Wang and Wen Journal of Inequalities and Applications (2015) 2015:73
DOI 10.1186/s13660-015-0595-6
RESEARCH
Open Access
On the partial finite sums of the reciprocals of
the Fibonacci numbers
Andrew YZ Wang* and Peibo Wen
*
Correspondence:
[email protected]
School of Mathematical Sciences,
University of Electronic Science and
Technology of China, Chengdu,
611731, P.R. China
Abstract
In this article, we obtain two interesting families of partial finite sums of the
reciprocals of the Fibonacci numbers, which substantially improve two recent results
involving the reciprocal Fibonacci numbers. In addition, we present an alternative and
elementary proof of a result of Wu and Wang.
MSC: 11B39
Keywords: Fibonacci numbers; partial sums; reciprocal
1 Introduction
The Fibonacci sequence [], Sequence A is defined by the linear recurrence relation
for n ≥ ,
Fn = Fn– + Fn–
where Fn is the nth Fibonacci number with F =  and F = . There exists a simple and
non-obvious formula for the Fibonacci numbers,
√ √  +  n
 –  n
–√
.
Fn = √




The Fibonacci sequence plays an important role in the theory and applications of mathematics, and its various properties have been investigated by many authors; see [–].
In recent years, there has been an increasing interest in studying the reciprocal sums
of the Fibonacci numbers. For example, Elsner et al. [–] investigated the algebraic relations for reciprocal sums of the Fibonacci numbers. In [], the partial infinite sums of the
reciprocal Fibonacci numbers were studied by Ohtsuka and Nakamura. They established
the following results, where · denotes the floor function.
Theorem . For all n ≥ ,
∞

Fk
– =
k=n
Fn– ,
if n is even;
Fn– – , if n is odd.
(.)
Theorem . For each n ≥ ,
∞


F
k=n k
– =
Fn Fn– – , if n is even;
if n is odd.
Fn Fn– ,
(.)
© 2015 Wang and Wen; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons
Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction
in any medium, provided the original work is properly credited.
Wang and Wen Journal of Inequalities and Applications (2015) 2015:73
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Further, Wu and Zhang [, ] generalized these identities to the Fibonacci polynomials
and Lucas polynomials and various properties of such polynomials were obtained.
Recently, Holliday and Komatsu [] considered the generalized Fibonacci numbers
which are defined by
Gn+ = aGn+ + Gn ,
n≥
with G =  and G = , and a is a positive integer. They showed that
– ∞

Gn – Gn– ,
if n is even and n ≥ ;
=
Gk
Gn – Gn– – , if n is odd and n ≥ 
k=n
(.)
∞
– 
aGn Gn– – , if n is even and n ≥ ;
=

G
if n is odd and n ≥ .
aG
n Gn– ,
k
k=n
(.)
and
More recently, Wu and Wang [] studied the partial finite sum of the reciprocal Fibonacci numbers and deduced that, for all n ≥ ,
– n

= Fn– .
Fk
(.)
k=n
Inspired by Wu and Wang’s work, we obtain two families of partial finite sums of the reciprocal Fibonacci numbers in this paper, which significantly improve Ohtsuka and Nakamura’s results, Theorems . and .. In addition, we present an alternative proof of (.).
2 Reciprocal sum of the Fibonacci numbers
We first present several well known results on Fibonacci numbers, which will be used
throughout the article. The detailed proofs can be found in [].
Lemma . Let n ≥ , we have
Fn – Fn– Fn+ = (–)n–
(.)
Fa Fb + Fa+ Fb+ = Fa+b+
(.)
and
if a and b are positive integers.
As a consequence of (.), we have the following result.
Corollary . For all n ≥ , we have
Fn = Fn– Fn + Fn Fn+ ,
(.)
Wang and Wen Journal of Inequalities and Applications (2015) 2015:73
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
Fn+ = Fn + Fn+
,
(.)
Fn+ = Fn– Fn+ + Fn Fn+ .
(.)
It is easy to derive the following lemma and we leave the proof as a simple exercise.
Lemma . For each n ≥ , we have
Fn+ Fn+ – Fn– Fn = Fn+ .
(.)
We now establish two inequalities on Fibonacci numbers which will be used later.
Lemma . If n ≥ , then
Fn– Fn– > Fn+ .
(.)
Proof It is easy to see that
Fn– Fn– – Fn+ = Fn– Fn– – (Fn– + Fn )
= Fn– Fn– – Fn– – (Fn– + Fn– )
= (Fn– – )Fn– – Fn– .
Since n ≥ , Fn– –  > . So
Fn– Fn– – Fn+ > Fn– – Fn– > ,
which completes the proof.
Lemma . For each n ≥ , we have
Fn– (Fn + Fn– ) > Fn– Fn– Fn Fn+ .
(.)
Proof Applying (.), we get
Fn– = Fn– Fn– + Fn Fn .
Thus
Fn– (Fn + Fn– ) ≥ (Fn– Fn– + Fn Fn )Fn > Fn Fn > Fn– Fn Fn .
Employing (.), we have
Fn > Fn Fn+ > Fn– Fn+ ,
which yields the desired equation (.).
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The following are some inequalities on the sum of reciprocal Fibonacci numbers.
Proposition . For all n ≥ , we have
n


