# On the Sum of Reciprocal Generalized Fibonacci Numbers

Authors: Pingzhi Yuan; Zilong He; Junyi Zhou
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We consider infinite sums derived from the reciprocals of the generalized Fibonacci numbers. We obtain some new and interesting identities for the generalized Fibonacci numbers.


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## Body

## 1. Introduction

For any integern≥0, the famous Fibonacci numbers Fn and Pell numbers are defined by the second-order linear recurrence sequences
(1)Fn+2=Fn+1+Fn,F0=0,F1=1,Pn+2=2Pn+1+Pn,P0=0,P1=1.
There are many interesting results on the properties of these two sequences; see [1–9]. In 2009, Ohtsuka and Nakamura [5] studied the properties of the Fibonacci numbers and proved the following two interesting identities:
(2)∑k=n∞‍1Fk-1=Fn-2,ifnisevenandn≥2;Fn-2-1,ifnisoddandn≥1,∑k=n∞‍1Fk2-1=Fn-1Fn-1,ifnisevenandn≥2;Fn-1Fn,ifnisoddandn≥1,
where x is the floor function; that is, it denotes the greatest integer less than or equal to x. Recently, Holliday and Komatsu [1] (Theorems 3 and 4) and Xu and Wang [7] proved the following interesting identities for the Pell numbers:
(3)∑k=n∞‍1Pk-1=Pn-1+Pn-2,ifnisevenandn≥2;Pn-1+Pn-2-1,ifnisoddandn≥1,∑k=n∞‍1Pk2-1=2Pn-1+Pn-1,ifnisevenandn≥2;2Pn-1+Pn,ifnisoddandn≥1,∑k=n∞‍1Pk3-1=Pn2Pn-1+3PnPn-12+-6182Pn-9182Pn-1,hhhhhhhhh.hhhifnisevenandn≥2;Pn2Pn-1+3PnPn-12+6182Pn+9182Pn-1,hhhhhhhhhhhhifnisoddandn≥1,
where providingP-1=P1=1. In [7, 8], the authors asked whether there exists a computational formula for
(4)∑k=n∞‍1Pks-1,
where s≥4 is a positive integer.Letp and q be integers such that p2+4q>0. Define the generalized Fibonacci sequence {Un(p,q)}, briefly {Un}, as shown: for n≥2(5)Un=pUn-1+qUn-2,
where U0=0, U1=1. The Binet formula for {Un} is
(6)Un=αn-βnα-β,
where α,β=(p±p2+4q)/2.The main purpose of this paper related to the computing problem of(7)U(s,n)=∑k=n∞‍1Uks-1
for s=3 and q=-1. For easy computation, we assume that p=a is a positive integer and q=-1 throughout the paper. We have the following.Theorem 1.
Leta≥3 be a positive integer, and let Gn be defined by the second-order linear recurrence sequence Gn+2=aGn+1-Gn,G0=0,G1=1. Then for all n≥2 one has
(8)∑k=n∞‍1Gk3-1=Gn3-Gn-13-3∑k=0n-4/5‍Gn-3-5k-2,hhhhhhhhha=3,n≡3(mod⁡5);Gn3-Gn-13-3∑k=0n-4/5‍Gn-3-5k-1,hhhhhhhhhhhhhHhhhhotherwise.


