International Journal of Mathematical Analysis
Vol. 10, 2016, no. 28, 1365 - 1373
HIKARI Ltd, www.m-hikari.com
https://doi.org/10.12988/ijma.2016.610118
On the Reciprocal Sums of Generalized
Fibonacci Numbers
Younseok Choo
Department of Electronic and Electrical Engineering
Hongik University
2639 Sejong-Ro, Sejong, 30016, Korea
c 2016 Younseok Choo. This article is distributed under the Creative ComCopyright mons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
Abstract
In this paper we obtain some identities for the infinite sum of the
reciprocal generalized Fibonacci numbers and the infinite sum of the
square of the reciprocal generalized Fibonacci numbers.
Mathematics Subject Classification: 11B39, 11B37
Keywords: Generalized Fibonacci numbers; floor function; infinite sum
1
Introduction
Throughout this paper we use the notation {Gn }∞
n=0 = S(G0 , G1 , x, y) to denote the generalized Fibonacci numbers {Gn }∞
generated
from the recurrence
n=0
relation
Gn = xGn−1 + yGn−2 , n ≥ 2,
with initial conditions G0 and G1 , where G0 is a nonnegative integer and
G1 , x and y are positive integers. Hence {Gn }∞
n=0 = S(0, 1, 1, 1) is the Fibonacci numbers, {Gn }∞
=
S(2,
1,
1,
1)
is
the
Lucas
numbers, and {Gn }∞
n=0
n=0 =
S(0, 1, 2, 1) is the Pell numbers etc.
Recently Ohtsuka and Nakamura [1] reported some interesting properties of
Fibonacci numbers. They proved that, for the Fibonacci numbers {Gn }∞
n=0 =
1366
Younseok Choo
S(0, 1, 1, 1), the following identities hold:
∞
X
$
1
k=n Gk
!−1 %
Gn − Gn−1 , if n is even and n ≥ 2;
Gn − Gn−1 − 1, if n is odd and n ≥ 1,
=
∞
X
$
(
1
2
k=n Gk
!−1 %
(
=
Gn−1 Gn − 1, if n is even and n ≥ 2;
Gn−1 Gn , if n is odd and n ≥ 1,
(1)
(2)
where b·c denotes the floor function.
Holliday and Komatsu [2] extended the above results as follows (see [3]
also): For {Gn }∞
n=0 = S(0, 1, x, 1),
$
∞
X
!−1 %
∞
X
!−1 %
1
k=n Gk
$
(
=
1
2
k=n Gk
(
=
Gn − Gn−1 , if n is even and n ≥ 2;
Gn − Gn−1 − 1, if n is odd and n ≥ 1,
(3)
xGn−1 Gn − 1, if n is even and n ≥ 2;
xGn−1 Gn , if n is odd and n ≥ 1.
(4)
Using a different approach, Zhang and Wang [4] independently proved (3)
for the Pell numbers {Gn }∞
n=0 = S(0, 1, 2, 1). Holliday and Komatsu [2] also
∞
showed that, for {Gn }n=0 = S(c, 1, 1, 1) (c ≥ 1), there exist positive integers
l0 , l1 , l2 and l3 such that
∞
X
$
$
1
k=n Gk
∞
X
1
2
k=n Gk
!−1 %
(
Gn − Gn−1 − 1, if n is even and n ≥ l0 ;
Gn − Gn−1 , if n is odd and n ≥ l1 ,
(5)
Gn−1 Gn + g(c) − 1, if n is even and n ≥ l2 ;
Gn−1 Gn − g(c), if n is odd and n ≥ l3 ,
(6)
=
!−1 %
(
=
where
( c(c+1)
, if c ≡ 0, 2 (mod 3);
if c ≡ 1 (mod 3).
3
c(c+1)+1
,
3
g(c) =
On the other hand, Wu and Zhang [3] considered the sums of the reciprocals
of Fibonacci polynomials and Lucas polynomials. In particular, they proved
that, for the Lucas polynomial {Gn }∞
n=0 = S(2, x, x, 1),
$
∞
X
1
k=n Gk
!−1 %
(
=
Gn − Gn−1 − 1, if n is even and n ≥ 2;
Gn − Gn−1 , if n is odd and n ≥ 3,
(7)
xGn−1 Gn + x2 + 1, if n is even and n ≥ 2;
xGn−1 Gn − x2 − 2, if n is odd and n ≥ 3.
(8)
where x ≥ 1, and
$
∞
X
1
2
k=n Gk
!−1 %
(
=
Reciprocal sums of generalized Fibonacci numbers
1367
where x ≥ 2.
For other interesting results, see [5], [6], [7], [8], [9] and references cited
therein.
