215
Acta Math. Univ. Comenianae
Vol. LXXVI, 2(2007), pp. 215–222
RELATIONS BETWEEN THE RECIPROCAL SUM AND THE
ALTERNATING SUM FOR GENERALIZED LUCAS NUMBERS
XIAOLI YE and ZHIZHENG ZHANG
Abstract. We establish relations between reciprocal and alternating sums involving generalized Lucas numbers.
1. Introduction
Let p be a nonzero real number. The generalized Fibonacci and Lucas numbers
are defined by
(1)
U0 = 0,
U1 = 1,
Un+2 = pUn+1 + Un ,
(n ≥ 0)
(2)
V0 = 2,
V1 = p,
Vn+2 = pVn+1 + Vn ,
(n ≥ 0)
respectively. Let α and β be the roots of x2 − px − 1 = 0. Then we have the
Binet’s formulas:
(3)
Un
(4)
Vn
αn − β n
,
α−β
= αn + β n .
=
For p = 1, {Un } and {Vn } are the well-known Fibonacci numbers Fn and Lucas
numbers Ln , respectively.
In [2], S. Rabinowitz discusses algorithmic aspects of certain finite reciprocal
sums and propose several open problems involving Fibonacci and Lucas numbers.
Melham [1] gave several relations between the reciprocal sum and the alternating
sum for generalized Fibonacci numbers, which we subsequently extended in [4].
In this note we establish some relations between the reciprocal sum and the alternating sum for generalized Lucas numbers. For Un and Vn , we have the following
Received March 3, 2006.
2000 Mathematics Subject Classification. Primary 11B39, 40A99.
Key words and phrases. Generalized Lucas number, reciprocal sum, alternating sum.
This research is supported by the National Natural Science Foundation of China (Grant
No. 10471016), the Natural Science Foundation of Henan Province (Grant No. 0511010300)
and the Natural Science Foundation of the Education Department of Henan Province (Grant
No. 200510482001).
216
XIAOLI YE and ZHIZHENG ZHANG
well known expansions:
Un+1 =
(5)
[ n2 ] X
n−k
k=0
k
pn−2k ,
(n ≥ 0),
and
(6)
Vn =
[ n2 ]
X
k=0
n − k n−2k
n
p
,
n−k
k
(n ≥ 1).
See [3, (2.7),(2.8)]. From (5) and (6), since p is a nonzero real number, it is easy
to obtain that Un > 0 (n ≥ 1) and Vn > 0 (n ≥ 0) for p > 0. For p < 0, if n is
an odd number, then we have Un+1 < 0 (n ≥ 0) and Vn < 0 (n ≥ 1). If n is an
even number, then we have Un+1 > 0 (n ≥ 0) and Vn > 0 (n ≥ 0). Hence we have
Un 6= 0 (n ≥ 1) and Vn 6= 0 (n ≥ 0). Our main result is the following.
Theorem 1.1. Let m and k be positive integers. Put
Sk (m) =
(7)
∞
X
1
V
V
V
. . . Vn+mk
n=1 n n+k n+2k
and
Tk (m) =
(8)
∞
X
(−1)n−1
.
V V
V
. . . Vn+mk
n=1 n n+k n+2k
Then
Sk (m) =
(9)
1
(−1)(m−1)k
1+
− V(m+1)k
" k
(m−1)k
X (−1)
Vi+(m+1)k − Vi+2(m+1)k
·
Vi Vi+k Vi+2k . . . Vi+(m+1)k
i=1
+ (p2 + 4)U(m+1)k U(m+2)k Tk (m + 2)
i
and
( k
X
1
(−1)i−1
Tk (m) =
1 + (−1)(m−1)k − (−1)k V(m+1)k i=1 Vi Vi+k Vi+2k . . . Vi+(m+1)k
h
i
(10)
· (−1)(m−1)k Vi+(m+1)k − (−1)k Vi+2(m+1)k
o
+(p2 + 4)U(m+1)k U(m+2)k Sk (m + 2) .
