International Journal of Contemporary Mathematical Sciences
Vol. 10, 2015, no. 8, 363 - 379
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ijcms.2015.5942
On Infinite Series Involving Fibonacci Numbers
Robert Frontczak1
Landesbank Baden-Wuerttemberg (LBBW)
Am Hauptbahnhof 2, 70173 Stuttgart, Germany
c 2015 Robert Frontczak. 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
The aim of the paper is to derive closed-form expressions for some
new types of infinite series involving Fibonacci numbers. For each type
numerous examples are presented. Some classical results are rediscovered as particular cases of the more general identities presented here.
Mathematics Subject Classification: 11B39, 11Y60
Keywords: Fibonacci number, Inverse tangent function, Infinite series
1
Introduction
The Fibonacci sequence (or Fibonacci numbers) is one of the most popular
and fascinating linear sequences in mathematics. It is defined recursively as
F0 = 0, F1 = 1, and Fn+2 = Fn+1 + Fn for n ≥ 0. Since its introduction
by Leonardo of Pisa at the beginning of the thirteenth century the sequence
attracted the interest of numerous mathematicians. As a result, hundreds of
formulas and identities involving Fibonacci numbers were developed (see [3]
and [5] for an excellent introduction).
The Binet form for the Fibonacci sequence is
Fn =
1
αn − β n
,
α−β
n ≥ 0,
(1)
Disclaimer: Statements and conclusions made in this article are entirely those of the
author. They do not necessarily reflect the views of LBBW.
364
Robert Frontczak
where α and β are roots of the quadratic equation x2 − x − 1 = 0. We have
√
√
1− 5
1+ 5
,
β=
.
(2)
α=
2
2
It follows that α2 = α + 1, β 2 = β + 1, αβ = −1, α + β = 1 and α2 + β 2 = 3.
The study of infinite series involving Fibonacci numbers starts with the following inverse tangent (i.e. arctangent) identity which was discovered in 1936
by Lehmer (Lehmer’s identity, see [3]):
∞
X
tan−1
n=1
1
F2n+1
=
π
.
4
(3)
Motivated by this striking result, many extensions and generalizations were
obtained in [4], [6], [7] and [8]. For instance, in the very recent paper [8] the
above inverse tangent formula was extended to
∞
X
n=1
−1
tan
1
F2n+2k−1
−1
= tan
1 ,
F2k
k ≥ 1.
(4)
On the other hand a well-known summation technique for infinite series is
employed by Brousseau in [2] to derive closed-form evaluations for a range of
infinite series involving Fibonacci numbers. As a classical example he proves
that
∞
X
Fn
= 1.
(5)
F F
n=1 n+1 n+2
The above inverse tangent evaluations are based on a telescoping property of
the inverse tangent function. The required trigonometric formula is
x−y tan−1 (x) − tan−1 (y) = tan−1
,
xy > −1.
(6)
1 + xy
However, in many situations the telescoping structure is hidden. In this paper
extending a theorem found in [1] new closed-form evaluations of infinite series
involving Fibonacci numbers are derived by elementary methods.
2
Main Result
Let tan−1 (x) denote the principal value of the inverse tangent function. Our
main result is the following statement:
Theorem 2.1 Let g(x) be a real function of one variable. Let h(x) be of
fixed sign and composite, h(x) = h(g(x)). Define H(x) by
H(x) =
h(g(x)) − h(g(x + 1))
.
1 + h(g(x))h(g(x + 1))
(7)
365
On infinite series involving Fibonacci numbers
Then we have
k
X
tan−1 H(n) = tan−1 h(g(1)) − tan−1 h(g(k + 1)),
(8)
tan−1 H(n) = tan−1 h(g(1)) − tan−1 ( lim h(g(k + 1))).
(9)
n=1
and
∞
X
k→∞
n=1
PROOF: By (6) we have that
k
X
tan−1 H(n) =
n=1
k X
tan−1 h(g(n)) − tan−1 h(g(n + 1)) .
n=1
2
The statements follow from telescoping.
Remark 2.2 If g(x) = x then the Theorem reduces to Theorem 2.1 in [1].
Also, as pointed out in [1] the condition on h(x) to be of fixed sign is included
in order to avoid a jump in Eq. (6) by π multiplied by the sign of x when
xy < −1.
In the following we are going to present several applications of the previous
theorem. The use the term ”type” for a specific form of the function h(x).
To provide a compact treatment, the presentation will be restricted to infinite
series only. In all applications g(n), n ≥ 1, will correspond to the Fibonacci
sequence Fn , a subsequence of it or a sequence closely related to Fn .
