Alternating Sums of the Reciprocal Fibonacci Numbers

Andrew Yezhou Wang School of Mathematical Sciences

University of Electronic Science and Technology of China

Chengdu 611731 P. R. China

              yzwang@uestc.edu.cn

Tingrui Yuan

School of Communication and Information Engineering University of Electronic Science and Technology of China

Chengdu 611731 P. R. China

               ytr2009@icloud.com

Abstract

In this paper, we investigate the alternating sums of the reciprocal Fibonacci numbers ∑_{k=n}^{mn} (−1)^{k}/F_{ak+b}, where a ∈ {1, 2, 3} and b < a. The integer parts of the reciprocals of these sums are expressed explicitly in terms of the Fibonacci numbers.

1 Introduction

For an integer n ≥ 0, the Fibonacci number F_{n} is defined recurrently by F_{n} = F_{n−1} + F_{n−2} with F_{0} = 0 and F_{1} = 1.

Recently, Ohtsuka and Nakamura [1] studied the infinite sums of the reciprocal Fibonacci numbers, and established the following result, where ⌊·⌋ denotes the floor function.

Theorem 1. For all n ≥ 2,

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

More recently, Wang and Wen [4] strengthened Theorem 1 to the finite sum case. Theorem 2. If m ≥ 3 and n ≥ 2, then

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

In this article, we focus on the alternating sums of the reciprocal Fibonacci numbers

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{ak+b}} ,

where a ∈ {1, 2, 3} and b < a. By evaluating the integer parts of these sums, we obtain several interesting families of identities concerning the Fibonacci numbers.

2 Results for a = 1

We first introduce several well-known results, which will be used throughout the article. The detailed proofs can be found in, for example, [3, Thm. 7, p. 9] and [2]. Lemma 3. For any positive intergers m and n, we have

F_{m}F_{n} + F_{m+1}F_{n+1} = F_{m+n+1}.

Lemma 4. For all n ≥ 1, we have

F_{2n+1} = F_{n+1}F_{n+2} − F_{n−1}F_{n}.

Lemma 5. Let a, b, c, d be positive integers with a + b = c + d and b ≥ max{c, d}. Then

F_{a}F_{b} − F_{c}F_{d} = (−1)^{a+1}F_{b−c}F_{b−d}.

For the sake of argument, we present four auxiliary functions

f_{1}(n) = \frac{1}{F_{n+1}} − \frac{(−1)^{n}}{F_{n}} − \frac{1}{F_{n+2}} ,
f_{2}(n) = \frac{1}{F_{n+1} − 1} − \frac{(−1)^{n}}{F_{n}} − \frac{1}{F_{n+2} − 1},
f_{3}(n) = \frac{−1}{F_{n+1} + 1} − \frac{(−1)^{n}}{F_{n}} + \frac{1}{F_{n+2} + 1},
f_{4}(n) = \frac{−1}{F_{n+1}} − \frac{(−1)^{n}}{F_{n}} + \frac{1}{F_{n+2}} .

It is clear that f_{i}(n) (1 ≤ i ≤ 4) is positive if n is odd, and negative otherwise. Lemma 6. If n ≥ 2 is even, then

f_{1}(n) + f_{1}(n + 1) < 0.

Proof. Since n is even, it is straightforward to see

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

where the last equality follows from Lemma 5 and the fact that n is even. Lemma 7. For all n ≥ 2, we have

f_{2}(n) + f_{2}(n + 1) > 0.

Proof. The statement is clearly true if n is odd. Thus, we focus on the case where n is even. It follows from the definition of f_{2}(n) and Lemma 5 that

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

which completes the proof.

Lemma 8. For all n ≥ 2, we have

\frac{F_{n+2}}{(F_{n+1} − 1) (F_{n+3} − 1)} − \frac{F_{n−1}}{F_{n}F_{n+1}} − \frac{1}{F_{2n+1} − 1} ≥ 0.

Proof. Applying Lemma 3, it is easy to see that, for n ≥ 2,

F_{2n+1} − 1 − 2F_{n}F_{n+1} = F_{n}^{2} + F_{n+1}^{2} − 2F_{n}F_{n+1} − 1 = (F_{n+1} − F_{n})^{2} − 1 ≥ 0, from which we derive the conclusion that

\frac{1}{F_{2n+1} − 1} ≤ \frac{1}{2F_{n}F_{n+1}} .

Therefore, we have

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

≥ \frac{F_{n+2}}{(F_{n+1} − 1) (F_{n+3} − 1)} − \frac{F_{n−1}}{F_{n}F_{n+1}} − \frac{1}{2F_{n}F_{n+1}}

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

whose numerator is

ψ(n) := 2F_{n}F_{n+1}F_{n+2} − (2F_{n−1} + 1) (F_{n+1} − 1) (F_{n+3} − 1) .

Applying Lemma 5 repeatedly and the fact F_{n+3} = 3F_{n+1} − F_{n−1}, we can obtain

ψ(n) = 2F_{n+1} (F_{n}F_{n+2} − F_{n−1}F_{n+3}) + 2F_{n−1}F_{n+1} + 2F_{n−1}F_{n+3} − F_{n+1}F_{n+3}
+ (F_{n+1} + F_{n+3}) − 2F_{n−1} − 1
= ((−1)^{n+1} + 1) 4F_{n+1} + 2F_{n−1}F_{n+1} + (2F_{n−1} − F_{n+1})F_{n+3} − 3F_{n−1} − 1
= ((−1)^{n+1} + 1) 4F_{n+1} + F_{n−1}(2F_{n+1} − F_{n+2}) + (F_{n−1}F_{n+2} − F_{n−2}F_{n+3})
− 3F_{n−1} − 1
= ((−1)^{n+1} + 1) 4F_{n+1} + F_{n−1}^{2} − 3F_{n−1} − 1 + (−1)^{n}3.