When p = 1, by special choices of k and m, we have the following interesting
results:
∞
∞
X
X
1
7
(−1)n−1
=
− 10
;
L L
12
L L
L
L
n=1 n n+1
n=1 n n+1 n+2 n+3
RECIPROCAL SUM AND GENERALIZED LUCAS NUMBERS
∞
X
(−1)n−1
L L
n=1 n n+1
=
∞
X
1
1
+2
;
4
L
L
L
L
n=1 n n+1 n+2 n+3
=
∞
3
15 X
(−1)n−1
−
;
28
2 n=1 Ln Ln+1 Ln+2 Ln+3 Ln+4
(−1)n−1
L L
L
n=1 n n+1 n+2
=
∞
11
15 X
1
+
;
168
2 n=1 Ln Ln+1 Ln+2 Ln+3 Ln+4
1
L
L
L
L
n=1 n n+1 n+2 n+3
=
∞
X
13
(−1)n−1
− 15
;
924
L L
L
L
L
L
n=1 n n+1 n+2 n+3 n+4 n+5
=
∞
29
25 X
1
+
;
2772
3 n=1 Ln Ln+1 Ln+2 Ln+3 Ln+4 Ln+5
=
∞
X
(−1)n−1
139
− 24
;
396
L L
L
L
n=1 n n+2 n+4 n+6
∞
X
1
n=1
Ln Ln+1 Ln+2
∞
X
∞
X
∞
X
(−1)n−1
L L
L
L
n=1 n n+1 n+2 n+3
∞
X
1
L
L
n
n+2
n=1
217
∞
∞
X
X
(−1)n−1
1
95
− 242
=
;
∞
∞ Ln+4 Ln+6
L
L
396
L
L
X
X
n
n+2
n
n+2
1 n=1
(−1)n−1
123
199 n=1 105
=
+
−
;
L L
L
16 · 11 · 29 84 · 18 · 47
2 n=1 Ln Ln+2 Ln+4 Ln+6 Ln+8
n=1 n n+2 n+4
∞
1
(−1)n−1
199
105 X
123
−
−
=
.
L L
L
16 · 11 · 29 84 · 18 · 47
2 n=1 Ln Ln+2 Ln+4 Ln+6 Ln+8
n=1 n n+2 n+4
∞
X
2. The proof of the results
Lemma 2.1. We have
(11)
Vn+k + (−1)k Vn−k = Vn Vk ;
(12)
Vn+k = Vn Uk+1 + Vn−1 Uk ;
(13)
Vn+k−1 =
(14)
U2k = Uk Vk ;
(15)
Un+k = Un Uk+1 + Un−1 Uk ;
(16)
2
U(m+1)k
− Umk U(m+2)k = (−1)mk Uk2 ;
(17)
2
Vn+(m+1)k
− Vn+mk Vn+(m+2)k = −(−1)n+mk (p2 + 4)Uk2 ;
Uk−1 Vn+k + (−1)k−1 Vn
;
Uk
218
XIAOLI YE and ZHIZHENG ZHANG
(18)
Uk Vn−mk = (−1)mk (Vn Umk+k − Vn+k Umk );
(19)
U(m+2)k Vi+(m+1)k − (−1)k U(m+1)k Vi+mk = Uk Vi+2(m+1)k ;
(20)
Un+k − (−1)k Un−k = Vn Uk ;
Vn Vn+(m+1)k = −
(21)
+
(−1)(m+1)k
U(m+1)k Vn+(m+1)k Vn+(m+2)k
Uk
(−1)(m+1)k
U(m+2)k Vn+mk Vn+(m+2)k
Uk
− (−1)n+k (p2 + 4)Uk U(m+2)k ;
Vn+mk Vn+(m+2)k − Vn Vn+(m+1)k
=
(22)
(−1)(m+1)k
U(m+1)k Vn+(m+1)k Vn+(m+2)k
Uk
(−1)(m+1)k
+ 1−
U(m+2)k Vn+mk Vn+(m+2)k
Uk
+ (−1)n+k (p2 + 4)Uk U(m+2)k ;
Vn+mk Vn+(m+2)k + (−1)k−1 Vn Vn+(m+1)k
=
(23)
(−1)mk
U(m+1)k Vn+(m+1)k Vn+(m+2)k
Uk
(−1)mk
+ 1−
U(m+2)k Vn+mk Vn+(m+2)k
Uk
+ (−1)n (p2 + 4)Uk U(m+2)k .
Proof. Use the Binet formulas (3) and (4) of Un and Vn .
