On one series of the reciprocals of the product of two Fibonacci numbers
whose indices differ by an even number
POTŮČEK R.
Department of Mathematics and Physics, University of Defence,
Kounicova 65, 662 10 Brno, CZECH REPUBLIC
https://fvt.unob.cz/fakulta/struktura/katedra-matematiky-a-fyziky-k-215/struktura-k-215/
ORCID iD: 0000-0003-4385-691X
Abstract: This paper is inspired by a very interesting YouTube video by Michael Penn, professor of mathematics
at Randolph College in Virginia, USA. He dedicated himself to the popularization of mathematics on his website
in addition to his teaching and scientific work at the university and in addition to his scientific work. First, we
deal with four specific series of the reciprocals of the product of two Fibonacci numbers whose indices differ by 2,
4, 6, and 8. Then, we generalize these four results to the series of the reciprocals of the product of two Fibonacci
numbers whose indices differ by an even number. Finally, we perform a numerical verification of the derived
formula using Maple 2020 software. Based on the derived formula, it can be concluded that the series we are
dealing with belong to infinite series whose sum can be expressed in closed form.
Key-Words: Fibonacci numbers, partial fractions decomposition, telescoping series, Maple 2020
Received: May 9, 2023. Revised: May 12, 2024. Accepted: June 17, 2024. Published: July 16, 2024.
1 Introduction
This paper, inspired by the YouTube video [1], has
been written to popularize Fibonacci numbers and, in
particular, the sums of the reciprocals of the expres-
sions with Fibonacci numbers. Articles dealing with
a similar topic include e.g. articles [2] and [3]. Some
results can also be found on Wikipedia [4]. All the
series discussed deal with in the paper are examples
of more complex numerical series with very simple
sums.
The Fibonacci numbers appeared for the first time
in ancient Indian mathematics before the beginning
of the era. In Europe, they were first described by
Italian mathematician Leonardo Pisano (Leonardo of
Pisa), also known as Fibonacci (about 1175–1250),
in the year 1202 in his book Liber Abaci. The Fi-
bonacci numbers include the rule of golden propor-
tions. In essence, this is an observation that the ra-
tio of any two sequential Fibonacci numbers approx-
imates to the value of the golden ratio.
The Fibonacci numbers, or the Fibonacci se-
quence, are one of the most famous and most widely
studied numbers in modern mathematics, with in-
teresting and amazing properties. The Fibonacci
numbers continue to be very popular and offer pro-
vide many numerous new topics for further research
work. For instance, several following articles have
been published in recent years, including [5], [6], [7],
and [8]. The Fibonacci numbers are a central topic in
various monographs, such as [9], [10], [11], [12], [13],
and [14].
2 Basic definitions and properties of
Fibonacci numbers
The Fibonacci numbers Fkare defined for an arbi-
trary integer k≥2by the recursive formula
Fk=Fk−1+Fk−2.(1)
The two initial values are F0= 0 and F1= 1.
Using formula (1), we obtain the well-known Fi-
bonacci sequence
{Fk}={0,1,1,2,3,5,8,13,21,34,55,89,
144,233,377,610,987, . . .}.
The ratio of two consecutive Fibonacci numbers con-
verges and approaches the golden ratio
φ=lim
k→∞
Fk+1
Fk
.
This limit has the value
φ=1 + √5
2≈1.618033988.
The Fibonacci numbers have a closed-form ex-
pression which is known as Binet’s formula
Fk=φk−ψk
√5, k ≥0,(2)
where ψis the conjugate of φ, i.e.
ψ=1−√5
2≈ −0.618033988,
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so for an arbitrary integer k≥0we have
Fk=1
√51 + √5
2k
−1−√5
2k
or
Fk=1
2k√51 + √5k−1−√5k.
The process of deriving formula (2) using generat-
ing functions can be found, for example, in the ar-
ticle [15]. Since 1−√5
2= 1 −1 + √5
2, we get that
ψ= 1 −φ.
