(n+1)t.
For an odd integer t, we have
(2.11)∑n=1m(1U2nt-1U2(n+1)t)=1U2t-1U2(m+1)t,
and taking s=2nt and r=2t in identity [5]:
(2.12)Us+rVs-UsVs+r=2(-1)sUr,
we get
(2.13)∑n=1m1U2ntU2(n+1)t=12U2t∑n=1m(V2ntU2nt-V2(n+1)tU2(n+1)t)=12U2t(V2tU2t-V2(m+1)tU2(m+1)t).
Substituting (2.11) and (2.13) in (2.10), we have the desired result.For example, if we taket=1 and p=1 in (2.6), we have
(2.14)∑n=1m1F2n+1+1=F2m+1-1F2(m+1).
Note that
(2.15)F2(m+1)∑n=1m1F2n+1+1=∑n=1mF2n.Corollary 2.4.
For an odd integert,
(2.16)∑n=1m1U(2n+1)t+Ut={12Vt(V(m+1)tU(m+1)t-VtUt),miseven,12Vt(ΔU(m+1)tV(m+1)t-VtUt),misodd,∑n=1m1U(2n+1)t-Ut={Δ2Vt(UtVt-U(m+1)tV(m+1)t),miseven,12Vt(ΔUtVt-V(m+1)tU(m+1)t),misodd.Proof.
Using the equalitiesV2n=Vn2-2(-1)n=ΔUn2+2(-1)n and U2n=UnVn in Theorem 2.3, the results are obtained.Corollary 2.5.
Lett be an odd integer. For  |β/α|<1,t>0 and  |α/β|<1,t<0,
(2.17)∑n=1∞1U(2n+1)t+Ut=12Vt(Δ-VtUt),∑n=1∞1U(2n+1)t-Ut=Δ2Vt(UtVt-1Δ),
and for |β/α|<1,t<0 and |α/β|<1,t>0,
(2.18)∑n=1∞1U(2n+1)t+Ut=-12Vt(Δ+VtUt),∑n=1∞1U(2n+1)t-Ut=Δ2Vt(UtVt+1Δ).Proof.
Since(2.19)lim⁡n→∞(Van+bUan+b)={Δ|βα|<1,a>0,|αβ|<1,a<0,-Δ|βα|<1,a<0,|αβ|<1,a>0,
the results are easily seen by equalities (2.16).Theorem 2.6.
For an integerm≥1 and an arbitrary nonzero integer  t,
(2.20)∑n=1m(-1)ntU(2n+1)t(V(2n+1)t-(-1)ntVt)2=14Δ2Ut(Vt2Ut2-V(m+1)t2U(m+1)t2).Proof.
By replacingn with (2n+1)t/2 and t with t/2 in (2.4), we have
(2.21)V(2n+1)t-(-1)ntVt=ΔUntU(n+1)t,
or
(2.22)1V(2n+1)t-(-1)ntVt=1ΔUntU(n+1)t.
Multiplying equality (2.22) by (-1)ntU(2n+1)t/UntU(n+1)t, we get
(2.23)(-1)ntU(2n+1)tUntU(n+1)t(V(2n+1)t-(-1)ntVt)=(-1)ntU(2n+1)tΔUnt2U(n+1)t2.
We have the sum:
(2.24)∑n=1m(-1)ntU(2n+1)tUntU(n+1)t(V(2n+1)t-(-1)ntVt)=1Δ∑n=1m(-1)ntU(2n+1)tUtn2Ut(n+1)2.
Using the equalities (2.1) and (2.21), the proof is obtained.Corollary 2.7.
For an arbitrary nonzero integert,
(2.25)∑n=1∞(-1)ntU(2n+1)t(V(2n+1)t-(-1)ntVt)2=14Δ2Ut(Vt2Ut2-Δ).Proof.
Takingm→∞ in Theorem 2.6 and using (2.19), the result is easily obtained.Theorem 2.8.
For an integerm≥0 and an arbitrary nonzero integer  t,
(2.26)∑n=0m(-1)ntU(2n+1)t(V(2n+1)t+(-1)ntVt)2=U(m+1)t24UtV(m+1)t2.Proof.
The proof of the theorem is similar to the proof of Theorem2.6.Corollary 2.9.
For an arbitrary nonzero integert,
(2.27)∑n=0∞(-1)ntU(2n+1)t(V(2n+1)t+(-1)ntVt)2=14ΔUt.Proof.
Takingm→∞ in Theorem 2.8 and using (2.19), the result is easily obtained.For example, if we taket=3 and p=1 in (2.27), we have
(2.28)∑n=0∞(-1)nF3(2n+1)(L3(2n+1)+(-1)n4)2=140.Theorem 2.10.
For an integerm≥1 and an arbitrary nonzero integer  t,
(2.29)∑n=1mU2(2n+1)t(U(2n+1)t2-Ut2)2=14U2t(V2t2U2t2-V2(m+1)t2U2(m+1)t2),∑n=1mU2(2n+1)t(V(2n+1)t2-Vt2)2=14Δ2U2t(V2t2U2t2-V2(m+1)t2U2(m+1)t2).Proof.
The proof of theorem is similar to the proof of Theorem2.6.Corollary 2.11.
For an arbitrary nonzero integert,
(2.30)∑n=1∞U2(2n+1)t(U(2n+1)t2-Ut2)2=14U2t(V2t2U2t2-Δ),∑n=1∞U2(2n+1)t(V(2n+1)t2-Vt2)2=14Δ2U2t(V2t2U2t2-Δ).Proof.
Takingm→∞ in Theorem 2.10 and using (2.19), the result is easily obtained.For example, if we taket=2 in the equality (2.30), we have
(2.31)∑n=1∞U4(2n+1)(V2(2n+1)2-V22)2=1Δ2U43.



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*Source: 684280-2012-01-05.xml*