## 2. Proof of the Main Result

In this section, we will prove our main result. We consider the case thatαβ=1 and s=3.Proof.
From the Taylor series expansion of(1-ε)-3 as ε→0, we have
(9)1-ε-3=1+∑n=1∞‍(n+1)(n+2)2εn=1+3ε+6ε2+O(ε3).
Using (6), we have
(10)1Gk3=α-β3α3k1-1α2k-3=α-β3α3k11-3/α2k+3/α4k+1/α6k=α-β3α3k1+3α2k+6α4k+10α4k-15α2k+6α4kα2k-13=α-β31α3k+3α5k+6α7k+10α4k-15α2k+6α7kα2k-13.
It is easy to check that(11)10α9k<10α4k-15α2k+6α7kα2k-13<11α9k
holds for a≥3 and k≥2.
Thus(12)∑k=n∞‍1Gk3=α-β3×1α3n·α3α3-1+3α5n·α5α5-1+6α7n·α7α7-1∑k=n∞‍10α4k-15α2k+6α7kα2k-13+∑k=n∞‍10α4k-15α2k+6α7kα2k-13=α-β3α3α3nα3-1×1+3α2nα2(α3-1)α5-1+6α4α4n(α3-1)α7-1+Rn,
where
(13)Rn=(α3-1)α3nα3∑k=n∞‍10α4k-15α2k+6α7kα2k-13.
Since ∑k=n∞‍1/α9k=α9/α9n(α9-1), we have
(14)10α6α6n(α6+α3+1)<Rn<11α6α6n(α6+α3+1)
holds for a≥3 and k≥2.
Taking reciprocal, we get(15)∑k=n∞‍1Gk3-1=α3-1α3nα-β3α31×1+3α2nα2α3-1α5-1+6α4α4nα3-1α7-1+Rn-3hhhhhhhhhhhhhhhhhh+6α4α4nα3-1α7-1+Rn-1<α3-1α3nα-β3α31×1+3α2nα2α3-1α5-1+6α4α4nα3-1α7-1hhhhhhhhhhhhhhhhhh+10α6α6nα6+α3+1-1<(α3-1)α3nα-β3α3-3α3-12αnα-β3αα5-1+δ1,
where
(16)δ1=-6αα3-12αnα-β3(α7-1)-10α3(α3-1)α3nα-β3(α6+α3+1)+9αα3-13αnα-β3α5-12+36α3α3-13α3nα-β3α5-1α7-1+36α5α3-13α5nα-β3α7-12+60α5α3-12α5nα-β3(α5-1)(α6+α3+1)+120α7α3-12α7nα-β3(α7-1)(α6+α3+1)+100α9α3-1α9nα-β3α6+α3+12
since
(17)11+ε=1-ε+ε2-ε31+ε.
An easy calculation shows that δ1≤4/αn+3 holds for a≥3 and k≥2. Therefore,
(18)∑k=n∞‍1Gk3-1<(α3-1)α3nα-β3α3-3α3-12αnα-β3α(α5-1)+δ1≤α3n-α3n-3α-β3-3α3-12αnα-β3α(α5-1)+4αn+3=Gn3-Gn-13-3αn+2α-βαα5-1+3α-1α-β3αn-α3-1α-β3α3n+4αn+3<Gn3-Gn-13-3αn+2α-βαα5-1+3(α-1)α-β3αn+4αn+3=Gn3-Gn-13-3αn+2(α-β)α(α5-1)+λ1,
where
(19)λ1=3(α-1)α-β3αn+4αn+3<0.1681
for a≥3 and n≥2.
Similarly, we have(20)∑k=n∞‍1Gk3-1=α3-1α3nα-β3α31×1+3α2nα2α3-1α5-1hhhhhhhhhhhhhhhhhhh+6α4α4n(α3-1)α7-1+Rn1+3α2nα2α3-1α5-1-1>α3-1α3nα-β3α31×1+3α2nα2α3-1α5-1+6α4α4nα3-1α7-1hhhhhhhhhhhhhhhhhhh+11α6α6nα6+α3+11+3α2nα2α3-1α5-1-1>Gn3-Gn-13-3αn+2(α-β)α(α5-1)+λ2.
Since
(21)11+ε=1-ε+ε2-ε3+ε41+ε,
and ε=3/α2nα2(α3-1)/α5-1+6α4/α4n(α3-1)/α7-1+11α6/α6n(α6+α3+1)<0.3 for a≥3 and n≥2, we have ε2-ε3>0.7ε2, whence we can take
(22)λ2=3(α-1)α-β3αn-α3-1α-β3α3n-6αα3-12αnα-β3(α7-1)-11α3α3-1α3nα-β3α6+α3+1+6.3αα3-13αnα-β3α5-12>0
for a≥3 and n≥2.
Consequently, we have shown that(23)Gn3-Gn-13-3αn+2(α-β)α(α5-1)+λ2<∑k=n∞‍1Gk3-1<Gn3-Gn-13-3αn+2(α-β)α(α5-1)+λ1,
where 0<λ2<λ1<0.1681 for a≥3 and n≥2, and λ1<0.0053 for a≥4 and n≥3.
Now the calculations show that(24)αn+2(α-β)α(α5-1)=Gn-3+Gn-8+⋯+G7+G2-α2+α3(α-β)(α5-1),hhhhhhhhhhhhhhhhhhhhhhhhhn≡0(mod⁡5);Gn-3+Gn-8+⋯+G8+G3-α2+α3(α-β)(α5-1),hhhhhhhhhhhhhhhhhhhhhhhhhn≡1(mod⁡5);Gn-3+Gn-8+⋯+G9+G4-α+α4(α-β)(α5-1),hhhhhhhhhhhhhhhhhhhhhhhhhn≡2(mod⁡5);Gn-3+Gn-8+⋯+G10+G5-1+α5(α-β)(α5-1),hhhhhhhhhhhhhhhhhhhhhhhhhn≡3(mod⁡5);Gn-3+Gn-8+⋯+G6+G1-α+α4(α-β)(α5-1),hhhhhhhhhhhhhhhhhhhhhlhhhn≡4mod⁡5.
The calculations also show that3(α2+α3)/(α-β)(α5-1)>λ1 for a≥3 and n≥2; 3(α+α4)/(α-β)(α5-1)>λ1 for a≥3 and n≥2; and 3(1+α5)/(α-β)(α5-1)>λ1+1 for a=3 and n≥3; 0.87<3(1+α5)/(α-β)(α5-1)<1 for a>3 and n≥3. Combining the calculations and (23), we obtain
(25)∑k=n∞‍1Gk3-1=Gn3-Gn-13-3∑k=0n-4/5‍Gn-3-5k-2,hhhh.hhha=3,n≡3(mod⁡5);Gn3-Gn-13-3∑k=0n-4/5‍Gn-3-5k-1,hhhhhhhhhhhhhhhhhotherwise.
Therefore we have proved Theorem1.Remark 2.
We can also compute the casess>3 or q=1; however, the computations are much more complicated. So we stop here.



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*Source: 402540-2014-12-10.xml*