By repeatedly using of the equality (1), i.e. by
means of the formulas
Fk+2 =Fk+1 +Fk
and
Fk+1 =Fk+2 −Fk,
where k≥0is an arbitrary integer, we obtain the fol-
lowing expressions for Fk+4,Fk+6 and Fk+8 in terms
of Fk+2 and Fk, which we will use in Sections 4, 5
and 6:
Fk+4 =Fk+3 +Fk+2 = (Fk+2 +Fk+1) + Fk+2
= 2Fk+2 +Fk+1 = 2Fk+2 + (Fk+2 −Fk)
= 3Fk+2 −Fk=F4Fk+2 −F2Fk,
(3)
Fk+6 =Fk+5 +Fk+4 = (Fk+4 +Fk+3) + Fk+4
= 2Fk+4 +Fk+3 = 2(Fk+3 +Fk+2) + Fk+3
= 3Fk+3 + 2Fk+2 = 3(Fk+2 +Fk+1)+2Fk+2
= 5Fk+2 + 3Fk+1 = 5Fk+2 + 3(Fk+2 −Fk)
= 8Fk+2 −3Fk=F6Fk+2 −F4Fk,
(4)
Fk+8 =Fk+7 +Fk+6 = (Fk+6 +Fk+5) + Fk+6
= 2Fk+6 +Fk+5 = 2(Fk+5 +Fk+4) + Fk+5
= 3Fk+5 + 2Fk+4 = 3(Fk+4 +Fk+3)+2Fk+4
= 5Fk+4 + 3Fk+3 = 5(Fk+3 +Fk+2)+3Fk+3
= 8Fk+3 + 5Fk+2 = 8(Fk+2 +Fk+1)+5Fk+2
= 13Fk+2 + 8Fk+1 = 13Fk+2 + 8(Fk+2 −Fk)
= 21Fk+2 −8Fk=F8Fk+2 −F6Fk.
(5)
There are plenty of other relationships for the Fi-
bonacci numbers, but here we present only those that
we will continue to use in the paper.
3 The series of the reciprocals of the
product of two Fibonacci numbers
whose indices differ by two
Let us consider the series
∞
k=1
1
FkFk+2
=1
F1F3
+1
F2F4
+1
F3F5
+···
=1
1·2+1
1·3+1
2·5+··· .
(6)
Denote the sum of the series (6) as S(2) and state, that
taking a sum is similar to taking an integral. If we in-
tegrate the reciprocal of the product of two functions,
then we often use the method of partial fraction de-
composition. Inspired by integration we use the de-
composition of a general kth term of the series (6) in
the form 1
FkFk+2
=A
Fk
+B
Fk+2
.
By multiplying this equation by the product FkFk+2
we get
1 = AFk+2 +BFk.
According to (1) we have Fk+2 =Fk+1 +Fk, so we
obtain an equation
1 = A(Fk+1 +Fk) + BFk,
from which we get
1 = AFk+1 + (A+B)Fk.
To obtain an equation with an unknown coefficient B,
we substitute A=1
Fk+1
. Then we obtain an equation
1 = 1 + 1
Fk+1
+BFk,
i.e.
0 = 1
Fk+1
+BFk,
from which we get
B=−1
Fk+1
.
So we have received the coefficients
A=1
Fk+1
and B=−1
Fk+1
.
That gives us the partial fraction decomposition in the
form
1
FkFk+2
=1
FkFk+1 −1
Fk+1Fk+2
.
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Now we can write the series (6) in the form
∞
k=1
1
FkFk+2
=
∞
k=11
FkFk+1 −1
Fk+1Fk+2 .
The sum S(2) we determine as a limit of the sequence
of partial sums {S(2)n}. We get
S(2) = lim
n→∞
S(2)n=lim
n→∞
n
k=1
1
FkFk+2
=lim
n→∞
n
k=11
FkFk+1 −1
Fk+1Fk+2 .
By writing out the terms of the nth partial sum, we
obtain
S(2) = lim
n→∞ 1
F1F2−1
F2F3
+1
F2F3−1
F3F4+1
F3F4−1
F4F5
+··· +1
Fn−1Fn−1
FnFn+1
+1
FnFn+1 −1
Fn+1Fn+2 .