3
3.1
Applications
First type: h(x) = a/x.
Corollary 3.1 For a 6= 0 let h(x) =
Then by Theorem 2.1
∞
X
−1
tan
a
x
and g(n) = Fmn+k , m ≥ 1, k ≥ 0.
a − Fmn+k ) −1
= tan
.
a2 + Fmn+k Fmn+m+k
Fm+k
a(F
n=1
mn+m+k
(10)
Especially, for each m and k
∞
X
n=1
−1
tan
m+k (Fmn+m+k − Fmn+k )
2
Fm+k
+ Fmn+k Fmn+m+k
F
=
π
.
4
(11)
366
Robert Frontczak
The case m = 1 and k = 0 gives
∞
X
−1
tan
n=1
aFn
= tan−1 (a).
2
a + Fn+1 Fn+2
(12)
√
√
√
From (12) inserting a = 1, a = 3/3, a = 3, a = 2 − 3 and a = α,
respectively, we obtain the infinite series evaluations
∞
X
tan−1
n=1
∞
X
π
Fn
= ,
1 + Fn+1 Fn+2
4
√
π
3Fn
= ,
1 + 3Fn+1 Fn+2
6
n=1
√
∞
π
X
3Fn
−1
tan
= ,
3
+
F
F
3
n+1
n+2
n=1
√
∞
X
π
(2 − 3)Fn
−1
√
tan
= ,
12
7 − 4 3 + Fn+1 Fn+2
n=1
and
∞
X
(13)
tan−1
tan−1
n=1
αFn
= tan−1 (α).
α + 1 + Fn+1 Fn+2
Also, observe that we have the ”symmetric” relation
√
√
∞
∞
X
X
3Fn
3Fn
−1
−1
tan
=2
tan
.
3 + Fn+1 Fn+2
1 + 3Fn+1 Fn+2
n=1
n=1
(14)
(15)
(16)
(17)
(18)
The case m = 2 and k = 0 in Eq. (10) gives
∞
X
tan−1
n=1
− F2n ) = tan−1 (a).
a2 + F2n F2n+2
a(F
2n+2
(19)
Using Cassini’s identity
Fn+1 Fn−1 = (−1)n + Fn2 ,
n≥1
and the relation
F2n = Fn2 + 2Fn−1 Fn ,
n ≥ 1,
(see [3]) the above expression may be restated as
∞
X
n=1
tan−1
a(F 2 + F 2 ) n+1
n
= tan−1 (a).
2
a2 − 1 + F2n+1
(20)
367
On infinite series involving Fibonacci numbers
For a = 1 we get a series for π/4 involving squared Fibonacci numbers:
∞
X
tan−1
F 2
n=1
2
n+1 + Fn
2
F2n+1
=
π
.
4
(21)
In a similar manner the case m = 2 and k = 1 gives
∞
X
tan−1
n=1
a(F 2 − F 2 ) a
n+2
n
= tan−1 ( ),
2
2
a + 1 + F2n+2
2
(22)
where we have used the relation (see [3])
2
F2n+1 = Fn+1
+ Fn2 ,
n ≥ 0.
The case m = k gives
∞
X
tan−1
a − Fm(n+1) ) −1
=
tan
.
a2 + Fm(n+1) Fm(n+2)
F2m
a(F
n=1
m(n+2)
(23)
From Corollary 3.1 the following summation identity is obtained straightforwardly by differentiation:
Corollary 3.2 For a 6= 0, m ≥ 1, k ≥ 0 it holds that
∞
X
n=1
(Fmn+k Fmn+m+k − a2 )(Fmn+m+k − Fmn+k )
Fm+k
= 2
.
2
2
2
2
(a + Fmn+k Fmn+m+k ) + a (Fmn+m+k − Fmn+k )
Fm+k + a2
(24)
Special cases are the next expressions. The case m = 1 and k = 0 gives
∞
X
n=1
1
Fn Fn+1 Fn+2 − a2 Fn
=
.
2
2
2
2
(a + Fn+1 Fn+2 ) + a Fn
1 + a2
(25)
Noticing that Eq. (25) (as well as (24)) are also well defined for a = 0, we
immediately get Brousseau’s series (5) as a special case of (25). Further explicit
evaluations of (25) are the identities
∞
X
n=1
4Fn Fn+1 Fn+2 − Fn
1
= ,
2
2
(1 + 4Fn+1 Fn+2 ) + 4Fn
5
∞
X
Fn Fn+1 Fn+2 − Fn
1
= ,
2
2
(1 + Fn+1 Fn+2 ) + Fn
2
(27)
Fn Fn+1 Fn+2 − α2 Fn
1
=
,
2
2
2
2
(α + Fn+1 Fn+2 ) + α Fn
α+2
(28)
n=1
∞
X
n=1
(26)
368
Robert Frontczak
as well as
∞
X
n=1
Fn Fn+1 Fn+2 − α4 Fn
1
.
