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Journal of Integer Sequences, Vol. 16 (2013),
Article 13.7.1
23 11
On the Sums of Reciprocal Generalized
Fibonacci Numbers
Kantaphon Kuhapatanakul1
Department of Mathematics
Faculty of Science
Kasetsart University
Bangkok 10900
Thailand
fscikpkk@ku.ac.th
Abstract
In this article we consider the infinite sums of reciprocal generalized Fibonacci
numbers and the infinite sums of reciprocal generalized Fibonacci sums. Applying the
floor function to the reciprocals of these sums, our results generalize some identities of
Holliday and Komatsu and extend some results of Liu and Zhao.
1
Introduction
Let a, b be two positive integers and c non-negative integer. Define the generalized Fibonacci
numbers Vn (c; a, b), briefly Vn , by the following relation
V0 = c,
V1 = 1
and
Vn+1 = aVn + bVn−1
(n ≥ 1).
Here note that Vn (0; 1, 1) = Fn are the Fibonacci numbers, Vn (2; 1, 1) = Ln are the Lucas
numbers and Vn (0; 2, 1) = Pn are the Pell numbers.
Ohtsuka and Nakamura [2] derived a formula for infinite sums of reciprocal Fibonacci
numbers, as follows,

!−1 
 (
 X

 ∞ 1
if n is even and n ≥ 2;
 = Fn−2 ,

(1)
Fk
Fn−2 − 1,
if n is odd and n ≥ 1,
k=n
1
Supported by the Kasetsart University Research and Development Institute (KURDI), Thailand.
1
where ⌊·⌋ is the floor function.
Wenpeng and Tingting [4] gave analogue of the identity (1) for the Pell numbers:

!−1 
 X
 (
∞


if n is even and n ≥ 2;
1

 = Pn − Pn−1 ,
Pk
Pn − Pn−1 − 1,
if n is odd and n ≥ 1.
k=n
(2)
Holliday and Komatsu [1] generalized (1) and (2) to the generalized Fibonacci numbers
Vn (0; a, 1), briefly un , and Vn (c; 1, 1) for c ≥ 1, briefly Gn . They showed that

!−1 
 X
 (
∞


if n is even and n ≥ 2;
1

 = un − un−1 ,
(3)
u
u
−
u
−
1,
if
n
is
odd
and
n
≥
1,
k
n
n−1
k=n
and

!−1 
 X
 (
∞


1

 = Gn−2 ,
Gk
Gn−2 − 1,
k=n
if n is odd and n ≥ n1 ;
if n is even and n ≥ n2 ,
where n1 and n2 are determined depending only on the value of c. For example, if Gn = Ln
or c = 2, then n1 = 2 and n2 = 3.
By the same proof as the one for (3), it is easy to verify that the identity (3) still holds
for Vn (0; a, b) in place of Vn (0; a, 1), provided that 1 ≤ b ≤ a.
In this paper, we first give the analogue of the identity (3) for the alternating sums of
reciprocals of Vn (0; a, b) and prove the following result in the next section.
Theorem 1. Let Un := Vn (0; a, b) with 1 ≤ b ≤ a. Then

!−1 

 X
∞
k


(−1)
 = (−1)n (Un + Un−1 ) − 1

U
k
k=n
We have a following corollary.
Corollary 2. For a positive integer n, we have

!−1 
 X

 ∞ (−1)k


 = (−1)n Fn+1 − 1.
(1)
F
k
k=n

!−1 

 ∞
k

 X (−1)
 = (−1)n (Pn + Pn−1 ) − 1.
(2) 
P
k
k=n
2
(n ≥ 1).
In 2012, Liu and Zhao [3] showed the formulas for the infinite sums of reciprocal hyperfibonacci numbers and hyperlucas numbers as:

!−1 
 X

∞


1

 = Fn − 1 (n ≥ 3),
(4)
Pk
F
i
i=0
k=n

!−1 
 X

 ∞

1

 = Ln − 1 (n ≥ 4).
(5)
Pk
L
i=0 i
k=n
Especially, they also gave a following general result for un = Vn (0; a, 1) as:

!−1 
 ∞

X


1

 = un − 1 (n ≥ 3).
Pk
i=0 ui
k=n
Next, we will extend the above results on the Fibonacci and Lucas numbers and Vn (0; a, 1)
to the generalized Fibonacci numbers Vn (c; 1, 1) and Vn (0; a, b) and prove the following results
in the next section.
Theorem 3. Let Gn = Vn (c; 1, 1) for c ≥ 1. We have