Due to the telescopic properties of this nth partial
sum, all summands will be canceled, with the ex-
ception of the first and last summands. Because
lim
n→∞
Fn=∞, we have lim
n→∞
1
Fn
= 0 and also
lim
n→∞
1
Fn+1Fn+2
= 0. So we get
S(2) =
∞
k=1
1
FkFk+2
=lim
n→∞1
F1F2−1
Fn+1Fn+2
=1
F1F2
=1
1·1= 1.
(7)
4 The series of the reciprocals of the
product of two Fibonacci numbers
whose indices differ by four
Let us consider the series
∞
k=1
1
FkFk+4
=1
F1F5
+1
F2F6
+1
F3F7
+···
=1
1·5+1
1·8+1
2·13 +··· .
(8)
Denote the sum of the series (8) as S(4) and as in
the previous section, we determine the sum S(4) by
using the decomposition of a general kth term of the
series (8) into the sum of partial fractions. So, we
suppose that we can write
1
FkFk+4
=A
Fk
+B
Fk+4
.
By multiplying by the product FkFk+4 we get
1 = AFk+4 +BFk.
According to (3) we obtain an equation
1 = A(3Fk+2 −Fk) + BFk,
from which we get
1 = 3AFk+2 + (B−A)Fk.
To obtain an equation with an unknown coefficient B,
we substitute A=1
3Fk+2
. Then we receive an equa-
tion
1 = 1 + B−1
3Fk+2 Fk,
i.e.
0 = B−1
3Fk+2 Fk,
whence we get
B=1
3Fk+2
.
So we have received the coefficients
A=1
3Fk+2
and also B=1
3Fk+2
.
That gives us the partial fraction decomposition in the
form
1
FkFk+4
=1
3FkFk+2
+1
3Fk+2Fk+4
.
Therefore we can write the series (6) in the form
∞
k=1
1
FkFk+4
=1
3
∞
k=11
FkFk+2
+1
Fk+2Fk+4 .
In Section 3 we show that the series
∞
k=1
1
FkFk+2
con-
verges to 1. The series
∞
k=1
1
Fk+2Fk+4
is clearly of the
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same type of convergent series that can be obtained by
suitable re-indexing. So now we can write
∞
k=1
1
FkFk+4
=1
3∞
k=1
1
FkFk+2
+
∞
k=1
1
Fk+2Fk+4
=1
3∞
k=1
1
FkFk+2
+
∞
k=3
1
FkFk+2 .
Because we can write the second series using the first
one in the form
∞
k=3
1
FkFk+2
=
∞
k=1
1
FkFk+2 −1
F1F3−1
F2F4
,
we have
∞
k=1
1
FkFk+4
=1
3∞
k=1
1
FkFk+2
+
∞
k=1
1
FkFk+2 −1
F1F3−1
F2F4
=1
32
∞
k=1
1
FkFk+2 −1
1·2−1
1·3.
According to (7) we obtain
S(4) =
∞
k=1
1
FkFk+4
=1
32
F1F2−1
F1F3−1
F2F4
=1
32
1−1
2−1
3=1
3·7
6=7
18 .
(9)
5 The series of the reciprocals of the
product of two Fibonacci numbers
whose indices differ by six
Let us consider the series
∞
k=1
1
FkFk+6
=1
F1F7
+1
F2F8
+1
F3F9
+···
=1
1·13 +1
1·21 +1
2·34 +···
(10)
and denote the sum of this series as S(6). We will
proceed similarly to the previous section, so we will
limit ourselves to only a brief comment on individual
steps.
We again assume that we can write
1
FkFk+6
=A
Fk
+B
Fk+6
,
i.e.
1 = AFk+6 +BFk.
According to (4) we obtain an equation
1 = A(8Fk+2 −3Fk) + BFk,
i.e.
1 = 8AFk+2 + (B−3A)Fk.