=
4
2
4
2
(α + Fn+1 Fn+2 ) + α Fn
3(α + 1)
(29)
The case m = 2 and k = 0 gives the summation formula
∞
X
n=1
2
2
+ Fn2 )
− 1 − a2 )(Fn+1
(F2n+1
1
=
,
2
2
+ Fn2 )2
)2 + a2 (Fn+1
(a2 − 1 + F2n+1
1 + a2
from which
∞
2
2
X
(F2n+1
− 2)(Fn+1
+ Fn2 )
n=1
and
∞
X
n=1
4
F2n+1
+
2
(Fn+1
+
Fn2 )2
1
= ,
2
2
2
+ Fn2 )
− 3)(Fn+1
(F2n+1
1
=
,
2
2
(1 + F2n+1 )2 + 2(Fn+1 + Fn2 )2
3
(30)
(31)
(32)
are two special cases. Finally, the pair m = 2 and k = 1 produces the identity
∞
X
n=1
3.2
2
2
− Fn2 )
+ 1 − a2 )(Fn+2
(F2n+2
2
=
.
2
2
(a2 + 1 + F2n+2 )2 + a2 (Fn+2 − Fn2 )2
4 + a2
(33)
Second type: h(x) = ax + b.
Corollary 3.3 For a > 0 let h(x) = ax + b. We will choose b such that
h(g(n)) ≥ 0 for n ≥ 0. Let g(n) = Fmn+k , m ≥ 1, k ≥ 0. Then by Theorem 2.1
∞
X
tan−1
n=1
π
a(Fmn+m+k − Fmn+k )
= −tan−1 (aFm+k +b).
a2 Fmn+k Fmn+m+k + ab(Fmn+m+k + Fmn+k ) + b2 + 1
2
(34)
The case m = 1 and k = 0 gives an identity containing four consecutive
Fibonacci numbers
∞
X
n=1
−1
tan
π
aFn
= − tan−1 (a + b),
2
2
a Fn+1 Fn+2 + abFn+3 + b + 1
2
(35)
from which the following evaluations are easily obtained:
The case a = 1 and b = 0 gives Eq. (13).
The case a = 2 and b = 0 gives
∞
X
n=1
tan−1
π
2Fn
= − tan−1 (2).
1 + 4Fn+1 Fn+2
2
(36)
369
On infinite series involving Fibonacci numbers
The case a = 1 and b = 1 gives
∞
π
X
Fn
= − tan−1 (2).
tan−1
F
F
+
F
+
2
2
n+1
n+2
n+3
n=1
The case a = α and b = 0 gives
∞
π
X
αFn
−1
tan
= − tan−1 (α).
2
1 + α Fn+1 Fn+2
2
n=1
(37)
(38)
Comparing with Eq. (17) we arrive at
∞
X
n=1
−1
tan
∞
π X
αFn
αFn
−1
= −
tan
.
1 + α2 Fn+1 Fn+2
2 n=1
α2 + Fn+1 Fn+2
The case a = α and b = β gives
∞
π
X
αFn
−1
= .
tan
α2 Fn+1 Fn+2 − Fn+3 + β + 2
4
n=1
As a final example in this subclass, take a = α2 and b = β 2 to get
∞
π
X
α 2 Fn
−1
tan
= − tan−1 (3).
4
2
α Fn+1 Fn+2 + Fn+3 + 3β
2
n=1
(39)
(40)
(41)
The case m = 2 and k = 0 gives
∞
π
2
X
+ Fn2 )
a(Fn+1
tan−1 2 2
= −tan−1 (a+b).
2 + F 2 ) − a2 + b 2 + 1
a
F
+
ab(4F
+
F
2
2n
2n+1
n+1
n
n=1
(42)
Choosing a = 1 and b = 0 in the above equation establishes Eq. (21).
The case m = 2 and k = 1 gives
∞
π
2
X
a(Fn+2
− Fn2 )
tan−1 2 2
= −tan−1 (2a+b),
2
2
2
2
2
a F2n+2 + ab(Fn + 2Fn+1 + Fn+2 ) + a + b + 1
2
n=1
(43)
from which
∞
F 2 − F 2 π
X
n+2
n
tan−1
= − tan−1 (2),
(44)
2
F
+
2
2
2n+2
n=1
and
∞
X
n=1
tan−1
2(F 2
2 n+2 − Fn )
2
F2n+2
+5
=
π
,
4
(45)
are two illustrative examples.
