Wang and Zhang Advances in Difference Equations (2016) 2016:27
DOI 10.1186/s13662-016-0761-2
RESEARCH
Open Access
The reciprocal sums of the Fibonacci
3-subsequences
Andrew YZ Wang* and Fan Zhang
*
Correspondence:
[email protected]
School of Mathematical Sciences,
University of Electronic Science and
Technology of China, Chengdu,
611731, P.R. China
Abstract
A Fibonacci 3-subsequence is a subsequence of the type Fn , Fn+3 , Fn+6 , . . . , where Fk
denotes the kth Fibonacci number. In this article, we investigate the reciprocal sums
of the Fibonacci 3-subsequences and obtain several interesting families of identities
involving the Fibonacci numbers.
MSC: 11B39
Keywords: Fibonacci 3-subsequence; reciprocal sum; floor function
1 Introduction
The Fibonacci sequence is defined by the linear recurrence relation
for n ≥ ,
Fn = Fn– + Fn–
where Fn is called the nth Fibonacci number with F =  and F = . There exists a simple
and nonobvious 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, Shimomura, and Shiokawa [–] investigated the algebraic relations for reciprocal sums of the Fibonacci numbers. Ohtsuka and
Nakamura [] studied the partial infinite sums of the reciprocal Fibonacci numbers. They
established the following results, where · denotes the floor function.
Theorem . For all n ≥ ,
∞

Fk
k=n
– =
Fn–
Fn– – 
if n is even,
if n is odd.
(.)
© 2016 Wang and Zhang. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License
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Wang and Zhang Advances in Difference Equations (2016) 2016:27
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Theorem . For all n ≥ ,
∞
– 
Fn Fn– – 
=

F
Fn Fn–
k=n k
if n is even,
if n is odd.
(.)
Recently, Wang and Wen [] considered the partial finite sum of the reciprocal
Fibonacci numbers and strengthened Theorem . and Theorem . to the finite sum
case.
Theorem .
(i) For all n ≥ ,
n

Fk
– = Fn– .
(.)
k=n
(ii) If m ≥  and n ≥ , then
– mn

Fn–
=
Fk
Fn– – 
k=n
if n is even,
if n is odd.
(.)
Theorem . For all m ≥  and n ≥ , we have
mn
– 
Fn Fn– – 
=

F
Fn Fn–
k=n k
if n is even,
if n is odd.
(.)
Furthermore, the present authors [] studied the reciprocal sums of even and odd terms
in the Fibonacci sequence and obtained the following main results.
Theorem . We have
mn
– 
Fn–
=
Fk
Fn– – 
k=n
if m =  and n ≥ ,
if m ≥  and n ≥ .
(.)
Theorem . For all n ≥  and m ≥ , we have
mn
k=n

– = Fn– .
Fk–
(.)
Theorem . If n ≥  and m ≥ , then
mn
– 
= Fn– – .
F
k=n k
(.)
Theorem . For all n ≥  and m ≥ , we have
mn
k=n