When we substitute A=1
8Fk+2
, we receive an equa-
tion
1 = 1 + B−3
8Fk+2 Fk,
whence we get
B=3
8Fk+2
.
These values of the coefficients Aand Blead to de-
composition into partial fractions of the form
1
FkFk+6
=1
8FkFk+2
+3
8Fk+2Fk+6
.
Therefore we can write the series (10) in the form
∞
k=1
1
FkFk+6
=1
8
∞
k=11
FkFk+2
+3
Fk+2Fk+6 .
In Section 4 we show that the series
∞
k=1
1
FkFk+4
con-
verges to 7
18. The series
∞
k=1
3
Fk+2Fk+6
is of the same
type of convergent series as the series
∞
k=1
1
FkFk+4
.
So we can write
∞
k=1
1
FkFk+6
=1
8∞
k=1
1
FkFk+2
+
∞
k=1
3
Fk+2Fk+6
=1
8∞
k=1
1
FkFk+2
+ 3
∞
k=3
1
FkFk+4 .
Because
∞
k=3
1
FkFk+4
=
∞
k=1
1
FkFk+4 −1
F1F5−1
F2F6
,
we have
∞
k=1
1
FkFk+6
=1
8∞
k=1
1
FkFk+2
+ 3∞
k=1
1
FkFk+4 −1
F1F5−1
F2F6
=1
8∞
k=1
1
FkFk+2
+3∞
k=1
1
FkFk+4 −1
1·5−1
1·8.
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According to (7) and (9) we obtain
∞
k=1
1
FkFk+6
=1
81+37
18 −1
5−1
8
=1
81+3·140 −72 −45
360 =1
8·360 + 69
360 ,
so
S(6) =
∞
k=1
1
FkFk+6
=1
81
F1F2
+ 31
32
F1F2−1
F1F3−1
F2F4−1
F1F5−1
F2F6
=1
83
F1F2−1
F1F3−1
F2F4−3
F1F5−3
F2F6
=1
8·143
120 =143
960 .
(11)
6 The series of the reciprocals of the
product of two Fibonacci numbers
whose indices differ by eight
Let us consider the series
∞
k=1
1
FkFk+8
=1
F1F9
+1
F2F10
+1
F3F11
=1
1·34 +1
1·55 +1
2·89 +···
(12)
and denote the sum of this series as S(8). We will
proceed briefly and similarly to the previous sections,
We again write
1
FkFk+8
=A
Fk
+B
Fk+8
,
i.e.
1 = AFk+8 +BFk.
According to (5) we get an equation
1 = A(21Fk+2 −8Fk) + BFk,
i.e.
1 = 21AFk+2 + (B−8A)Fk.
When we substitute A=1
21Fk+2
, we receive an
equation
1 = 1 + B−8
21Fk+2 Fk,
whence we get
B=8
21Fk+2
.
These values of the coefficients Aand Blead to de-
composition into partial fractions of the form
1
FkFk+8
=1
21FkFk+2
+8
21Fk+2Fk+8
.
Therefore we can write the series (12) in the form
∞
k=1
1
FkFk+8
=1
21
∞
k=11
FkFk+2
+8
Fk+2Fk+8 .
In Section 5 we show that the series
∞
k=1
1
FkFk+6
converges to 143
960 and the series
∞
k=1
8
Fk+2Fk+8
is
of the same type of convergent series as the series
∞
k=1
1
FkFk+6
. So we can write
∞
k=1
1
FkFk+8
=1
21∞
k=1
1
FkFk+2
+
∞
k=1
8
Fk+2Fk+8
=1
21∞
k=1
1
FkFk+2
+ 8
∞
k=3
1
FkFk+6 .
Because
∞
k=3
1
FkFk+6
=
∞
k=1
1
FkFk+6 −1
F1F7−1
F2F8
,
we have
∞
k=1
1
FkFk+8
=1
21∞
k=1
1
FkFk+2
+ 8∞
k=1
1
FkFk+6 −1
F1F7−1
F2F8
=1
21∞
k=1
1
FkFk+2
+ 8∞
k=1
1
FkFk+6 −1
1·13 −1
1·21.