Differentiation of Eq. (34) with respect to a and b will result in the following
summations:
370
Robert Frontczak
Corollary 3.4 Let a and b be defined as above. We set
x1 = Fmn+m+k − Fmn+k
x2 = Fmn+k Fmn+m+k
x3 = Fmn+m+k + Fmn+k .
Then
∞
X
n=1
and
∞
X
n=1
a2 x1 x2 − (b2 + 1)x1
Fm+k
=
,
2
2
2
2
2
(a x2 + abx3 + b + 1) + a x1
1 + (aFm+k + b)2
(46)
a2 x1 x3 + 2abx1
1
=
.
2
2
2
2
2
(a x2 + abx3 + b + 1) + a x1
1 + (aFm+k + b)2
(47)
In case m = 1 and k = 0 the sums reduce to
∞
X
n=1
and
∞
X
n=1
1
a2 Fn Fn+1 Fn+2 − (b2 + 1)Fn
=
,
2
2
2
2
2
(a Fn+1 Fn+2 + abFn+3 + b + 1) + a Fn
1 + (a + b)2
(48)
a2 Fn Fn+3 + 2abFn
1
.
=
2
2
2
2
2
(a Fn+1 Fn+2 + abFn+3 + b + 1) + a Fn
1 + (a + b)2
(49)
We continue by stating a few explicit examples of the above identities:
∞
X
Fn Fn+1 Fn+2 − 2Fn
1
= ,
2
2
(Fn+1 Fn+2 + Fn+3 + 2) + Fn
5
(50)
1
2Fn Fn+1 Fn+2 − 3Fn
=
,
(2Fn+1 Fn+2 + 2Fn+3 + 3)2 + 2Fn2
9
(51)
3Fn Fn+1 Fn+2 − 4Fn
1
= ,
2
2
(3Fn+1 Fn+2 + 3Fn+3 + 4) + 3Fn
13
(52)
n=1
∞
X
n=1
∞
X
n=1
∞
X
2Fn Fn+1 Fn+2 − Fn
1
= ,
2
2
(2Fn+1 Fn+2 + 1) + 2Fn
3
(53)
α4 Fn Fn+1 Fn+2 − 3β 2 Fn
1
= ,
4
2
2
4
2
(α Fn+1 Fn+2 + Fn+3 + 3β ) + α Fn
10
(54)
n=1
∞
X
n=1
∞
X
n=1
Fn Fn+3
1
= .
2
2
(Fn+1 Fn+2 + 1) + Fn
2
(55)
On infinite series involving Fibonacci numbers
371
It is also worth to mention that the choice b = −a in (48) and (49) gives
∞
X
n=1
and
∞
X
n=1
a2 Fn Fn+1 Fn+2 − (a2 + 1)Fn
= 1,
(a2 Fn+1 Fn+2 − a2 Fn+3 + a2 + 1)2 + a2 Fn2
(56)
a2 Fn Fn+3 − 2a2 Fn
= 1.
(a2 Fn+1 Fn+2 − a2 Fn+3 + a2 + 1)2 + a2 Fn2
(57)
Note that both series are independent of a.
Explicit evaluations in case m = 2 are limited to three examples:
∞
X
n=1
∞
X
n=1
and
∞
X
n=1
2
2
(Fn+2
− Fn2 )(4F2n+2
+ 3)
2
=
,
2
2
(4F2n+2 + 5)2 + 4(Fn+2 − Fn2 )2
17
(58)
2
2
+ 1)
− Fn2 )(2F2n+2
(Fn+2
2
= ,
2
2
2
2
2
(2F2n+2 + 3) + 2(Fn+2 − Fn )
9
(59)
2
2
− 2β)
− Fn2 )(α2 F2n+2
(Fn+2
2
=
.
2
2
2
2
(α2 F2n+2 − Fn+2 − 2Fn+1 − Fn2 + 4)2 + α2 (Fn+2 − Fn2 )2
3α2
(60)
Remark 3.5 It is clear that considering h(x) = ax2 + bx + c more advanced
series involving tan−1 (a + b + c) may be derived from the Theorem. We do not
go in this direction, however. Instead, we focus on functional forms of h(x)
and g(n) which will be able to produce alternating series. This is done next.
3.3
Third type: h(x) = axr .