In this paper we present unified results in the same direction for a generalized Fibonacci numbers {Gn }∞
n=0 = S(c, d, x, 1). The results obtained here are
quite general and include (1)–(8) as special cases.
2
Main Results
The following lemma [2] below will be used to prove Theorem 2.1 and Theorem
2.2.
Lemma 2.1 For {Gn }∞
n=0 = S(c, d, x, 1), we have
G2n − Gn−1 Gn+1 = (−1)n (c2 + cdx − d2 ).
Theorem 2.1 For the generalized Fibonacci numbers {Gn }∞
n=0 = S(c, d, x, 1),
there exist positive integers n0 , n1 , n2 and n3 such that (a) and (b) below hold:
(a) If c2 + cdx − d2 > 0, then
$
∞
X
1
k=n Gk
!−1 %
(
=
Gn − Gn−1 − 1, if n is even and n ≥ n0 ;
Gn − Gn−1 , if n is odd and n ≥ n1 .
(9)
(b) If c2 + cdx − d2 < 0, then
$
∞
X
1
k=n Gk
!−1 %
(
=
Gn − Gn−1 , if n is even and n ≥ n2 ;
Gn − Gn−1 − 1, if n is odd and n ≥ n3 .
(10)
Proof. Firstly we consider the case where c2 + cdx − d2 > 0. Using Lemma
2.1, we have
1
1
1
1
−
−
−
Gn − Gn−1 Gn Gn+1 Gn+2 − Gn+1
=
X1
,
(Gn − Gn−1 )Gn Gn+1 (Gn+2 − Gn+1 )
where
X1 = Gn+2 (Gn−1 Gn+1 − G2n ) + Gn−1 (Gn Gn+2 − G2n+1 )
= (−1)n−1 Gn+2 (c2 + cdx − d2 ) + (−1)n Gn−1 (c2 + cdx − d2 )
= (−1)n−1 (c2 + cdx − d2 )(Gn+2 − Gn−1 ).
1368
Younseok Choo
If n is even, then X1 < 0, and
1
1
1
1
−
<
+
.
Gn − Gn−1 Gn+2 − Gn+1
Gn Gn+1
Repeatedly applying the above inequality, we have
∞
X
1
1
<
, if n is even.
Gn − Gn−1 k=n Gk
(11)
Similarly we obtain
∞
X
1
1
<
, if n is odd.
Gn − Gn−1
k=n Gk
(12)
On the other hand, again using Lemma 2.1, we obtain
1
1
1
1
−
−
−
Gn − Gn−1 − 1 Gn Gn+1 Gn+2 − Gn+1 − 1
=
X2
,
(Gn − Gn−1 − 1)Gn Gn+1 (Gn+2 − Gn+1 − 1)
where
X2 = G2n − Gn−1 Gn+1 − Gn−1 (G2n+1 − Gn Gn+2 ) − (G2n+1 − Gn Gn+2 )
−Gn+2 (G2n − Gn−1 Gn+1 ) + Gn+1 Gn+2 − Gn−1 Gn − Gn − Gn+1
= (−1)n−1 (c2 + cdx − d2 )(Gn+2 − Gn−1 − 2) + Gn+1 Gn+2 − Gn−1 Gn
−Gn − Gn+1 .
If n is even, then there exists a positive integer n0 such that X2 > 0 for n ≥ n0
and
1
1
1
1
+
<
−
.
Gn Gn+1
Gn − Gn−1 − 1 Gn+2 − Gn+1 − 1
Repeatedly applying the above inequality, we have
∞
X
1
1
<
, if n is even and n ≥ n0 .
Gn − Gn−1 − 1
k=n Gk
Now, by Lemma 2.1,
1
1
1
1
−
−
−
Gn − Gn−1 + 1 Gn Gn+1 Gn+2 − Gn+1 + 1
=
X3
,
(Gn − Gn−1 + 1)Gn Gn+1 (Gn+2 − Gn+1 + 1)
(13)
Reciprocal sums of generalized Fibonacci numbers
1369
where
X3 = Gn−1 Gn+1 − G2n + Gn−1 (Gn Gn+2 − G2n+1 ) + G2n+1 − Gn Gn+2
+Gn+2 (Gn−1 Gn+1 − G2n ) + Gn−1 Gn − Gn+1 Gn+2 − Gn − Gn+1
= (−1)n−1 (c2 + cdx − d2 )(Gn+2 − Gn−1 + 2) + Gn−1 Gn − Gn+1 Gn+2
−Gn − Gn+1 .
If n is odd, then there exists a positive integer n1 such that X3 < 0 for n ≥ n1
and
1
1
1
1
−
<
+
.
Gn − Gn−1 + 1 Gn+2 − Gn+1 + 1
Gn Gn+1
Repeatedly applying the above inequality, we have
∞
X
1
1
<
, if n is odd and n ≥ n1 .