Fk–
– = Fn– .
(.)
Wang and Zhang Advances in Difference Equations (2016) 2016:27
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In this article, applying elementary methods, we investigate the reciprocal sums of
the Fibonacci -subsequences, by which we mean the subsequences of the type Fn , Fn+ ,
Fn+ , . . . and obtain several interesting families of identities involving the Fibonacci numbers.
2 Main results I: the reciprocal sums
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 . For any positive integers a and b, we have
Fa Fb + Fa+ Fb+ = Fa+b+ .
(.)
As a consequence of (.), we have the following conclusion.
Corollary . If n ≥ , then
Fn+ = Fn– Fn+ + Fn Fn+ ,
(.)
Fn+ = Fn+ Fn+ – Fn– Fn .
(.)
The following interesting identity concerning the Fibonacci numbers plays a central role
in the proofs of our main results.
Lemma . Assume that a, b, c are given nonnegative integers with a ≥ b. For n ≥ a + c,
we have
Fn+a Fn–a–c – Fn+b Fn–b–c = (–)n–a–c+ Fa+b+c Fa–b .
(.)
Proof We proceed by induction on n. It is clearly true for n = a + c. Assuming that the
result holds for any integer n ≥ a + c, we show that the same is true for n + .
First, it is easy to check that
Fn–+a Fn–a–c + Fn+a Fn––a–c = Fn–+a Fn–a–c + Fn+a (Fn+–a–c – Fn–a–c )
= Fn–+a Fn–a–c + Fn+a Fn+–a–c – Fn+a Fn–a–c
= Fn–c – Fn+a Fn–a–c ,
where the last equality follows from (.).
Now we have
Fn++a Fn+–a–c = (Fn–+a + Fn+a )(Fn––a–c + Fn–a–c )
= Fn–+a Fn–a–c + Fn+a Fn––a–c + Fn+a Fn–a–c + Fn–+a Fn––a–c
= Fn–c + Fn–+a Fn––a–c .
Similarly, we get Fn++b Fn+–b–c = Fn–c + Fn–+b Fn––b–c .
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Therefore, by the induction hypothesis, we arrive at
Fn++a Fn+–a–c – Fn++b Fn+–b–c = F(n–)+a F(n–)–a–c – F(n–)+b F(n–)–b–c
= (–)n–a–c Fa+b+c Fa–b ,
which completes the induction on n.
Before introducing our main results, we first establish an inequality.
Lemma . If n ≥ , then
Fn– Fn Fn+ > Fn+ .
(.)
Proof It follows from (.) that
Fn+ Fn+ > Fn+ .
Since n ≥ , we have
Fn– Fn – Fn+ = Fn– Fn – (Fn + Fn+ )
= Fn– Fn – (Fn– + Fn )
= Fn (Fn– – ) – Fn–
> Fn – Fn–
> .
Thus, Fn– Fn Fn+ > Fn+ Fn+ > Fn+ .
Theorem . For all n ≥ ,
– n

= Fn– .
Fk
(.)
k=n
Proof Setting n = k, a = , b = , and c =  in (.), we obtain
Fk+ Fk– – Fk Fk– = (–)k–  = (–)k .
(.)
It is straightforward to check that, for k ≥ ,



Fk Fk+ – Fk– Fk+ – Fk Fk–
–
–
=
Fk– Fk Fk+
Fk– Fk Fk+
=
Fk+ (Fk – Fk– ) – Fk Fk–
Fk– Fk Fk+
=
Fk+ Fk– – Fk Fk–
(–)k
=
,
Fk– Fk Fk+
Fk– Fk Fk+
where the last equality follows from (.).
(.)
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Therefore, we get that
n
n

(–)k–


=
–
+
.
Fk Fn– Fn+
Fk– Fk Fk+
k=n
k=n
If n is even, it is clear that
n
k=n
(–)k–
< ;
Fk– Fk Fk+
thus,
n


<
.
Fk Fn–
k=n
Now we consider the case where n is odd. It follows from (.) and the condition n ≥ 
that
Fn– Fn Fn+ > Fn+ ,
from which we derive
n
n




(–)k–
=
–
+
+
Fk Fn– Fn+ Fn– Fn Fn+
Fk– Fk Fk+
k=n
k=n+
<
<



–
+
Fn– Fn+ Fn– Fn Fn+

Fn–
.
Hence, we always have
n


<
,
Fk Fn–
(.)
k=n
irrespective of the parity of n.
If we let n = k, a = , b = , and c =  in (.), then
Fk+ Fk– – Fk Fk– = (–)k– .
(.)
By elementary manipulations and (.) we deduce that, for k ≥ ,

Fk– + 
–

(–)k  – Fk– – (Fk+ + )

=
–
Fk Fk+ + 
(Fk– + )Fk (Fk+ + )
=
(–)k  – Fk–

–
(Fk– + )Fk (Fk+ + ) (Fk– + )Fk
<–

,
(Fk– + )Fk
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which implies that
n
n




–
+
>
Fk Fn– +  Fn+ + 
(Fk– + )Fk
k=n
k=n
>



.
–
+
Fn– +  Fn+ +  (Fn– + )Fn
It follows from (.) that
Fn+ = Fn– Fn+ + Fn Fn+
> Fn– Fn + Fn
= (Fn– + )Fn ,
from which we get that
n