Corollary 3.6 For r an integer, r ≥ 1, and a 6= 0 let h(x) = axr . Let
further g(n) = Fn+k /Fn , k ≥ 1. Then limn→∞ g(n) = αk and
∞
X
n=1
tan−1
a(F r F r
r
r
− Fn+1
Fn+k
)
r
= tan−1 (aαkr ) − tan−1 (aFk+1
), (61)
r
r
r
Fnr Fn+1
+ a2 Fn+k
Fn+1+k
n
n+1+k
or equivalently
∞
a(F r F r
r
r
X
π
n n+1+k − Fn+1 Fn+k )
tan−1
= tan−1 (aαkr ) − · sgn(a),
r
r
r
r
2
Fn Fn+1 + a Fn+k Fn+1+k
2
n=0
(62)
where sgn(x) denotes the sign function of x. Especially, for each k ≥ 1 and
r ≥ 1 we have the remarkable identities
∞
−αkr (F r F r
π
r
r
X
n n+1+k − Fn+1 Fn+k )
= ,
(63)
tan−1
r
r
2kr F r F r
α
+
F
F
4
n+1
n
n+k
n+1+k
n=0
372
Robert Frontczak
∞
X
−1
−√3αkr (F r F r
−1
−√3αkr (F r F r
tan
n=0
∞
X
tan
n=0
and
∞
X
n=0
tan−1
r
r
n n+1+k − Fn+1 Fn+k )
r
r
r
+ 3Fn+k
Fn+1+k
α2kr Fnr Fn+1
π
,
6
(64)
r
r
) π
Fn+k
− Fn+1
= ,
r
r
r
Fn+1+k
+ Fn+k
3α2kr Fnr Fn+1
3
(65)
n
=
n+1+k
−(2 − √3)αkr (F r F r
r
r
) 5π
Fn+k
− Fn+1
√
.
=
r
r
r
12
α2kr Fnr Fn+1
+ (7 − 4 3)Fn+k
Fn+1+k
n
n+1+k
(66)
A direct consequence of the Corollary is
Remark 3.7 Let S(r, k, a) denote the sum on the LHS of Eq. (62). Then
S(r, k, a) possesses the following symmetry property: If kr = q and d is a
divisor of q including 1 and q, then S(d, d∗ , a) = S(d∗ , d, a), where d∗ = q/d.
For instance, we have that S(1, 2, a) = S(2, 1, a) or S(3, 1, a) = S(1, 3, a).
Though the above results may be interesting, the alternating structure of
the series is still hidden and not obvious. To make the structure more evident
we prove the following Lemma:
Lemma 3.8 For n ≥ 0 and k ≥ 1
Fn Fn+1+k − Fn+1 Fn+k = (−1)n+1 Fk .
Furthermore, for each r ≥ 2 we have the relation
r X
r
r r
r
r
r−i r−i
Fn Fn+1+k − Fn+1 Fn+k =
(−1)i(n+1) Fki Fn+1
Fn+k .
i
(67)
(68)
i=1
PROOF: To prove the first identity we use induction on k. For k = 1 the
statement follows from Cassini’s Theorem. Let the statement be true for a
fixed k > 1. Then
Fn Fn+2+k − Fn+1 Fn+1+k = Fn (Fn+1+k + Fn+k ) − Fn+1 (Fn+k + Fn+k−1 )
= Fn Fn+1+k − Fn+1 Fn+k + Fn Fn+k − Fn+1 Fn+k−1
= (−1)n+1 Fk + (−1)n+1 Fk−1 = (−1)n+1 Fk+1 .
The second identity follows from applying the binomial formula.
2
Remark 3.9 The first identity of the Lemma is a non-symmetric generalization of Cassini’s Theorem. It may be seen as a companion of the symmetric
generalization due to Catalan (see [3], p. 109):
Fn−k Fn+k − Fn2 = (−1)n+k+1 Fk2 .
373
On infinite series involving Fibonacci numbers
Equipped with the findings of the Lemma, the case r = 1 can be stated as
∞
X
tan−1
n=0
a(−1)n+1 Fk
π
−1
k
=
tan
(aα
)
−
· sgn(a).
Fn Fn+1 + a2 Fn+k Fn+1+k
2
(69)
The subcase k = 1 and a = 1/α gives
∞
X
tan−1
n=0
π
α(−1)n
= .
α2 Fn Fn+1 + Fn+1 Fn+2
4
(70)
Also, comparing the results for a = 1 with that one for a = 1/α:
∞
X
∞
X
(−1)n+1
α(−1)n+1
π
tan
=
tan−1
= tan−1 (α)− .
