On the Reciprocal Sums of Products of Fibonacci Numbers

Younseok Choo Department of Electronic and Electrical Engineering Hongik University Sejong 30016 Republic of Korea

                      yschoo@hongik.ac.kr

Abstract In this paper we study the reciprocal sums of products of two different Fibonacci numbers. We obtain some identities related to the numbers ⌊(∑_{k=n}^{∞} 1/F_{k}F_{k+m})^{−1}⌋, m ≥ 1, where ⌊·⌋ indicates the floor function.

1 Introduction

As is well known, the Fibonacci numbers F_{n} are generated from the recurrence relation

F_{n} = F_{n−1} + F_{n−2} (n ≥ 2),

with initial condition F_{0} = 0 and F_{1} = 1.

Recently Ohtsuka and Nakamura [7] found interesting properties of the Fibonacci numbers and proved Theorem 1 below. Theorem 1. For the Fibonacci numbers, the following identities hold:

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}} )^{−1}⌋ = \begin{cases}F_{n} − F_{n−1}, & if n ≥ 2 and n is even; \\ F_{n} − F_{n−1} − 1, & if n ≥ 3 and n is odd, \end{cases}  (1)

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}^{2}} )^{−1}⌋ = \begin{cases}F_{n−1}F_{n} − 1, & if n ≥ 2 and n is even; \\ F_{n−1}F_{n}, & if n ≥ 3 and n is odd. \end{cases}  (2)

Following the paper of Ohtsuka and Nakamura [7], diverse results in the same direction have been reported in the literature [1, 2, 3], [5], [8, 9, 10, 11, 12, 13]. Among them, Liu and Wang [5] considered the product of two reciprocal Fibonacci numbers, and obtained several interesting results. For example, they proved Theorem 2 below for the products of two consecutive Fibonacci numbers.

Theorem 2. Let m ≥ 2. Then

⌊( ∑_{k=n}^{mn} \frac{1}{F_{k}F_{k+1}} )^{−1}⌋ = \begin{cases}F_{n}^{2}, & if n ≥ 2 and n is even; \\ F_{n}^{2} − 1, & if n ≥ 3 and n is odd. \end{cases}  (3)

Motivated by Theorem 2, we study the reciprocal sums of products of two different Fibonacci numbers in this paper. We obtain some identities related to the numbers

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} )^{−1}⌋, m ≥ 1.

Remark 3. The following identity was conjectured by Ohtsuka and proved by Bruckman [6]:

( ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} )^{−1} = ∑_{k=1}^{n−1} F_{k}F_{k+m} − \frac{1}{3}F_{m−2(−1)^{n}} + O(\frac{1}{F_{n}^{2}} ), m ≥ 0.

For the case where m = 0 and n is large, (2) also can be derived from the above result.

2 Main results

We will use Lemma 4 below to prove our main results.

Lemma 4 (Koshy [4]). For the Fibonacci numbers, we have

F_{m}F_{n} − F_{m+k}F_{n−k} = (−1)^{n−k}F_{m−n+k}F_{k}.

Our main results are stated in the following theorem.

Theorem 5. For the Fibonacci numbers, (a), (b) and (c) below hold:

(a) Let m ≥ 1. If

\frac{2F_{m} − F_{m+1}}{3} ∉ ℤ,

then there exist positive integers n_{0} and n_{1} such that

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} )^{−1}⌋ = \begin{cases}F_{n+m−1}F_{n} + g_{m} − 1, & if n ≥ n_{0} and n is even; \\ F_{n+m−1}F_{n} − g_{m}, & if n ≥ n_{1} and n is odd, \end{cases}  (4)

where

g_{m} = ⌊\frac{2F_{m} − F_{m+1}}{3} ⌋ + 1.

(b) For m = 2,

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} )^{−1}⌋ = F_{n+m−1}F_{n}, for n ≥ 1.  (5)

(c) Let m ≥ 3. If

\frac{2F_{m} − F_{m+1}}{3} ∈ ℤ,

then there exist positive integers n_{2} and n_{3} such that

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} )^{−1}⌋ = \begin{cases}F_{n+m−1}F_{n} + ĝ_{m} − 1, & if n ≥ n_{2} and n is even; \\ F_{n+m−1}F_{n} − ĝ_{m} − 1, & if n ≥ n_{3} and n is odd, \end{cases}  (6)

where

ĝ_{m} = \frac{2F_{m} − F_{m+1}}{3} .