>
.
Fk Fn– + 
(.)
k=n
Combining (.) and (.), we have

Fn– + 
<
n


<
,
Fk Fn–
k=n
which yields the desired identity.
We now study a generalization of Theorem . and start with an inequality.
Lemma . If m ≥  and n ≥ , then



–
>
.
Fn– Fn Fn+ Fn+ Fn+ Fn+ Fmn+
(.)
Proof Letting a = b = n –  in (.), we obtain Fn Fn < Fn– . If we set a = n –  and b = n + 
in (.), then Fn– Fn+ < Fn+ . Therefore,
Fn Fn Fn+ < Fn+ ,
from which we derive that






–
–
≥
–
–
Fn– Fn Fn+ Fn+ Fn+ Fn+ Fmn+ Fn– Fn Fn+ Fn+ Fn+ Fn+ Fn+
>



–
–
Fn– Fn Fn+ Fn+ Fn+ Fn+ Fn Fn Fn+
=
Fn–
Fn–
–
Fn– Fn Fn Fn+ Fn– Fn+ Fn+ Fn+
> ,
which completes the proof.
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Theorem . If m ≥  and n ≥ , then
mn
– 
Fn–
if n is even,
=
Fk
Fn– –  if n is odd.
k=n
(.)
Proof Applying the same argument as in the proof of Theorem ., it is easy to see that if
n is even, then

Fn– + 
<
mn


<
,
Fk Fn–
k=n
and thus the statement is true when n is even. We now concentrate ourselves on the case
where n is odd.
Invoking (.), a direct calculation shows that, for k ≥ ,

Fk (Fk+ – Fk– ) – (Fk– – )(Fk+ – )


–
=
–
Fk– –  Fk Fk+ – 
(Fk– – )Fk (Fk+ – )
=
(Fk Fk– – Fk+ Fk– ) + Fk– + Fk+ – 
(Fk– – )Fk (Fk+ – )
=
(–)k  + Fk– + Fk+ – 
(Fk– – )Fk (Fk+ – )
> .
Hence, we arrive at
mn




–
<
.
<
Fk Fn– –  Fmn+ –  Fn– – 
(.)
k=n
Employing (.) and the fact that n is odd, we conclude that
mn
mn

(–)k–


=
–
+
Fk Fn– Fmn+
Fk– Fk Fk+
k=n
k=n
>
>




–
+
–
Fn– Fmn+ Fn– Fn Fn+ Fn+ Fn+ Fn+

Fn–
,
(.)
where the last inequality follows from Lemma ..
It follows from (.) and (.) that if m ≥  and n is odd, then

Fn–
<
mn


,
<
Fk Fn– – 
k=n
from which the desired identity follows immediately.
Applying a similar analysis to the subsequences {Fk+ } and {Fk+ }, we obtain the following results, whose proofs are omitted here.
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Theorem . For all n ≥ ,
n
k=n

– = Fn– .
Fk+
(.)
Theorem . If m ≥  and n ≥ , then
mn
k=n

– =
Fk+
Fn– – 
Fn–
if n is even,
if n is odd.
(.)
Theorem . For all n ≥ ,
n
k=n

– = Fn .
Fk+
(.)
Theorem . If m ≥  and n ≥ , then
mn
k=n

– Fk+
=
Fn
Fn – 
if n is even,
if n is odd.
(.)
3 Main results II: the reciprocal square sums
In the rest of the article, we study the reciprocal square sums of the Fibonacci -subsequences.
Lemma . Let a and b be two given integers with a ≥ b ≥ . For all n ≥ a, we have




Fn+a
Fn–a
– Fn+b
Fn–b
= Fa+b Fa–b Fa + (–)a–b Fb + (–)n–a+ Fn .
(.)
Proof Employing (.), we derive that




Fn+a
Fn–a
– Fn+b
Fn–b
= (Fn+a Fn–a – Fn+b Fn–b )(Fn+a Fn–a + Fn+b Fn–b )
= (–)n–a+ Fa+b Fa–b Fn + Fn+a Fn–a – Fn + Fn+b Fn–b – Fn
= (–)n–a+ Fa+b Fa–b Fn + (–)n–a+ Fa + (–)n–b+ Fb
= Fa+b Fa–b Fa + (–)a–b Fb + (–)n–a+ Fn .
The proof is complete.
Lemma . If n ≥ , we have





Fn+ – Fn
– Fn
> Fn
+ .
Fn+
(.)
Proof First, it is easy to see that


– Fn
= (Fn+ + Fn )(Fn+ – Fn ) = Fn+ Fn+ .
Fn+
(.)
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Similarly, we have


– Fn
= Fn+ Fn+ .
Fn+
(.)