2
Fn Fn+1 + Fn+1 Fn+2
α Fn Fn+1 + Fn+1 Fn+2
4
n=1
n=1
(71)
Further examples in this subcase are
√
∞
X
√
π
3(−1)n+1
−1
tan
= tan−1 ( 3α) − ,
(72)
F
F
+
3F
F
3
n
n+1
n+1
n+2
n=1
−1
∞
X
√
π
3(−1)n+1
3α
−1
= tan (
)− ,
3Fn Fn+1 + Fn+1 Fn+2
3
6
√
−1
tan
n=1
(73)
as well as
∞
X
n=1
−1
tan
√
√
(2 − 3)(−1)n+1
π
√
= tan−1 ((2 − 3)α) − , (74)
12
Fn Fn+1 + (7 − 4 3)Fn+1 Fn+2
The subcase k = 2 gives very similar results. Two explicit evaluations are
∞
X
tan−1
n=0
and
∞
X
π
α2 (−1)n
= .
α4 Fn Fn+1 + Fn+2 Fn+3
4
(75)
√
−1
tan
n=0
π
3α2 (−1)n
= .
3α4 Fn Fn+1 + Fn+2 Fn+3
3
(76)
For a > 0, the subcase k = 3 may be stated as
∞
X
n=0
tan−1
π
2a(−1)n
= − tan−1 (aα3 ).
Fn Fn+1 + a2 Fn+3 Fn+4
2
(77)
In case r = 2 we obtain
a(F 2 + 2(−1)n+1 F F F ) π
k n+1 n+k
k
tan−1
= tan−1 (aα2k ) − · sgn(a). (78)
2
2
2
2
2
Fn Fn+1 + a Fn+k Fn+1+k
2
n=0
∞
X
374
Robert Frontczak
Again, for each pair k and a numerous special cases are immediately available.
We restrict the presentation of explicit examples to a few series: Take k = 1
and a = 1 to get
∞
X
−1
tan
n=0
−1 + 2(−1)n F 2 ) π
n+1
= − tan−1 (α + 1).
2
2
2
2
Fn Fn+1 + Fn+1 Fn+2
2
(79)
In a similar manner we get
∞
X
tan−1
n=0
α2 (−1 + 2(−1)n F 2 ) π
n+1
= ,
2
2
2
4
2
α Fn Fn+1 + Fn+1 Fn+2
4
√3α2 (−1 + 2(−1)n F 2 ) n+1
=
tan−1
2
2
2
4
2
α Fn Fn+1 + 3Fn+1 Fn+2
n=0
∞
√3α2 (−1 + 2(−1)n F 2 ) X
n+1
−1
tan
=
2
2
4F 2F 2
3α
+
F
F
n+1
n+1
n+2
n
n=0
∞
X
(80)
π
,
6
(81)
π
.
3
(82)
Also, for each a:
∞
X
∞
X
a(1 + 2(−1)n+1 F 2 ) a(−1)n+1
n+1
−1
tan
=
tan
.
2
2
2
2
2
2
Fn Fn+1 + a Fn+2 Fn+3
Fn Fn+1 + a Fn+1 Fn+2
n=0
n=0
(83)
Inserting k = 3 and a = 1 and a = 3, respectively, results in the evaluations
−1
∞
X
tan−1
n=0
4(−1 + (−1)n F
2
Fn2 Fn+1
+
n+1 Fn+3 )
2
2
Fn+3 Fn+4
=
π
− tan−1 ((α + 1)3 ),
2
(84)
and
∞
X
tan−1
12(−1 + (−1)n F
n=0
2
Fn2 Fn+1
+
n+1 Fn+3 )
2
2
9Fn+3 Fn+4
=
π
− tan−1 (3(α + 1)3 ).
2
(85)
We continue with presenting series identities that are derived from Corollary 3.6 by differentiation with respect to a.
Corollary 3.10 For r, k ≥ 1 set
r
r
r
x1 = Fnr Fn+1+k
− Fn+1
Fn+k
r
x2 = Fnr Fn+1
r
r
x3 = Fn+k
Fn+1+k
.
Then
∞
X
n=1
r
Fk+1
x1 (x2 − a2 x3 )
αkr
=
−
.
2r
(x2 + a2 x3 )2 + a2 x21
1 + a2 α2kr 1 + a2 Fk+1
(86)
375
On infinite series involving Fibonacci numbers
Especially for a = 0 we get the identity
∞
r
r
r
X
Fn+k
− Fn+1
Fnr Fn+1+k
r
Fnr Fn+1
n=1
r
= αkr − Fk+1
.