Proof. (a) To prove (4), consider

X_{1} = \frac{1}{F_{n+m−1}F_{n} + (−1)^{n}g_{m}} − \frac{1}{F_{n+m+1}F_{n+2} + (−1)^{n}g_{m}} − \frac{1}{F_{n}F_{n+m}} − \frac{1}{F_{n+1}F_{n+m+1}}
= \frac{X̂_{1}}{\{F_{n+m−1}F_{n} + (−1)^{n}g_{m}\}\{F_{n+m+1}F_{n+2} + (−1)^{n}g_{m}\}F_{n}F_{n+m}F_{n+1}F_{n+m+1}} ,

where, by the identity F_{n+m+1}F_{n+2} − F_{n+m−1}F_{n} = F_{n}F_{n+m} + F_{n+1}F_{n+m+1}

X̂_{1} = (F_{n}F_{n+m} + F_{n+1}F_{n+m+1})X̃_{1},

with

X̃_{1} = F_{n}F_{n+1}F_{n+m}F_{n+m+1} − F_{n+m−1}F_{n+m+1}F_{n}F_{n+2}
−(−1)^{n}g_{m}(F_{n+m−1}F_{n} + F_{n+m+1}F_{n+2}) − g_{m}^{2}.

From Lemma 4, we have

F_{n+1}F_{n+m} − F_{n+m+1}F_{n} = (−1)^{n}F_{m},
F_{n+m+1}F_{n} − F_{n+m−1}F_{n+2} = (−1)^{n}(F_{m} − F_{m+1}),
F_{n+m+1}F_{n−1} − F_{n+m}F_{n} = (−1)^{n}F_{m+1}.

Then

F_{n}F_{n+1}F_{n+m}F_{n+m+1} − F_{n+m−1}F_{n+m+1}F_{n}F_{n+2}
= F_{n+m+1}F_{n}{F_{n+m+1}F_{n} + (−1)^{n}F_{m}}
−F_{n+m+1}F_{n}{F_{n+m+1}F_{n} + (−1)^{n}(F_{m+1} − F_{m})}
= (−1)^{n}F_{n+m+1}F_{n}(2F_{m} − F_{m+1}),

and

F_{n+m−1}F_{n} + F_{n+m+1}F_{n+2} = 3F_{n+m+1}F_{n} + F_{n+m+1}F_{n−1} − F_{n+m}F_{n}
= 3F_{n+m+1}F_{n} + (−1)^{n}F_{m+1}.

Hence

X̃_{1} = (−1)^{n}F_{n+m+1}F_{n}(2F_{m} − F_{m+1} − 3g_{m}) − g_{m}F_{m+1} − g_{m}^{2}.

Assume that n is even. Since g_{m} > 0 and 2F_{m} − F_{m+1} − 3g_{m} < 0, then X_{1} < 0 and

\frac{1}{F_{n+m−1}F_{n} + g_{m}} − \frac{1}{F_{n+m+1}F_{n+2} + g_{m}} < \frac{1}{F_{n}F_{n+m}} + \frac{1}{F_{n+1}F_{n+m+1}} .

Repeatedly applying the above inequality, we have

\frac{1}{F_{n+m−1}F_{n} + g_{m}} < ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} , if n ≥ 2 and n is even.  (7)

Similarly, if n is odd, then there exists a positive integer m_{1} such that, for n ≥ m_{1}, X_{1} > 0 and

\frac{1}{F_{n}F_{n+m}} + \frac{1}{F_{n+1}F_{n+m+1}} < \frac{1}{F_{n+m−1}F_{n} − g_{m}} − \frac{1}{F_{n+m+1}F_{n+2} − g_{m}} ,

from which we obtain

∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} < \frac{1}{F_{n+m−1}F_{n} − g_{m}} , if n ≥ m_{1} and n is odd.  (8)