According to (7) and (11) we obtain
∞
k=1
1
FkFk+8
=1
211+8143
960 −1
13 −1
21=
=1
211+8·13013 −6720 −4160
87360
=1
21 ·87360 + 17064
87360 ,
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so
S(8) =
∞
k=1
1
FkFk+8
=1
211
F1F2
+ 81
83
F1F2−1
F1F3−1
F2F4
−3
F1F5−3
F2F6−1
F1F7−1
F2F8
=1
214
F1F2−1
F1F3−1
F2F4−3
F1F5
−3
F2F6−8
F1F7−8
F2F8
=1
21 ·104424
87360 =4351
76440 .
(13)
7 The series of the reciprocals of the
product of two Fibonacci numbers
whose indices differ in the even
number
Let us consider the series
∞
k=1
1
FkFk+2n
=1
F1Fk+2n
+1
F2F2+2n
+1
F3F3+2n
+1
F4F4+2n
+··· ,
(14)
where nis an arbitrary positive integer, and denote the
sum of this series as S(2n). From the above formulas
(7), (9), (11), (13), which can be written in the forms
S(2) =
∞
k=1
1
FkFk+2
=1
F1F2
,
S(4) =
∞
k=1
1
FkFk+4
=1
32
F1F2−1
F1F3−1
F2F4,
S(6) =
∞
k=1
1
FkFk+6
=1
83
F1F2−1
F1F3−1
F2F4
−3
F1F5−3
F2F6,
S(8) =
∞
k=1
1
FkFk+8
=1
214
F1F2−1
F1F3−1
F2F4
−3
F1F5−3
F2F6−8
F1F7−8
F2F8,
i.e. the formulas that have the forms
S(2) =
∞
k=1
1
FkFk+2
=1
F2·1
F1F2
,
S(4) =
∞
k=1
1
FkFk+4
=1
F42
F1F2−F2
F1F3−F2
F2F4,
S(6) =
∞
k=1
1
FkFk+6
=1
F63
F1F2−F2
F1F3−F2
F2F4
−F4
F1F5−F4
F2F6,
S(8) =
∞
k=1
1
FkFk+8
=1
F84
F1F2−F2
F1F3−F2
F2F4
−F4
F1F5−F4
F2F6−F6
F1F7−F6
F2F8,
it can be concluded that the formula for the sum of the
reciprocals of two Fibonacci numbers whose indices
differ in the even number 2nwill have the form
S(2n) =
∞
k=1
1
FkFk+2n
=1
F2nn
F1F2−F2
2
k=1
1
FkFk+2
−F4
2
k=1
1
FkFk+4 − ··· − F2n−2
2
k=1
1
FkFk+2n−2.
Now we can formulate the main result of this paper:
Theorem 1. For each positive integer n,
S(2n) =
∞
k=1
1
FkFk+2n
=1
F2nn
F1F2−
n−1
i=1 F2i
2
k=1
1
FkFk+2i.
(15)
8 Numerical verification
We solve the problem of determining the values of
the sum S(2n)for n= 1,2, . . . , 10, for n=
20,30, . . . , 100, for n= 500 and for n= 1000. We
use on the one hand an approximate direct evaluation
of the sum
S(2n, t) =
t
k=1
1
FkFk+2n
,
where the upper index t= 103, using the basic pro-
gramming language of the computer algebra system
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Maple 2020, and on the other hand the formula (15)
for evaluating the sum S(2n)for the same values of
the variable n. We compare 21 pairs of these ways
obtained sums S(2n, t)and S(2n)to verify the for-
mula (15). We use the following simple procedure
recfiboeven.