It follows from (.) that Fn+ > Fn
+  and Fn+ > Fn+ Fn+ + . Therefore,


Fn+ Fn+ + Fn
.
Fn+ Fn+ > Fn
(.)
The desired result follows immediately from (.), (.), and (.).
Lemma . For n ≥ , we have

(Fn

Fn
– 



– Fn– )Fn (Fn+

– Fn
)
–

(Fn+

Fn+
+ 



– Fn )Fn+ (Fn+

– Fn+
)
>

 
Fn
Fn+
.






–  > Fn
– Fn–
and Fn+
+  < Fn+
– Fn+
, so
Proof It is easy to see that Fn


– 
Fn
>  






(Fn – Fn– )Fn (Fn+ – Fn ) Fn (Fn+ – Fn
)
and


Fn+
+ 
.
>








(Fn+
– Fn
)Fn+
(Fn+
– Fn
)Fn+
(Fn+
– Fn+
)
Hence,

(Fn
>

– 
Fn



– Fn– )Fn
(Fn+



Fn
(Fn+

– Fn
)

– Fn
)
–

(Fn+
–

(Fn+

Fn+
+ 



– Fn )Fn+
(Fn+

– Fn+
)


=   ,


– Fn )Fn+ Fn Fn+
which completes the proof.
We now present the first reciprocal square sum of the Fibonacci -subsequence.
Theorem . If n ≥  and m ≥ , then
mn
– 


Fn
– Fn–
=


F
– Fn–
–
Fn
k=n k
if n is even,
if n is odd.
(.)
Proof We first consider the case where n is even. Applying (.), by a direct calculation we
get that, for k ≥ ,

Fk
=



–  – 


– Fk– Fk Fk+ – Fk








(Fk+
– Fk
+ Fk–
) – (Fk
– Fk–
)(Fk+
– Fk
)
Fk





(Fk – Fk– )Fk (Fk+ – Fk )
Wang and Zhang Advances in Difference Equations (2016) 2016:27
=



Fk–
– Fk
Fk+





(Fk
– Fk–
)Fk
(Fk+
– Fk
)
=

 + (–)k Fk
,





(Fk
– Fk–
)Fk
(Fk+
– Fk
)
Page 10 of 15
(.)
from which we get
mn
mn

(–)k Fk
+



.
= 
– 
–








F
F
–
F
F
–
F
(F
–
F
)F
(F
n
n–
mn+
mn
k
k– k k+ – Fk )
k=n k
k=n
Since n is even, it is easy to see that
mn
k=n

(–)k Fk
+
> ,





(Fk
– Fk–
)Fk
(Fk+
– Fk
)
which implies that
mn




< 
– 
< 
.




F
Fn – Fn– Fmn+ – Fmn Fn – Fn–
k=n k
(.)
A similar manipulation yields that, for k ≥ ,

Fk

F  – F  +  + (–)k Fk



–  – 
=  k–  k+  



– Fk– +  Fk Fk+ – Fk +  (Fk – Fk– + )Fk (Fk+ – Fk + )
≤–



– Fk–
–  – Fk
Fk+





(Fk
– Fk–
+ )Fk
(Fk+
– Fk
+ )
<–

,



Fk
(Fk+
– Fk
+ )
where the last inequality follows from the easily checked fact





– Fk–
–  – Fk
> Fk
– Fk–
+ .
Fk+
Now we have
mn
mn




–
+
>








F
Fn – Fn– +  Fmn+ – Fmn +  k=n Fk (Fk+ – Fk
+ )
k=n k
>
≥
>



– 
+  




Fn
– Fn–
+  Fmn+
– Fmn
+  Fn
(Fn+ – Fn
+ )