.
>
Fk Fn– + 
(.)
k=n
Proof For all k ≥ ,

Fk– + 
–

Fk – Fk– – 


=
–
–
Fk Fk– + 
Fk (Fk– + ) Fk– + 
=
(Fk– – )(Fk– + ) – Fk Fk– – Fk
(Fk– + )(Fk– + )Fk
=

– Fk Fk– –  – Fk
Fk–
.
(Fk– + )(Fk– + )Fk

– Fk Fk– = (–)k . Therefore,
Invoking (.), we obtain Fk–

Fk– + 
–

(–)k –  – Fk

=
–
.
Fk Fk– +  (Fk– + )(Fk– + )Fk
Now we have
n
n



(–)k– +  + Fk
–
+
=
Fk Fn– +  Fn– + 
(Fk– + )(Fk– + )Fk
k=n
k=n
>

Fn– + 
–

Fn– + 
+
n
k=n

(Fk– + )(Fk– + )



>
+
–
.
Fn– +  (Fn– + )(Fn– + ) Fn– + 
Because of (.), we have
Fn– +  – (Fn– + )(Fn– + ) = Fn– – Fn– Fn– – Fn– – Fn–

= Fn–
+ Fn– Fn+ – Fn
> .
Thus, we arrive at
n


>
.
Fk Fn– + 
k=n
This completes the proof.
Proposition . Assume that m ≥ . Then, for all even integers n ≥ , we have
mn


<
.
Fk Fn–
k=n
(.)
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Proof By elementary manipulations and (.), we deduce that

Fk–
–


(–)k
–
=
,
Fk Fk– Fk– Fk– Fk
k ≥ .
Hence, for n ≥ , we have
mn
mn

(–)k–


=
–
+
.
Fk Fn– Fmn–
Fk– Fk– Fk
k=n
(.)
k=n
Since n is even,
mn
k=n
(–)k–
< ,
Fk– Fk– Fk
from which we conclude that
mn


<
.
Fk Fn–
k=n
The proof is complete.
Proposition . If n ≥  is odd, then
n


<
.
Fk Fn–
(.)
k=n
Proof It is straightforward to check that the statement is true when n = .
Now we assume that n ≥ . Since n is odd, we have
n
k=n+
(–)k–
< .
Fk– Fk– Fk
Applying (.) and (.) yields




–
<
–
Fn– Fn– Fn Fn– Fn Fn+ Fn–
=
Fn– – Fn Fn+
Fn Fn+ Fn–
=–
Fn– Fn–
Fn Fn+ Fn–
< .
Employing (.) and the above two inequalities, (.) follows immediately.
Proposition . Let m ≥  be given. If n ≥  is odd, we have
mn


>
.
Fk Fn–
k=n
(.)
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Proof It is easy to see that
m


> ,
Fk F
k=
thus (.) holds for n = . Now we assume that n ≥ .
Based on (.) and using the fact n is odd, we have
mn
mn





(–)k–
=
+
–
–
+
.
Fk Fn– Fn– Fn– Fn Fn– Fn Fn+ Fmn–
Fk– Fk– Fk
k=n
k=n+
It is clear that
mn
k=n+
(–)k–
> .
Fk– Fk– Fk
Since m ≥  and invoking (.), we obtain
Fmn– (Fn– + Fn ) ≥ Fn– (Fn– + Fn ) > Fn– Fn– Fn Fn+ ,
which implies


Fn+ – Fn–


–
–
=
–
Fn– Fn– Fn Fn– Fn Fn+ Fmn– Fn– Fn– Fn Fn+ Fmn–
=
Fn– + Fn
Fn– + Fn
–
Fn– Fn– Fn Fn+ Fmn– (Fn– + Fn )
> .
Therefore, (.) also holds for n ≥ .
Now we state our main results on the sum of reciprocal Fibonacci numbers.
Theorem . For all n ≥ , we have
– n

= Fn– .
Fk
(.)
k=n
Proof Combining (.), (.), and (.), we conclude that, for all n ≥ ,