(87)
The case r = 1 simplifies to
∞
X
(−1)n+1 Fk (Fn Fn+1 − a2 Fn+k Fn+1+k )
n=1
(Fn Fn+1 + a2 Fn+k Fn+1+k )2 + a2 Fk2
=
αk
Fk+1
.
−
2
2
2k
1+a α
1 + a2 Fk+1
(88)
Setting a = 0 we get for each k
∞
X
(−1)n+1
n=1
Fn Fn+1
=
αk − Fk+1
,
Fk
(89)
which in view of the formula αk = αFk + Fk−1 may be written in the form
∞
X
(−1)n+1
n=1
Fn Fn+1
= α − 1.
(90)
For k = 1 and a = 1 we get
∞
X
n=1
2
(−1)n+1 Fn+1
12−α
=
.
2
Fn+1 (Fn + Fn+2 )2 + 1
22+α
(91)
For k = 1 and a = 1/2 we get
∞
X
(−1)n+1 (4Fn Fn+1 − Fn+1 Fn+2 )
n=1
(4Fn Fn+1 + Fn+1 Fn+2
)2
+4
=
1 4α − 5
.
5 α+5
(92)
√
For k = 1 and a = 1/ α we get
∞
X
(−1)n+1 (αFn Fn+1 − Fn+1 Fn+2 )
n=1
(αFn Fn+1 + Fn+1 Fn+2 )2 + α
= 0.
(93)
The case r = 2 produces
∞
X
n=1
=
2
2
2
(Fk2 + 2(−1)n+1 Fk Fn+1 Fn+k )(Fn2 Fn+1
− a2 Fn+k
Fn+1+k
)
2
2
2
(Fn2 Fn+1
+ a2 Fn+k
Fn+1+k
)2 + a2 (Fk2 + 2(−1)n+1 Fk Fn+1 Fn+k )2
2
Fk+1
α2k
−
.
4
1 + a2 α4k 1 + a2 Fk+1
(94)
376
Robert Frontczak
For a = 0 this simplifies to
∞
X
Fk + 2(−1)n+1 Fn+1 Fn+k
n=1
2
Fn2 Fn+1
= (α−1)(αk +Fk+1 ) = Fk α+2Fk−1 (α−1). (95)
Inserting k = 1, 2, 3 we get
∞
2
X
1 + 2(−1)n+1 Fn+1
= (α − 1)(α + 1) = α,
(96)
= (α − 1)(α + 3) = 3α − 2,
(97)
= (α − 1)(α + 2) = 2α − 1.
(98)
2
Fn2 Fn+1
n=1
∞
X
1 + 2(−1)n+1 Fn+1 Fn+2
2
Fn2 Fn+1
n=1
and
∞
X
1 + (−1)n+1 Fn+1 Fn+3
2
Fn2 Fn+1
n=1
As a final example for this type of evaluations we take the pair k = 1 and
a = 1 and get after some algebra
∞
X
n=1
3
2
Fn+1
(Fn+1 + 2Fn )(1 + 2(−1)n+1 Fn+1
)
1
= .
2
2
2
2
2
2
n+1
2
(Fn Fn+1 + Fn+1 Fn+2 ) + (1 + 2(−1) Fn+1 )
6
(99)
In our final application of Theorem 2.1 we consider series that contain the
exponential function and therefore exhibit as very fast convergence.
3.4
Fourth type: h(x) = ae−bx .
Corollary 3.11 For a 6= 0 and b > 0 let h(x) = ae−bx . Let further g(n) =
Fn+k , k ≥ 0. Then
∞
X
−1
tan
n=1
a(ebFn+k+1 − ebFn+k ) = tan−1 (ae−bFk+1 ).
a2 + ebFn+k+2
(100)
Especially, for each b and k
∞
X
tan−1
n=1
∞
X
−1
tan
n=1
∞
X
n=1
−1
tan
eb(Fk+1 +Fn+k+1 ) − eb(Fk+1 +Fn+k ) e2bFk+1
+
ebFn+k+2
=
√3(eb(Fk+1 +Fn+k+1 ) − eb(Fk+1 +Fn+k ) ) 3e2bFk+1
+
ebFn+k+2
√3(eb(Fk+1 +Fn+k+1 ) − eb(Fk+1 +Fn+k ) ) e2bFk+1
+
3ebFn+k+2
π
,
4
(101)
=
π
,
3
(102)
=
π
,
6
(103)
377
On infinite series involving Fibonacci numbers
and
(2 − √3)(eb(Fk+1 +Fn+k+1 ) − eb(Fk+1 +Fn+k ) ) π
√
= .