Fn



+  
– 



– Fn– +  Fn (Fn+ – Fn + ) Fn+ – Fn
+

,


Fn
– Fn–
+
where the last inequality follows from (.).
(.)
Wang and Zhang Advances in Difference Equations (2016) 2016:27
Page 11 of 15
Combining (.) and (.), we get that



< 
,
<




Fn – Fn– +  k=n Fk Fn – Fn–
mn
which means that the statement is true when n is even.
We now consider the case where n is odd. It is not hard to derive that



–  – 



Fk
– Fk–
–  Fk
Fk+ – Fk
–
=



+ Fk+
– Fk–
+ 
(–)k Fk





(Fk – Fk– – )Fk (Fk+ – Fk
– )
> ,
where the last inequality follows from the fact that Fk+ = Fk + Fk– .
Therefore, we arrive at
mn




– 
< 
.
< 




F
Fn – Fn– –  Fmn+ – Fmn –  Fn – Fn–
–
k=n k
(.)
Invoking (.) and Lemma ., we derive that if n is odd, then
mn



= 
– 



F
F
–
F
F
n
n–
mn+ – Fmn
k=n k
+
>
=
>

(Fk
k=n

Fn
–
>
mn

Fn
– 

+ 





– Fn– (Fn – Fn– )Fn
(Fn+
– Fn
)

(Fn+

Fn

(–)k– Fk
– 




– Fk– )Fk (Fk+
– Fk
)

+ 
Fn+



– Fn )Fn+
(Fn+

– Fn+
)
–



Fn+
– Fn



+   – 


– Fn– Fn Fn+ Fn+ – Fn



+   –


Fn
– Fn–
Fn Fn+ Fn+ Fn+

Fn

,

– Fn–
(.)
where the last inequality follows from the fact that Fn+ > Fn Fn+ and Fn+ > Fn+ .
It follows from (.) and (.) that



,
<
< 




Fn – Fn– k=n Fk Fn – Fn–
–
mn
which yields the desired result.
For the subsequences {Fk+ } and {Fk+ }, we have similar results.
Wang and Zhang Advances in Difference Equations (2016) 2016:27
Page 12 of 15
Theorem . If n ≥  and m ≥ , then
mn

– =
F
k=n k+


Fn+
– Fn–
–  if n is even,


if n is odd.
Fn+ – Fn–
(.)
Theorem . If n ≥  and m ≥ , then
mn

F
k=n k+
– =


– Fn–
Fn+


–
Fn+ – Fn–
if n is even,
if n is odd.
(.)
Remark The proof of Theorem . is similar to that of Theorem . and is omitted here.
Since the telescoping technique for the proof of Theorem . is very different from that
for Theorem ., we give a detailed proof of Theorem . in the next section.
4 Proof of Theorem 3.6
We first present a preliminary result, which plays a central role in the later proof.
Lemma . For all n ≥ , we have




> Fn+
 Fn+
– Fn+
Fn+




> Fn+
– Fn–
+  Fn+
– Fn+
+ .
Proof The second inequality is obvious, so we concentrate ourselves on the first one. It is
easy to see that



=  · Fn+ Fn+ > Fn+
– Fn+
.
 Fn+
(.)
Employing (.), we get Fn+ = Fn+ (Fn+ + Fn+ ); therefore,


= Fn+
(Fn+ + Fn+ ) .
Fn+
(.)
It follows from Fn+ > Fn+ = Fn+ – Fn+ that (Fn+ + Fn+ ) > Fn+ ; thus,

.
(Fn+ + Fn+ ) > Fn+
Combining (.), (.), and (.) yields the desired result.
(.)
Proof of Theorem . We first consider the case where n is even. Employing (.) again,
by a direct calculation we deduce that, for k ≥ ,

Fk+



–  – 


– Fk– Fk+ Fk+ – Fk+
=








Fk+
(Fk+
– Fk+
+ Fk–
) – (Fk+
– Fk–
)(Fk+
– Fk+
)





(Fk+ – Fk– )Fk+ (Fk+ – Fk+ )
=



Fk+
– Fk+
Fk–





(Fk+
– Fk–
)Fk+
(Fk+
– Fk+
)
=

 + (–)k Fk+
,





(Fk+
– Fk–
)Fk+
(Fk+
– Fk+
)
(.)
Wang and Zhang Advances in Difference Equations (2016) 2016:27
Page 13 of 15
from which we obtain
mn

F
k=n k+
=


– 


– Fn– Fmn+ – Fmn+

Fn+
–
mn

(Fk+
k=n

(–)k Fk+
+
.