Gn − Gn−1 + 1 k=n Gk
(14)
Then, (9) follows from (11), (12), (13) and (14). The proof of (10) can be
proceeded similarly, and is omitted.
Remark 2.1 For {Gn }∞
n=0 = S(0, 1, x, 1), it is easily seen that X2 ≥ 0 for n
odd and X3 < 0 for n even. Then, (10) reduces to (3).
Remark 2.2 For {Gn }∞
n=0 = S(c, 1, 1, 1) (c ≥ 1), we have that and X2 > 0 if
n is odd and large, and X3 < 0 if n is odd and large. Then, (9) reduces to (5).
Remark 2.3 Consider {Gn }∞
n=0 = S(2, x, x, 1). If n is even and n ≥ 2, then
X2 =
=
≥
>
Gn+1 Gn+2 − Gn−1 Gn − Gn − Gn+1 − (x2 + 4)(Gn+2 − Gn−1 − 2)
Gn+1 (xGn+1 − x3 − 4x − 1) + Gn (xGn − x2 − 5) + (x2 + 4)(Gn−1 + 2)
x4 + x3 + 4x2 + 2x + 2
0.
Similarly, if n is odd and n ≥ 3, then
X3 =
=
≤
<
(x2 + 4)(Gn+2 − Gn−1 + 2) + Gn−1 Gn − Gn+1 Gn+2 − Gn − Gn+1
Gn+1 (x3 + 4x − 1 − xGn+1 ) + Gn (x2 + 3 − xGn ) − (x2 + 4)(Gn−1 − 2)
−x5 − 2x4 − 3x3 − 6x2 + 2x + 2
0.
Consequently, (7) is a special case of (9).
Theorem 2.2 For the generalized Fibonacci numbers {Gn }∞
n=0 = S(c, d, x, 1),
(a) and (b) below hold:
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Younseok Choo
(a) If x2 (c2 + cdx − d2 ) is not an integer multiple of x2 + 2, define
g := g(c, d, x) = c2 + cdx − d2 −
h 2(c2
+ cdx − d2 ) i
,
x2 + 2
(15)
where [·] denotes the integer part. Then there exist positive integers n4 , n5 , n6
and n7 such that
(i) if c2 + cdx − d2 > 0, then
$
∞
X
1
2
k=n Gk
!−1 %
(
xGn−1 Gn + g − 1, if n is even and n ≥ n4 ;
xGn−1 Gn − g, if n is odd and n ≥ n5 ,
=
(16)
(ii) if c2 + cdx − d2 < 0, then
$
∞
X
1
2
k=n Gk
!−1 %
(
=
xGn−1 Gn + g, if n is even and n ≥ n6 ;
xGn−1 Gn − g − 1, if n is odd and n ≥ n7 .
(17)
(b) If x2 (c2 + cdx − d2 ) is an integer multiple of x2 + 2, define
g := g(c, d, x) =
x2 (c2 + cdx − d2 )
.
x2 + 2
(18)
Then there exist positive integers n8 and n9 such that
$
∞
X
1
2
k=n Gk
!−1 %
(
=
xGn−1 Gn + g, if n is even and n ≥ n8 ;
xGn−1 Gn − g, if n is odd and n ≥ n9 .
(19)
Proof. Firstly we consider the case where x2 (c2 + cdx − d2 ) is not an integer
multiple of x2 + 2 and c2 + cdx − d2 > 0. In this case, (15) implies
x2 (c2 + cdx − d2 ) − g(x2 + 2) < 0.
By Lemma 2.1, we have
Y1 =
1
1
1
−
− 2
n
n+1
xGn−1 Gn + (−1) g − 1 xGn Gn+1 + (−1) g − 1 Gn
x2 G2n + 2(−1)n+1 g
1
− 2
n
n+1
(xGn−1 Gn + (−1) g − 1)(xGn Gn+1 + (−1) g − 1) Gn
Z1
=
,
n
(xGn−1 Gn + (−1) g − 1)(xGn Gn+1 + (−1)n+1 g − 1)G2n
=
where
Z1 = x2 G2n (G2n − Gn−1 Gn+1 ) + xGn (Gn−1 + Gn+1 )
+2(−1)n+1 gG2n + (−1)n+1 gxGn (Gn+1 − Gn−1 ) + g 2 − 1
n
o
= (−1)n G2n x2 (c2 + cdx − d2 ) − g(x2 + 2) + xGn (Gn−1 + Gn+1 ) + g 2 − 1.
1371
Reciprocal sums of generalized Fibonacci numbers
There exists a positive integer n̂4 such that Y1 > 0 if n is even and n ≥ n̂4 and
1
1
1
−
.