!−1 
 X

 ∞

1

 = Gn − 1
Pk
G
i=0 i
k=n
(n ≥ n0 ),
where n0 is determined depending only on the value of c.
For example, we can determine n0 for a fixed c as follows:
c
n0
1
2
2
4
3-5
6
6-13
8
14-34
10
35-89
12
90-233
14
234-610
16
611-1597
18
1598-4181
20
Put c = 1, 2 in Theorem 3, we can, respectively, deduce the identities (4) and (5).
Theorem 4. Let a ≥ b ≥ 1 and Un = Vn (0; a, b). We have

!−1 

 X
∞


1
 = Un − 1 (n ≥ Na ),

Pk
i=0 Ui
k=n
where Na = 3 for a = 1 and Na = 2 for a ≥ 2.
We have the following corollary.
Corollary 5. We have

 X
 ∞
1

Pk
k=n
i=0 Pi
!−1 


 = Pn − 1
3
(n ≥ 2).
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Proofs of Theorems
We begin with some identities of the generalized Fibonacci numbers Vn (c; a, b) whose induction proofs are omitted.
Lemma 6. Let a, b be positive integers and c non-negative integer. Then for n ≥ 1 we have
(1) Vn+1 Vn−1 − Vn2 = (−1)n bn−1 (1 − ac − bc2 ).
(2)
n
X
i=0
Vi =
1
(Vn+1 + bVn + ac − c − 1).
a+b−1
Proof of Theorem 1 : By Lemma 6, we have
Un+1 Un−1 − Un2 = (−1)n bn−1
and
n
X
i=0
Ui =
1
(Un+1 + bUn − 1) .
a+b−1
Since a ≥ b ≥ 1, we have Un+1 − Un−1 > an−1 ≥ bn−1 , so
(−1)n
(−1)n+1
(−1)n
−
−
Un + Un−1 − (−1)n Un+1 + Un − (−1)n+1
Un
n+1
(−1)n+1
(−1) Un−1 + 1
−
=
Un (Un + Un−1 − (−1)n ) Un+1 + Un − (−1)n+1
(−1)n+1 Un−1 Un+1 + (−1)n Un2 + Un+1 − Un−1 + (−1)n
=
Un (Un + Un−1 − (−1)n ) (Un+1 + Un − (−1)n+1 )
(−1)n−1 bn−1 + Un+1 − Un−1 + (−1)n
=
Un (Un + Un−1 − (−1)n ) (Un+1 + Un − (−1)n+1 )
> 0.
By applying the above inequality repeatedly, we obtain
1
(−1)n
1
>
+
n
n+1
(−1) (Un + Un−1 ) − 1
Un
(−1) (Un+1 + Un ) − 1
n
(−1)n+1
1
(−1)
+
+
>
Un
Un+1
(−1)n+2 (Un+2 + Un+1 ) − 1
..
.
(−1)n (−1)n+1 (−1)n+2
>
+
+
+ ··· .
Un
Un+1
Un+2
4
Therefore,
∞
X
(−1)k
k=n
Similarly, we have
Uk
<
1
.
(−1)n (Un + Un−1 ) − 1
(−1)n
(−1)n+1
(−1)n
−
+
= (−1)n
Un
Un + Un−1 Un+1 + Un
(6)
1
Un−1
+
Un (Un + Un−1 ) Un+1 + Un
Un−1 Un+1 − Un2
n
= (−1)
Un (Un + Un−1 )(Un+1 + Un )
bn−1
=
Un (Un + Un−1 )(Un+1 + Un )
> 0,
so
(−1)n
(−1)n
(−1)n+1
<
+
.
Un + Un−1
Un
Un+1 + Un
Repeating the above inequality, we obtain
∞
X
(−1)k
Uk
k=n
>
(−1)n
.
Un + Un−1
(7)
Combining the (6) and (7), we get
1
(−1)n (Un
it is equivalent to
+ Un−1 )
<
∞
X
(−1)k
k=n
Uk
<
(−1)n (Un
1
,
+ Un−1 ) − 1

!−1 
 X

∞
k


(−1)

