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.
1. Introduction
Let p be a nonzero integer such that Δ=p2+4≠0. The generalized Fibonacci and Lucas sequences are defined by the following recurrences: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).
If α and β are the roots of equation x2-px-1=0, the Binet formulas of the sequences {Un} and {Vn} have the forms:Un=αn-βnα-β,Vn=αn+βn,
respectively.
In [1], Backstrom developed formulas for closely related series of the form:∑n=0∞1Fan+b+c,
for certain values of a, b, and c. For example, he obtained the following series:∑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:∑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,
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:∑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: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:∑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.
Let n be an arbitrary nonzero integer. For integer m≥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,
∑k=0m(-1)knU(2k+1)nV(k+1)n2Vkn2=U(m+1)n24UnV(m+1)n2.
Proof.
We give the proof of Lemma 2.1 as the proofs of the sums in [4], using the following equalities:
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 integers n and t,
V2n-(-1)n-tV2t=ΔUn-tUn+t,V2n+(-1)n-tV2t=Vn-tVn+t,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 integer t,
∑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 replacing n with (2n+1)t in (2.5), we have
U(2n+1)t2-Ut2=U2ntU2(n+1)t,
or
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:
1U(2n+1)t+Ut=1Vt(1U2nt+(-1)tU2(n+1)t)-UtU2ntU2(n+1)t.
We have the sum
∑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
∑n=1m(1U2nt-1U2(n+1)t)=1U2t-1U2(m+1)t,
and taking s=2nt and r=2t in identity [5]:
Us+rVs-UsVs+r=2(-1)sUr,
we get
∑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 take t=1 and p=1 in (2.6), we have
∑n=1m1F2n+1+1=F2m+1-1F2(m+1).
Note that
F2(m+1)∑n=1m1F2n+1+1=∑n=1mF2n.
Corollary 2.4.
For an odd integer t,
∑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 equalities V2n=Vn2-2(-1)n=ΔUn2+2(-1)n and U2n=UnVn in Theorem 2.3, the results are obtained.
Corollary 2.5.
Let t be an odd integer. For |β/α|<1,t>0 and |α/β|<1,t<0,
∑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,
∑n=1∞1U(2n+1)t+Ut=-12Vt(Δ+VtUt),∑n=1∞1U(2n+1)t-Ut=Δ2Vt(UtVt+1Δ).
Proof.
Since
limn→∞(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 integer m≥1 and an arbitrary nonzero integer t,
∑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 replacing n with (2n+1)t/2 and t with t/2 in (2.4), we have
V(2n+1)t-(-1)ntVt=ΔUntU(n+1)t,
or
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
(-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:
∑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 integer t,
∑n=1∞(-1)ntU(2n+1)t(V(2n+1)t-(-1)ntVt)2=14Δ2Ut(Vt2Ut2-Δ).
Proof.
Taking m→∞ in Theorem 2.6 and using (2.19), the result is easily obtained.
Theorem 2.8.
For an integer m≥0 and an arbitrary nonzero integer t,
∑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 Theorem 2.6.
Corollary 2.9.
For an arbitrary nonzero integer t,
∑n=0∞(-1)ntU(2n+1)t(V(2n+1)t+(-1)ntVt)2=14ΔUt.
Proof.
Taking m→∞ in Theorem 2.8 and using (2.19), the result is easily obtained.
For example, if we take t=3 and p=1 in (2.27), we have
∑n=0∞(-1)nF3(2n+1)(L3(2n+1)+(-1)n4)2=140.
Theorem 2.10.
For an integer m≥1 and an arbitrary nonzero integer t,
∑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 Theorem 2.6.
Corollary 2.11.
For an arbitrary nonzero integer t,
∑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.
Taking m→∞ in Theorem 2.10 and using (2.19), the result is easily obtained.
For example, if we take t=2 in the equality (2.30), we have
∑n=1∞U4(2n+1)(V2(2n+1)2-V22)2=1Δ2U43.
BackstromR. P.On reciprocal series related to Fibonacci numbers with subscripts in arithmetic progression1981191421ZBL0446.10011PopovB. S.On certain series of reciprocals of Fibonacci numbers19842232612652-s2.0-0035600655ZBL0539.10012PopovB. S.Summation of reciprocal series of numerical functions of second order19862411721ZBL0597.10008GauthierN.Solution to problem H-68020114919092VajdaS.1989New York, NY, USAJohn Wiley & Sons