Fn– + 
<
n


<
,
Fk Fn–
k=n
from which (.) follows immediately.
Remark Identity (.) was first conjectured by Professor Ohtsuka, the first author of [].
Based on the formula of Fn and using analytic methods, Wu and Wang [] presented a
proof of (.). In contrast to Wu and Wang’s work, the techniques we use here are more
elementary.
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Theorem . If m ≥  and n ≥ , then
mn
– 
Fn– ,
if n is even;
=
Fk
Fn– – , if n is odd.
k=n
(.)
Proof It is clear that
m
– 
= F .
Fk
(.)
k=
Combining (.) and (.), we find that, for all even integers n ≥ ,

Fn– + 
<
mn


<
.
Fk Fn–
(.)
k=n
Thus (.) and (.) show that, for all m ≥ ,
mn
– 
= Fn– ,
Fk
k=n
provided that n ≥  is even.
Next we aim to prove that, for m ≥  and all odd integers n ≥ ,
– mn

= Fn– – .
Fk
(.)
k=n
If n = , we can readily see that
m

> ,
Fk
k=
thus (.) holds for n = . So in the rest of the proof we assume that n ≥ .
It is not hard to derive that, for all k ≥ ,

Fk– – 
–

(–)k –  + Fk

=
> .
–
Fk Fk– –  Fk (Fk– – )(Fk– – )
Hence, we get
mn




–
<
.
<
Fk Fn– –  Fmn– –  Fn– – 
(.)
k=n
Finally, combining (.) with (.) yields (.).
Remark As m → ∞, (.) becomes (.). Hence our result, Theorem ., substantially
improves Theorem ..
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3 Reciprocal square sum of the Fibonacci numbers
We first give several preliminary results which will be used in our later proofs.
Lemma . For all n ≥ ,
Fn Fn+ – Fn– Fn+ = (–)n– .
(.)
Proof It is easy to show that
Fn Fn+ – Fn– Fn+ = Fn Fn+ – Fn– (Fn + Fn+ )
= Fn Fn+ – Fn Fn– – Fn– Fn+
= Fn (Fn+ – Fn– ) – Fn– Fn+
= Fn – Fn– Fn+ .
Employing (.), the desired result follows.
Proposition . Given an integer m ≥  and let n ≥  be odd, we have
mn


<
.

F
F
n– Fn
k=n k
Proof It is straightforward to check that, for each k ≥ ,




–
–  –
Fk– Fk Fk Fk+
Fk+ Fk+
=


Fk+ – Fk– Fk+
Fk+ – Fk– Fk Fk+ – Fk– Fk Fk+
Fk Fk+

Fk– Fk Fk+
Fk+
=

+ Fk )
Fk Fk+ (Fk+ Fk+ – Fk– Fk ) – Fk– Fk+ (Fk+

Fk– Fk Fk+
Fk+
=
(Fk Fk+ – Fk– Fk+ )Fk+

Fk– Fk Fk+
Fk+
=
(–)k– Fk+
,

Fk– Fk Fk+
Fk+
where the last equality follows from (.).
Since n is odd, we have




–
–  –
> .
Fn– Fn Fn Fn+
Fn+ Fn+
If m is even, then
mn




<
–
<
.

F
F
F
F
F
Fn
F
n–
n
mn
mn+
n–
k=n k
(.)
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If m is odd, then
mn





<
–
+  <
.

Fn– Fn Fmn– Fmn Fmn Fn– Fn
F
k=n k
Thus, (.) always holds.
Proposition . Let n be odd, then we have
n


.
>

Fn– Fn + 
F
k=n k
(.)
Proof Invoking (.), we can readily derive that
⎧
⎨ –  Fk– Fk +
, if k is odd;



Fk (Fk– Fk +)(Fk Fk+ +)
– –
=
Fk Fk+ +
Fk– Fk +  Fk Fk Fk+ +  ⎩ – 
, if k is even.
F (F F +)(F F +)
k
k– k
k k+
Now we have
n


Fn– Fn + 
+
=


Fn– Fn + 
Fn (Fn– Fn + )(Fn Fn+ + )
F
k=n k
Fn+ Fn+ + 
+ ···

Fn+
(Fn Fn+ + )(Fn+ Fn+ + )

Fn Fn+ + 
–
+ 
Fn Fn+ + 
Fn (Fn– Fn + )(Fn Fn+ + )
+
>
Fn– Fn + 


+
–
.
Fn– Fn +  Fn (Fn– Fn + )(Fn Fn+ + ) Fn Fn+ + 
It is obvious that Fk– Fk ≥ Fk– Fk+ . From (.) and the fact that n is odd, we obtain
Fn– Fn +  Fn– Fn+ +  Fn
≥
=  = ,
Fn
Fn
Fn
which implies that
n