 = (−1)n (Un + Un−1 ) − 1.
U
k
k=n
Proof of Theorem 3 : By Lemma 6, we have
Gn+1 Gn−1 − G2n = (−1)n (1 − c − c2 )
and
n
X
Gn = Gn+2 − 1.
i=0
5
Suppose c ≥ 1, we have
1
1
1
1
1
1
−
− Pn
−
−
=
Gn − 1 Gn+1 − 1
Gn − 1 Gn+1 − 1 Gn+2 − 1
i=0 Gi
Gn+1
1
=
−
(Gn − 1)(Gn+2 − 1) Gn+1 − 1
2Gn − 1 + (−1)n+1 (c2 + c − 1)
=
.
(Gn − 1)(Gn+1 − 1)(Gn+2 − 1)
Since Gn is monotone increasing with n, we can take n so large that 2Gn ≥ (−1)n (c2 + c)
for a fixed c. Hence, the numerator of the right-hand side of the above identity is positive if
n ≥ N1 for some positive integer N1 , so we get
Thus,
1
1
1
≥ Pn
+
Gn − 1
Gn+1 − 1
i=0 Gi
1
1
1
+
≥ Pn
+ Pn+1
Gn+2 − 1
i=0 Gi
i=0 Gi
1
1
1
+ Pn+2
+ ··· .
≥ Pn
+ Pn+1
G
G
i
i
i=0 Gi
i=0
i=0
∞
X
1
i=0 Gi
k=n
On the other hand, we have
1
Pn
i=0
Gi
−
Pk
≤
1
Gn − 1
(n ≥ N1 ).
(8)
1
1
1
1
1
−
+
=
+
Gn Gn+1
Gn+2 − 1 Gn Gn+1
1
1 − Gn+1
+
=
Gn (Gn+2 − 1) Gn+1
Gn−1 + (−1)n (c2 + c − 1)
=
.
Gn Gn+1 (Gn+2 − 1)
Similarly, we can take n so large that Gn−1 + (−1)n (c2 + c − 1) > 0 for a fixed c. Hence, the
numerator of the right-hand side of the above identity is positive if n ≥ N2 for some positive
integer N2 , so we get
1
1
1
< Pn
+
Gn
Gn+1
i=0 Gi
1
1
1
< Pn
+ Pn+1
+
Gn+2
i=0 Gi
i=0 Gi
1
1
1
+ Pn+2
+ ··· .
< Pn
+ Pn+1
i=0 Gi
i=0 Gi
i=0 Gi
6
Thus,
∞
X
k=n
1
Pk
i=0 Gi
>
1
Gn
(n ≥ N2 ).
(9)
Combining the two inequalities (8) and (9) we obtain
∞
X
1
1
1
,
<
≤
Pk
Gn k=n i=0 Gi
Gn − 1
where n ≥ n0 = max{N1 , N2 }, which completes the proof.
Proof of Theorem 4 : The
P case a = 1 has already been proved, it suffices to show the
case a ≥ 2. Defining Sn = ni=0 Ui . We have
1
1
1
Sn−1
1
−
+
=
−
Sn Un Un+1
Un+1 Un Sn
Un Sn − Un+1 Sn−1
=
Un Un+1 Sn
Un+1 − Un + (−1)n+1
> 0,
=
aUn Un+1 Sn
and for n ≥ 2
1
1
Sn−1 + 1
1
1
−
−
=
−
Un − 1 Un+1 − 1 Sn
(Un − 1)Sn Un+1 − 1
Un+1 Sn−1 − Un Sn + aSn
=
(Un − 1)(Un+1 − 1)Sn
2Un + (−1)n − 1
=
> 0.
a(Un − 1)(Un+1 − 1)Sn
Then, we obtain
∞
which completes the proof.
3
X 1
1
1
<
≤
Un k=n Sk
Un − 1
(n ≥ 2),
Acknowledgement
The author is grateful to the anonymous referee for his/her helpful comments.
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References
[1] S. H. Holliday and T. Komatsu, On the sum of reciprocal generalized Fibonacci numbers,
Integers 11A (2011).
[2] H. Ohtsuka and S. Nakamura, On the sum of reciprocal sums of Fibonacci numbers,
Fibonacci Quart. 46/47 (2008/2009), 153–159.
[3] R. Liu and F. Z. Zhao, On the sums of reciprocal hyperfibonacci numbers and hyperlucas
numbers, J. Integer Seq. 15 (2012), Article 12.4.5.
[4] Z. Wenpeng and W. Tingting, The infinite sum of reciprocal Pell numbers, Appl. Math.
Comput. 218 (2012), 6164–6167.
2010 Mathematics Subject Classification: Primary 11B37; Secondary 11B39.
Keywords: second-order recurrence, Fibonacci number, reciprocal sum.
(Concerned with sequences A000032, A000045, A000071, A000129, and A001610.)
Received April 26 2013; revised version received July 3 2013. Published in Journal of Integer
Sequences, July 30 2013.
Return to Journal of Integer Sequences home page.
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