Lemma 2.2. We have
∞
X
1
V
V
.
.
.
V
Vn+(m+1)k
n
n+k
n+(m−1)k
n=1
(24)
=
k
Umk + (−1)mk Uk
(−1)mk Uk X
1
Sk (m) −
U(m+1)k
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
219
RECIPROCAL SUM AND GENERALIZED LUCAS NUMBERS
and
∞
X
(−1)n−1
V V
. . . Vn+(m−1)k Vn+(m+1)k
n=1 n n+k
(25)
=
k
(−1)(m+1)k Uk X
(−1)i−1
Umk + (−1)(m+1)k Uk
Tk (m) −
.
U(m+1)k
U(m+1)k
VV
. . . Vi+mk
i=1 i i+k
Proof. We give only the proof of (24), the proof of (25) is similar. Using (11),
we have
∞
X
1
V
V
.
.
.
V
n+(m−1)k Vn+(m+1)k
n=1 n n+k
=
∞
X
Vn+(m+1)k + (−1)k Vn+(m−1)k
Vn Vn+k . . . Vn+(m−1)k Vn+(m+1)k
n=1
=
∞ X
n=1
=
!
1
Vk Vn+mk
(−1)k
1
+
Vn Vn+k . . . Vn+(m−1)k
Vn Vn+k . . . Vn+(m−2)k Vn+(m+1)k
1
Vk Vn+mk
∞
1
(−1)k X
1
.
Sk (m) +
Vk
Vk n=1 Vn Vn+k . . . Vn+(m−2)k Vn+mk Vn+(m+1)k
Using (12), (13) and (15), we have
∞
X
1
V
V
.
.
.
V
Vn+mk Vn+(m+1)k
n
n+k
n+(m−2)k
n=1
=
=
∞
X
Vn+(m+1)k + (−1)k UU2k
Vn+(m−2)k
k
n=1
Vn Vn+k . . . Vn+(m−2)k Vn+mk Vn+(m+1)k
∞ X
n=1
Uk
U3k Vn+(m−1)k
1
Vn Vn+k . . . Vn+(m−2)k Vn+mk
+
=
!
(−1)k UU2k
k
Vn Vn+k . . . Vn+(m−3)k Vn+mk Vn+(m+1)k
!
·
Uk
U3k Vn+(m−1)k
Uk
Sk (m)
U3k
+
∞
(−1)k U2k X
1
.
U3k
V
V
.
.
.
V
V
n
n+k
n+(m−3)k n+(m−1)k Vn+mk Vn+(m+1)k
n=1
220
XIAOLI YE and ZHIZHENG ZHANG
Similarly, we obtain
∞
X
1
V V
. . . Vn+(m+1)k
n=1 n n+2k
∞
(−1)k Umk X
1
Sk (m) +
=
U(m+1)k
U(m+1)k n=1 Vn+k . . . Vn+(m+1)k
Uk
=
Uk
U(m+1)k
Sk (m) +
k
(−1)k Umk
(−1)k Umk X
1
Sk (m) −
.
U(m+1)k
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
Hence, using (16) and (17), we have
∞
X
1
V
V
.
.
.
V
n
n+k
n+(m−1)k Vn+(m+1)k
n=1
1
(−1)k Uk
(−1)2k Uk U2k
(−1)mk Uk U2k
=
1+
+
+ ... +
Vk
U3k
U3k U4k
Umk U(m+1)k
k
1
(−1)mk U2k
(−1)mk U2k X
+
Sk (m) −
U(m+1)k
Vk U(m+1)k i=1 Vi Vi+k . . . Vi+mk
U(m−1)k
Uk
1
Uk U2k
U2k
U2k
U3k
Umk
−
=
1+
+
−
+
− ... −
+
Vk
Uk2
U2k
U3k
U3k
U4k
Umk
U(m+1)k
k
1
(−1)mk U2k
(−1)mk Uk X
+
· Sk (m) −
U(m+1)k
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
(−1)mk U2k
1
Umk U2k
+
Sk (m)
=
1−1+
Vk
Uk U(m+1)k
U(m+1)k
−
k
(−1)mk Uk X
1
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
k
1
Umk + (−1)mk Uk
(−1)mk Uk X
=
Sk (m) −
.