– Fk– )Fk+ (Fk+
– Fk+
)
Since n is even, it is easy to see that
mn
k=n

(–)k Fk+
+
> ,





(Fk+
– Fk–
)Fk+
(Fk+
– Fk+
)
which implies that
mn

F
k=n k+
<

Fn+

.

– Fn–
(.)
Similarly, we can derive that, for k ≥ ,



–  – 



Fk+
– Fk–
+  Fk+
Fk+ – Fk+
+
=



– Fk+
+  + (–)k Fk+
Fk–




– Fk– + )Fk+ (Fk+ – Fk+
+ )

(Fk+
< .
We now have
mn
k=n



– 
= 



Fk+
Fn+ – Fn–
+  Fmn+
– Fmn+
+
+
>



Fn+
– Fn–
–  – Fn+

+ 





– Fn– +  (Fn+ – Fn– + )Fn+ (Fn+ – Fn+
+ )

Fn+


– Fn+
+


+ 





Fn+
– Fn–
+  (Fn+
– Fn–
+ )(Fn+
– Fn+
+ )
–
>



Fk+
– Fk–
–  – (–)k Fk+




– Fk– + )Fk+ (Fk+ – Fk+
+ )

(Fk+
k=n

Fn+
–
>
mn

Fn+


– Fn+
+

,


Fn+
– Fn–
+
where the last inequality follows from Lemma ..
(.)
Wang and Zhang Advances in Difference Equations (2016) 2016:27
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Combining (.) and (.) yields that



< 
,
<




Fn+ – Fn– +  k=n Fk+ Fn+ – Fn–
mn
which shows that the statement is true when n is even.
Next we turn to consider the case where n is odd. It is not hard to derive that

Fk+
=



–  – 


– Fk– –  Fk+ Fk+ – Fk+
–



+ Fk+
– Fk–
+ 
(–)k Fk+




– Fk– – )Fk+ (Fk+ – Fk+
– )

(Fk+
> ,
where the last inequality follows from the fact that Fk+ = Fk+ + Fk+ .
Therefore, we obtain
mn

F
k=n k+
<

Fn+



– 
< 
.



– Fn– –  Fmn+ – Fmn+ –  Fn+ – Fn–
–
(.)
Applying (.) and the fact that n is odd, we derive that
mn

F
k=n k+
=

Fn+
+
>
>

(Fk+
k=n

(–)k– Fk+
– 




– Fk– )Fk+ (Fk+
– Fk+
)

+ 
Fn+

– 






(Fn+ – Fn+ )Fn+ (Fn+ – Fn+ ) Fn+ – Fn+



+ 
– 






Fn+
– Fn–
Fn+ (Fn+
– Fn+
) Fn+
(Fn+
– Fn+
)
–
=
mn

Fn+
– 

+







Fn+ – Fn– (Fn+ – Fn– )Fn+ (Fn+
– Fn+
)
–
>


– 


– Fn– Fmn+ – Fmn+

Fn+


– Fn+



+ 
– 




Fn+
– Fn–
Fn+ Fn+
Fn+ – Fn+

Fn+

,

– Fn–
(.)
where the last inequality follows from Lemma ..
It follows from (.) and (.) that



,
<
< 




Fn+ – Fn– k=n Fk+ Fn+ – Fn–
–
mn
which yields the desired result.
Wang and Zhang Advances in Difference Equations (2016) 2016:27
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5 Conclusions
In this paper, we investigate in two ways the reciprocal sums of the Fibonacci -subsequences, where a Fibonacci -subsequence is a subsequence of the type Fn , Fn+ , Fn+ , . . . .
One is focused on the ordinary sums, and the other is concerned with the square sums.
By evaluating the integer parts of the reciprocals of these sums we get several interesting families of identities. The results are new and important to those with closely related
research interests.
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
The authors would like to thank the anonymous reviewers for their helpful comments. This work was supported by the
National Natural Science Foundation of China (No. 11401080).
Received: 7 December 2015 Accepted: 18 January 2016
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