Next, consider

X_{2} = \frac{1}{F_{n+m−1}F_{n} + (−1)^{n}g_{m} − 1} − \frac{1}{F_{n+m}F_{n+1} + (−1)^{n+1}g_{m} − 1} − \frac{1}{F_{n}F_{n+m}}
= \frac{X̂_{2}}{\{F_{n+m−1}F_{n} + (−1)^{n}g_{m} − 1\}\{F_{n+m}F_{n+1} + (−1)^{n+1}g_{m} − 1\}F_{n}F_{n+m}} ,

X̂_{2} = F_{n}F_{n+m}^{2}F_{n+1} − F_{n+m}F_{n+m−1}F_{n}F_{n+1} − F_{n}^{2}F_{n+m−1}F_{n+m}
−(−1)^{n}g_{m}(2F_{n}F_{n+m} − F_{n+m−1}F_{n} + F_{n+m}F_{n+1})
+F_{n+m−1}F_{n} + F_{n+m}F_{n+1} + g_{m}^{2} − 1.

From Lemma 4, we have

F_{n+m−1}F_{n} − F_{n+m−2}F_{n+1} = (−1)^{n+1}F_{m−2} = (−1)^{n}(F_{m+1} − 2F_{m}).

Then

F_{n}F_{n+m}F_{n+1}F_{n+m} − F_{n+m}F_{n+m−1}F_{n}F_{n+1} − F_{n}^{2}F_{n+m−1}F_{n+m}
= F_{n}F_{n+m}(F_{n+1}F_{n+m−2} − F_{n}F_{n+m−1})
= (−1)^{n}F_{n}F_{n+m}(2F_{m} − F_{m+1}),

and

2F_{n}F_{n+m} + F_{n+m}F_{n+1} − F_{n+m−1}F_{n}
= 3F_{n}F_{n+m} + F_{n+m}F_{n−1} − F_{n+m−1}F_{n}
= 3F_{n}F_{n+m} + (−1)^{n}(2F_{m+2} − F_{m+3}).

Hence

X̂_{2} = (−1)^{n}F_{n}F_{n+m}(2F_{m} − F_{m+1} − 3g_{m}) + F_{n+m−1}F_{n} + F_{n+m}F_{n+1}
−g_{m}(2F_{m+2} − F_{m+3}) + g_{m}^{2} − 1.

Suppose that n is even. Since

−2 ≤ 2F_{m} − F_{m+1} − 3g_{m} ≤ −1,

then

F_{n}F_{n+m}(2F_{m} − F_{m+1} − 3g_{m}) + (F_{n+m−1}F_{n} + F_{n+m}F_{n+1})
≥ −2F_{n}F_{n+m} + F_{n}F_{n+m−1} + F_{n+1}F_{n+m}
= (F_{n−1} − F_{n})(F_{n+m−1} + F_{n−m−2}) + F_{n}F_{n+m−1}
= F_{n−1}F_{n+m−1} − F_{n−2}F_{n+m−2}
> 0,

and there exists a positive integer m_{2} such that, for n ≥ m_{2}, X_{2} > 0 and

\frac{1}{F_{n}F_{n+m}} < \frac{1}{F_{n+m−1}F_{n} + (−1)^{n}g_{m} − 1} − \frac{1}{F_{n+m}F_{n+1} + (−1)^{n+1}g_{m} − 1}.

Repeatedly applying the above inequality, we have

∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} < \frac{1}{F_{n+m−1}F_{n} + g_{m} − 1}, if n ≥ m_{2} and n is even.  (9)

On the other hand,

X_{3} = \frac{1}{F_{n+m−1}F_{n} + (−1)^{n}g_{m} + 1} − \frac{1}{F_{n+m}F_{n+1} + (−1)^{n+1}g_{m} + 1} − \frac{1}{F_{n}F_{n+m}}
= \frac{X̂_{3}}{\{F_{n+m−1}F_{n} + (−1)^{n}g_{m} + 1\}\{F_{n+m−1}F_{n+1} + (−1)^{n+1}g_{m} + 1\}F_{n}F_{n+m}} ,

where

X̂_{3} = X̂_{2} − 2(F_{n+m−1}F_{n} + F_{n+m}F_{n+1})
= (−1)^{n}F_{n}F_{n+m}(2F_{m} − F_{m+1} − 3g_{m}) − F_{n+m−1}F_{n} − F_{n+1}F_{n+1}
−g_{m}(2F_{m+2} − F_{m+3}) + g_{m}^{2} − 1.