<
2
n
Gn
xGn−1 Gn + (−1) g − 1 xGn Gn+1 + (−1)n+1 g − 1
Repeatedly applying the above inequality, we have
∞
X
1
1
<
, if n is even and n ≥ n̂4 .
2
xGn−1 Gn + (−1)n g − 1
k=n Gk
(20)
Similarly, using Lemma 2.1,
Y2 =
1
1
1
−
−
xGn−1 Gn + (−1)n g + 1 xGn Gn+1 + (−1)n+1 g + 1 G2n
1
x2 G2n + 2(−1)n+1 g
− 2
n
n+1
(xGn−1 Gn + (−1) g + 1)(xGn Gn+1 + (−1) g + 1) Gn
Z2
=
,
n
(xGn−1 Gn + (−1) g + 1)(xGn Gn+1 + (−1)n+1 g + 1)G2n
=
where
Z2 = x2 G2n (G2n − Gn−1 Gn+1 ) − xGn (Gn−1 + Gn+1 )
+2(−1)n+1 gG2n − (−1)n gxGn (Gn+1 − Gn−1 ) + g 2 − 1
n
o
= (−1)n G2n x2 (c2 + cdx − d2 ) − g(x2 + 2) − xGn (Gn−1 + Gn+1 ) + g 2 − 1.
There exists a positive integer n̂5 such that Y2 < 0 if n is odd and n ≥ n̂5 and
1
1
1
−
< 2.
n
n+1
xGn−1 Gn + (−1) g + 1 xGn Gn+1 + (−1) g + 1
Gn
Repeatedly applying the above inequality, we have
∞
X
1
1
<
, if n is odd and n ≥ n5 .
xGn−1 Gn + (−1)n g + 1 k=n G2k
On the other hand,
Y3 =
1
1
1
1
−
− 2 − 2
n
n
xGn−1 Gn + (−1) g xGn+1 Gn+2 + (−1) g Gn Gn+1
=
(G2n + G2n+1 )Z3
,
(xGn−1 Gn + (−1)n g)(xGn Gn+1 + (−1)n+1 g)G2n G2n+1
(21)
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Younseok Choo
where
Z3 = (−1)n Gn Gn+1 x(x2 (c2 + cdx − d2 ) − g(x2 + 2)) + gx2 (c2 + cdx − d2 ) − g 2 .
There exists a positive integer ñ4 such that Y3 < 0 if n is even and n ≥ ñ4 and
1
1
1
1
−
< 2 + 2 .
n
n+1
xGn−1 Gn + (−1) g xGn+1 Gn+2 + (−1) g
Gn Gn+1
Repeatedly applying the above inequality, we have
∞
X
1
1
<
, if n is even and n ≥ ñ4 .
xGn−1 Gn + g k=n G2k
(22)
Similarly there exists a positive integer ñ5 such that
∞
X
1
1
, if n is odd and n ≥ ñ5 .
<
2
xGn−1 Gn − g
k=n Gk
(23)
Then (16) follows from (20), (21), (22) and (23). The proof of (17) can be
proceeded similarly, and is omitted.
Now suppose that x2 (c2 + cdx − d2 ) is an integer multiple of x2 + 2. Then
there exist positive integers n̂8 and n̂9 such that Y2 < 0 if n is even and n ≥ n̂8 ,
or if n is odd and n ≥ n̂9 . Then
∞
X
1
1
<
, if n ≥ n̂8 is even or if n ≥ n̂9 is odd.
n
xGn−1 Gn + (−1) g + 1 k=n G2k
(24)
Similarly, there exist positive integers ñ8 and ñ9 such that
∞
X
1
1
<
, if n ≥ ñ8 is even or if n ≥ ñ9 is odd.
2
xGn−1 Gn + (−1)n g
k=n Gk
(25)
Then, (19) follows from (24), (25), and the proof is completed.
Remark 2.4 For {Gn }∞
n=0 = S(0, 1, x, 1), g = −1 and (17) reduces to (4).
Remark 2.5 For {Gn }∞
n=0 = S(c, 1, 1, 1) (c ≥ 1), g in (15) is given by
g = c2 + c − 1 −
h 2(c2
+ c − 1) i
,
3
which is equivalent to
( c(c+1)
g=
, if c ≡ 0, 2 (mod 3);
if c ≡ 1 (mod 3).
3
c(c+1)+1
,
3
Reciprocal sums of generalized Fibonacci numbers
1373
Hence (6) is a special case of (16).
Remark 2.6 Consider {Gn }∞
n=0 = S(2, x, x, 1). Then, for x ≥ 2,
g = x2 + 4 −
h 2x2
+ 8i
x2 + 2
= x2 + 2,
and (16) reduces to (8).
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Received: November 9, 2016; Published: December 17, 2016