+
–
.
>

Fn– Fn +  (Fn– Fn + )(Fn Fn+ + ) Fn Fn+ + 
F
k=n k
By (.) and (.), we have

Fn Fn+ +  > (Fn– Fn + Fn Fn+ ) Fn + Fn+
> (Fn– Fn + )(Fn Fn+ + ),
from which we conclude that
n


.
>

F
F
F
n– n + 
k=n k
The proof is complete.
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Proposition . Suppose that m ≥  and n >  is even. Then
mn


.
<

Fn– Fn – 
F
k=n k
(.)
Proof Applying (.), we can rewrite Fk as
Fk = Fk– Fk Fk+ + (–)k– Fk .
(.)
In addition,
(Fk– Fk – )(Fk Fk+ – ) = Fk– Fk Fk+ – Fk– Fk – Fk Fk+ + 
= Fk– Fk Fk+ – Fk– Fk – Fk + .
(.)
Combining (.) and (.) yields
Fk




– –
=
– 
Fk– Fk –  Fk Fk Fk+ –  (Fk– Fk – )(Fk Fk+ – ) Fk
=
Fk– Fk –  + Fk + (–)k– Fk
(Fk– Fk – )(Fk Fk+ – )Fk
≥
Fk– Fk – 
(Fk– Fk – )(Fk Fk+ – )Fk
> .
Therefore,
mn




–
<
,
<

Fn– Fn –  Fmn Fmn+ –  Fn– Fn – 
F
k=n k
which completes the proof.
Proposition . If n >  is even, then
n


>
.

Fn– Fn
F
k=n k
Proof Employing (.), we can deduce that

Fk– Fk
–


(–)k–
–
=
.

Fk Fk Fk+ Fk– Fk Fk+
Hence, since n is even, we have
n
n



(–)k
=
+
–


F
F
F
F
F
F
F
F
n– n
n n+
k=n k
k=n k– k k+




=
+
–
–

Fn– Fn
Fn– Fn Fn+ Fn Fn+
Fn+ Fn Fn+
+
n–

(–)k
+
.


F
F
F
F
F
k–
k+
n–
n Fn+
k
k=n+
(.)
Wang and Wen Journal of Inequalities and Applications (2015) 2015:73
Page 11 of 13
It is easy to see that
n–
(–)k–
> ,
F F F
k=n+ k– k k+
thus
n





>
+
–
–
.

Fn– Fn Fn+ Fn Fn+
F  Fn– Fn
Fn+ Fn Fn+
k=n k
We claim that



–
–
> .


Fn– Fn Fn+ Fn Fn+ Fn+ Fn Fn+
First, by (.), we have


Fn+
Fn+

–
–
=
–
.





Fn– Fn Fn+ Fn Fn+ Fn+ Fn Fn+ Fn– Fn Fn+ Fn+ Fn Fn+
It follows from (.), (.), and (.) that
Fn > Fn– Fn ,

Fn+ > Fn+
,
Fn+ > Fn Fn+ ,
which implies that


> Fn– Fn Fn+
Fn+ .
Fn Fn+
Thus we obtain



–
–
> ,


Fn– Fn Fn+ Fn Fn+ Fn+ Fn Fn+
which yields the desired (.).
Now we introduce our main result on the square sum of reciprocal Fibonacci numbers.
Theorem . For all n ≥  and m ≥ , we have
– mn

Fn Fn– ,
=

F
F
n Fn– – ,
k=n k
if n is odd;
if n is even.
(.)
Proof We first consider the case when n is odd. If n = , the result is clearly true. So we
assume that n ≥ .
It follows from (.) that
mn
n



.
≥
>


F
F
F
F
n–
n+
k=n k
k=n k
(.)
Wang and Wen Journal of Inequalities and Applications (2015) 2015:73
Page 12 of 13
Employing (.) and (.) yields



<
<
,

Fn– Fn + 
Fk Fn– Fn
mn
k=n
which implies that, if n >  is odd, we have
– mn

= Fn Fn– .
F
k=n k
We now consider the case where n >  is even. It follows from (.) that
mn
n



≥
>
.


Fn– Fn
F
F
k=n k
k=n k
(.)
Combining (.) and (.), we arrive at

Fn– Fn
<
mn


<
,

Fn– Fn – 
F
k=n k
from which we find that, if n >  is even,
mn
– 
= Fn Fn– – .
F
k=n k
This completes the proof.
Remark Theorem . can be regarded as the limiting case as m → ∞ in (.).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to deriving all the results of this article, and read and approved the final manuscript.
Acknowledgements
This work was supported by the National Natural Science Foundation of China (No. 11401080) and the Research Project
of Education Teaching Reform of University of Electronic Science and Technology of China (No. 2013XJYEL032). The
authors would like to thank the referees for helpful comments leading to an improvement of an earlier version.
Received: 14 November 2014 Accepted: 12 February 2015
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