U(m+1)k
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
The proof of the theorem: We give only the proof of (9), the proof of (10) is
similar. Since
∞ X
1
1
−
Vn Vn+k . . . Vn+(m−1)k Vn+(m+1)k
Vn+k Vn+2k . . . Vn+mk Vn+(m+2)k
n=1
(26)
=
k
X
i=1
1
,
Vi Vi+k . . . Vi+(m−1)k Vi+(m+1)k
RECIPROCAL SUM AND GENERALIZED LUCAS NUMBERS
221
using (22) and Lemma 2.1, then the left side of (26) can be written as
∞
X
Vn+mk Vn+(m+2)k − Vn Vn+(m+1)k
n=1
=
=
=
Vn Vn+k . . . Vn+(m+1)k Vn+(m+2)k
(−1)(m+1)k
U(m+1)k Sk (m) − (−1)k (p2 + 4)Uk U(m+2)k Tk (m + 2)
Uk
X
∞
(−1)(m+1)k
1
+ 1−
U(m+2)k
Uk
V
V
.
.
.
V
n+(m−1)k Vn+(m+1)k
n=1 n n+k
(−1)(m+1)k
U(m+1)k Sk (m) − (−1)k (p2 + 4)Uk U(m+2)k Tk (m + 2)
Uk
Umk +(−1)mk Uk
(−1)(m+1)k
Sk (m)
U(m+2)k
+ 1−
Uk
U(m+1)k
#
k
1
(−1)mk Uk X
−
U(m+1)k i=1 Vi Vi+k . . . Vi+mk
(−1)k +(−1)mk −(−1)k V(m+1)k
Uk Sk (m)−(−1)k (p2 +4)Uk U(m+2)k Tk (m + 2)
U(m+1)k
(27)
k
(−1)mk Uk − (−1)k U(m+2)k X
1
.
U(m+1)k
V
V
.
. . Vi+mk
i
i+k
i=1
−
Considering the right sides of (26) and (27), we have
(−1)k + (−1)mk − (−1)k V(m+1)k
Uk Sk (m)
U(m+1)k
=
k X
(−1)mk Uk − (−1)k U(m+2)k Vi+(m+1)k + U(m+1)k Vi+mk
U(m+1)k Vi Vi+k . . . Vi+(m+1)k
i=1
+(−1)k (p2 + 4)Uk U(m+2)k Tk (m + 2)
=
k
X
(−1)mk Vi+(m+1)k − (−1)k Vi+2(m+1)k
i=1
U(m+1)k Vi Vi+k . . . Vi+(m+1)k
Uk + (−1)k (p2 + 4)Uk U(m+2)k Tk (m + 2).
Hence
Sk (m)
=
1
1 + (−1)(m−1)k − V(m+1)k
"
k
X
(−1)(m−1)k Vi+(m+1)k − Vi+2(m+1)k
i=1
Vi Vi+k Vi+2k . . . Vi+(m+1)k
i
+(p2 + 4)U(m+1)k U(m+2)k Tk (m + 2) .
222
XIAOLI YE and ZHIZHENG ZHANG
The proof of (9) is completed.
Acknowledgment. The authors would like to thank the anonymous referee
for his/her valuable suggestions that have significantly improved the presentation
of this paper.
References
1. Melham R. S., On some reciprocal sums of Brousseau: an alternative approach to that of
Carlitz, The Fibonacci Quarterly, 41(1) (2003), 59–62.
2. Rabinowitz S., Algorithmic summation of reciprocals of products of Fibonacci numbers, The
Fibonacci Quarterly, 37(2) (1999), 122–127.
3. Sun Z. H., Expansions and identities concerning Lucas sequences, The Fibonacci Quarterly,
44(2) (2006), 145–153.
4. Ye X. L. and Zhang Z. Z., On some formulas of the reciprocal sum and the alternating sum
for generalized Fibonacci numbers, Advanced Studies in Contemporary Mathematics, 10(2)
(2005), 143–148.
Xiaoli Ye, Department of Mathematics, Luoyang Teachers’ College, Luoyang 471022, P. R. China
Zhizheng Zhang, Department of Mathematics, Luoyang Teachers’ College, Luoyang 471022, P.
R. China, e-mail: zhzhzhang-yang@163.com