tan−1
bF
2bF
12
(7 − 4 3)e k+1 + e n+k+2
n=1
∞
X
(104)
For k = 0 and k = 1 Eq. (100) simplifies to
∞
X
tan−1
n=1
a(ebFn+1 − ebFn ) = tan−1 (ae−b ),
a2 + ebFn+2
(105)
from which the following identities are special cases:
∞
X
eFn+1 − eFn −1 1
=
tan
( ),
F
1 + e n+2
e
(106)
e2Fn+1 − e2Fn −1 1
=
tan
( 2 ),
2F
1 + e n+2
e
(107)
α(eFn+1 − eFn ) α
= tan−1 ( ),
2
F
n+2
α +e
e
(108)
tan−1
n=1
∞
X
tan−1
n=1
∞
X
tan−1
n=1
∞
X
−1
tan
e(1+Fn+1 ) − e(1+Fn ) e2
n=1
∞
X
−1
tan
−1
tan
3e2 + eFn+2
√3(e(1+Fn+1 ) − e(1+Fn ) ) n=1
as well as
∞
X
−1
tan
n=1
=
√3(e(1+Fn+1 ) − e(1+Fn ) ) n=1
∞
X
+
eFn+2
e2
+
3eFn+2
π
,
4
(109)
=
π
,
3
(110)
=
π
,
6
(111)
(2 − √3)(e(1+Fn+1 ) − e(1+Fn ) ) π
√
= .
12
(7 − 4 3)e2 + eFn+2
(112)
For k = 2 Eq. (100) reduces to
∞
X
n=1
tan−1
aebFn+2 (ebFn+1 − 1) = tan−1 (ae−2b ).
a2 + ebFn+4
Differentiation of Eq. (100) with respect to the parameter a gives
(113)
378
Robert Frontczak
Corollary 3.12 Set
x1 = ebFn+k+1 − ebFn+k ,
Then
∞
X
n=1
x2 = ebFn+k+2 .
x1 (x2 − a2 )
e−bFk+1
=
.
(x2 + a2 )2 + a2 x21
1 + a2 e−2bFk+1
(114)
Especially for a = 0 we get the identity
∞
X
ebFn+k+1 − ebFn+k
= e−bFk+1 .
bFn+k+2
e
n=1
(115)
We finish the study by presenting a few special cases of the Corollary in case
k = 0. For a = 1 and b = 1 we get
∞
X
n=1
(eFn+1 − eFn )(eFn+2 − 1)
e
=
.
(eFn+2 + 1)2 + (eFn+1 − eFn )2
e2 + 1
(116)
For a = e and b = 1 we get
∞
X
n=1
(eFn+1 − eFn )(eFn+2 − e2 )
1
=
.
F
2
2
2
F
F
2
(e n+2 + e ) + e (e n+1 − e n )
2e
(117)
For a = 1/e and b = 1 the result is
∞
X
n=1
4
e
(eFn+1 − eFn )(e2+Fn+2 − 1)
=
.
(e2+Fn+2 + 1)2 + e2 (eFn+1 − eFn )2
e4 + 1
(118)
Conclusion
In this paper we have derived closed-form expressions for various types of infinite series involving Fibonacci numbers. As a direct consequence new series
representations for π have been established. The approach presented is very
general in the sense that alternating and non-alternating series may be studied. Also, it is not limited to a specific sequence. Thus similar results may be
derived for Lucas numbers, Pell numbers or other sequences. This is left for
future research.
References
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On infinite series involving Fibonacci numbers
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[2] B.A. Brousseau, Fibonacci-Lucas Infinite Series Research Topic, The Fibonacci Quarterly, 7 (1968), no. 2, 211 - 217.
[3] R.P. Grimaldi, Fibonacci and Catalan numbers: An Introduction, John
Wiley & Sons, New York, 2012. http://dx.doi.org/10.1002/9781118159743
[4] A.R. Guillot, Problem XXX, The Fibonacci Quarterly, 7 (1977), no. 2,
257.
[5] T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley
& Sons, New York, 2001. http://dx.doi.org/10.1002/9781118033067
[6] R.S. Melham, Sums involving Fibonacci and Pell numbers, Portugaliae
Mathematica, 56 (1999), no. 3, 309 - 317.
[7] R.S. Melham and A.G. Shannon, Inverse trigonometric and hyperbolic
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[8] K. Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, Notes on Number Theory and Discrete Mathematics, 21 (2015), no.
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Received: September 21, 2015; Published: November 11, 2015