>with(combinat,fibonacci):
>recfiboeven:=proc(n)
local i,k,s,st,s1; st:=0;
s1:=sum(fibonacci(2*i)*sum(1/(fibonacci(k)
*fibonacci(k+2*i)),k=1..2),i=1..n-1);
s:= (1/fibonacci(2*n))*(n/(fibonacci(1)
*fibonacci(2))-s1);
print(”n=”,n,”s=”,evalf[12](s));
for k from 1 to 1000 do
st:=st+1/(fibonacci(k)*fibonaci(k+2*n));
end do;
print(”st=”,evalf[12](st));
print(”diff=”,evalf[12](abs(st-s)));
end proc:
>for n from 1 to 10 do
recfiboeven(n);
end do;
for n from 20 by 10 to 100 do
recfiboeven(n);
end do;
recfiboeven(500); recfiboeven(1000);
The approximate values of the sum S(2n)obtained by
this procedure are stated in Table 1. Computation of
21 pairs of the sums S(2n)and S(2n, 103)took about
7seconds. The absolute errors, i.e. the differences
|S(2n)−S(2n, 103)|, range between about 10−418
(for n= 1) and 10−836 (for n= 1000).
S(2) S(4) S(6)
1.000000 0.388889 0.148958
S(8) S(10) S(12)
5.692046 ·10−22.174299 ·10−28.305156 ·10−3
S(14) S(16) S(18)
3.172291 ·10−31.211708 ·10−34.628312 ·10−4
S(20) S(40) S(60)
1.767858 ·10−41.168677 ·10−87.725769 ·10−13
S(80) S(100) S(120)
5.107271 ·10−17 3.376262 ·10−21 2.231944 ·10−25
S(140) S(160) S(180)
1.475470 ·10−29 9.753886 ·10−34 6.447998 ·10−38
S(200) S(500) S(1000)
4.262575 ·10−42 2.751439 ·10−209 2.830868 ·10−418
Table 1: Some approximate values of the sums S(2n)
[source: own modeling in Maple 2020]
9 Conclusion
In this paper we dealt with the sum of the series of
reciprocals of the product of two Fibonacci numbers
whose indices differ by an even number, i.e. with the
series ∞
k=1
1
FkFk+2n
,
where nis an arbitrary positive integer. We derived
that the sum S(2n)of this series is given by the for-
mula
S(2n) =
∞
k=1
1
FkFk+2n
=1
F2nn
F1F2−
n−1
i=1 F2i
2
k=1
1
FkFk+2i.
We verified this result by computing 20 sums using
the computer algebra system Maple 2020. The se-
ries above so belong to special types of infinite series,
such as geometric and telescoping series, whose sums
are given analytically by means of a simple formula
in closed form.
Area of Further Development
The results of the paper can be generalized and further
extended to similar series of reciprocals of the expres-
sions containing the Fibonacci numbers Fkor Lucas
numbers Lk, Note that the Lucas numbers are defined
by the analogous recurrence relation
Lk=Lk−1+Lk−2, k ≥2,
as the Fibonacci numbers, but the first two Lucas
numbers are
L0= 2 and L1= 1.
Series and their sums that can be studied further can
be, for example, series of the form
∞
k=1
1
LkLk+2n
,
where nis an arbitrary positive integer, or
∞
k=1
1
FkFk+2Fk+4
.
EQUATIONS
DOI: 10.37394/232021.2024.4.4
Potůček R.
E-ISSN: 2732-9976
30
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Acknowledgement
This work was supported by the Project for the De-
velopment of the Organization DZRO ”Military au-
tonomous and robotic systems” under Ministry of De-
fence and Armed Forces of Czech Republic.
Contribution of Individual Authors to the
Creation of a Scientific Article (Ghostwriting
Policy)
The author contributed in the present research, at all
stages from the formulation of the problem to the
final findings and solution.
Sources of Funding for Research Presented in a
Scientific Article or Scientific Article Itself
No funding was received for conducting this study.
Conflict of Interest
The author has no conflict of interest to declare that
is relevant to the content of this article.
Creative Commons Attribution License 4.0
(Attribution 4.0 International, CC BY 4.0)
This article is published under the terms of the
Creative Commons Attribution License 4.0
https://creativecommons.org/licenses/by/4.0/deed.en
_US
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DOI: 10.37394/232021.2024.4.4
Potůček R.
E-ISSN: 2732-9976
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