Suppose that n is odd. As shown above, we have

−F_{n}F_{n+m}(2F_{m} − F_{m+1} − 3g_{m}) − F_{n+m−1}F_{n} − F_{n+m}F_{n+1} < F_{n−2}F_{n+m−2} − F_{n−1}F_{n+m−1}. Hence there exists a positive integer m_{3} such that, for n ≥ m_{3}, X_{3} < 0 and

\frac{1}{F_{n+m−1}F_{n} + (−1)^{n}g_{m} + 1} − \frac{1}{F_{n+m}F_{n+1} + (−1)^{n+1}g_{m} + 1} < \frac{1}{F_{n}F_{n+m}} ,

from which we have

\frac{1}{F_{n+m−1}F_{n} − g_{m} + 1} < ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} , if n ≥ m_{3} and n is odd.  (10)

Then, (4) follows from (7), (8), (9) and (10).

(b) Since F_{n+2}F_{n+3} − F_{n}F_{n+1} = F_{n}F_{n+2} + F_{n+1}F_{n+3}, we have

\frac{1}{F_{n}F_{n+1}} − \frac{1}{F_{n+2}F_{n+3}} − \frac{1}{F_{n}F_{n+2}} − \frac{1}{F_{n+1}F_{n+3}} = \frac{F_{n+2}F_{n+3}−F_{n}F_{n+1}−(F_{n}F_{n+2}+F_{n+1}F_{n+3})}{F_{n}F_{n+1}F_{n+2}F_{n+3}} =0,

i.e.,

\frac{1}{F_{n}F_{n+1}} − \frac{1}{F_{n+2}F_{n+3}} = \frac{1}{F_{n}F_{n+2}} + \frac{1}{F_{n+1}F_{n+3}} .

Repeatedly applying the above equality, we obtain (5).

(c) Let m ≥ 3 and assume that

ĝ_{m} = \frac{2F_{m} − F_{m+1}}{3} ∈ ℤ.

We recall the proof of (a). Replacing g_{m} by ĝ_{m} in X̃_{1}, we have

X̃_{1} = −ĝ_{m}F_{m+1} − ĝ_{m}^{2} < 0.

Then X_{1} < 0 if n ≥ 2 and n is even or if n ≥ m_{4} and n is odd for some positive integer m_{4}, and we have

\frac{1}{F_{n+m−1}F_{n} + (−1)^{n}ĝ_{m}} < ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} , if n ≥ 2 (n is even) or if n ≥ m_{4} (n is odd). (11)

Similarly there exist positive integers m_{5} and m_{6} such that X_{2} > 0 if n ≥ m_{5} and n is even, or if n ≥ m_{6} and n is odd, from which we have

∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+m}} < \frac{1}{F_{n+m−1}F_{n} + (−1)^{n}ĝ_{m} − 1}, if n ≥ m_{5} (n is even) or if n ≥ m_{6} (n is odd).

(12)

Then, (6) follows from (11) and (12).

Remark 6. From Theorem 5, we have

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+1}} )^{−1}⌋ = \{ \begin{matrix}F_{n}^{2}, & if n ≥ 2 and n is even; \\ F_{n}^{2} − 1, & if n ≥ 1 and n is odd, \end{matrix}

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}F_{k+6}} )^{−1}⌋ = \{ \begin{matrix}F_{n+5}F_{n}, & if n ≥ 2 and n is even; \\ F_{n+5}F_{n} − 2, & if n ≥ 1 and n is odd, \end{matrix}

etc.

3 Acknowledgments

The author is thankful to the editor-in-chief and to the anonymous referee for their helpful comments which led to the improved presentation of the paper.

References

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2010 Mathematics Subject Classification: Primary 11B39; Secondary 11B37. Keywords: Fibonacci number, reciprocal, floor.

(Concerned with sequence \underline{A000108}.)

Received July 17 2017; revised versions received December 29 2017; January 22 2018. Published in Journal of Integer Sequences, March 8 2018.

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