If n is even, we have ψ(n) = (F_{n−1} − 1)(F_{n−1} − 2) ≥ 0. If n is odd, we have

ψ(n) = (F_{n−1} + 1)(F_{n−1} + 4) + 8(F_{n} − 1) > 0.

Therefore, ψ(n) ≥ 0 always holds. This completes the proof. Lemma 9. If n ≥ 2 and m ≥ 2, then

f_{2}(n) + f_{2}(n + 1) + f_{2}(mn) + \frac{1}{F_{mn+2} − 1} > 0.

Proof. If mn is odd, then the result follows from Lemma 7 and the fact f_{2}(mn) > 0. So we assume that mn is even. Now we have

f_{2}(mn) + \frac{1}{F_{mn+2} − 1} = \frac{1}{F_{mn+1} − 1} − \frac{1}{F_{mn}} = \frac{−(F_{mn−1} − 1)}{F_{mn}(F_{mn+1} − 1)} > \frac{−1}{F_{mn+1} − 1}.

From the proof of Lemma 7 we know that whether n is even or odd,

f_{2}(n) + f_{2}(n + 1) ≥ \frac{F_{n+2}}{(F_{n+1} − 1) (F_{n+3} − 1)} − \frac{F_{n−1}}{F_{n}F_{n+1}} .

Therefore,

f_{2}(n) + f_{2}(n + 1) + f_{2}(mn) + \frac{1}{F_{mn+2} − 1} > \frac{F_{n+2}}{(F_{n+1} − 1) (F_{n+3} − 1)} − \frac{F_{n−1}}{F_{n}F_{n+1}} − \frac{1}{F_{mn+1} − 1}

≥ \frac{F_{n+2}}{(F_{n+1} − 1) (F_{n+3} − 1)} − \frac{F_{n−1}}{F_{n}F_{n+1}} − \frac{1}{F_{2n+1} − 1}

≥ 0,

where the last inequality follows from Lemma 8.

Employing the fact 2(F_{2n+2} + 1) ≥ (F_{n+1} + 1)(F_{n+3} + 1) and similar arguments in the proof of Lemma 8, we have the following result, whose proof is omitted here.

Lemma 10. If n ≥ 5 is odd, then

f_{3}(n) + f_{3}(n + 1) > \frac{1}{F_{2n+2} + 1}.

Now we establish two properties about f_{4}(n).

Lemma 11. For n ≥ 1, we have

f_{4}(n) + f_{4}(n + 1) < 0.

Proof. If n is even, the result follows from the definition of f_{4}(n). Next we consider the case where n is odd. Applying the argument in the proof of Lemma 6, we can easily deduce that

f_{4}(n) + f_{4}(n + 1) = \frac{−2}{F_{n+1}} + \frac{1}{F_{n}} + \frac{1}{F_{n+3}} = \frac{−2}{F_{n}F_{n+1}F_{n+3}} < 0.

This completes the proof.

Lemma 12. If n ≥ 1 and m ≥ 2, then

f_{4}(n) + f_{4}(n + 1) + f_{4}(mn) < 0.

Proof. If mn is even, the result follows from Lemma 11 and the fact f_{4}(mn) < 0. So we assume that mn is odd, which implies that m ≥ 3 and n is odd. Since mn is odd, we have

f_{4}(mn) = \frac{−1}{F_{mn+1}} + \frac{1}{F_{mn}} + \frac{1}{F_{mn+2}} < \frac{1}{F_{mn}} ≤ \frac{1}{F_{3n}} .

Now we have

f_{4}(n) + f_{4}(n + 1) + f_{4}(mn) < \frac{−2}{F_{n}F_{n+1}F_{n+3}} + \frac{1}{F_{3n}} .

To complete the proof, we only need to show that 2F_{3n} > F_{n}F_{n+1}F_{n+3}.

It follows from Lemma 3 that F_{2n+2} = F_{n−1}F_{n+2} + F_{n}F_{n+3}, which implies F_{n}F_{n+1}F_{n+3} < F_{n+1}F_{2n+2}. Furthermore, employing Lemma 3 again, we can conclude that

F_{n+1}F_{2n+2} = (F_{n−1} + F_{n})(F_{2n} + F_{2n+1})
= (F_{n−1}F_{2n} + F_{n}F_{2n+1}) + F_{n−1}F_{2n+1} + F_{n}F_{2n}
= F_{3n} + F_{n−1}F_{2n+1} + F_{n+1}F_{2n} − F_{n−1}F_{2n}
= F_{3n} + (F_{n−1}F_{2n−1} + F_{n+1}F_{2n})
< 2F_{3n},

which completes the proof.

Theorem 13. If n ≥ 4 and m ≥ 2, then

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

Proof. We first consider the case where n is even. It follows from Lemma 6 that

∑_{k=n}^{mn−1} f_{1}(k) < 0.

It is clear that mn is even, which ensures that

f_{1}(mn) + \frac{1}{F_{mn+2}} < 0.

With the help of f_{1}(n) and the above two inequalities, we can obtain

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{k}} = \frac{1}{F_{n+1}} − (\frac{1}{F_{mn+2}} + f_{1}(mn)) − ∑_{k=n}^{mn−1} f_{1}(k) > \frac{1}{F_{n+1}} .

Applying Lemma 7 and Lemma 9, we have

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{k}} = \frac{1}{F_{n+1} − 1} − (f_{2}(n) + f_{2}(n + 1) + f_{2}(mn) + \frac{1}{F_{mn+2} − 1}) − ∑_{k=n+2}^{mn−1} f_{2}(k)
< \frac{1}{F_{n+1} − 1}.

Therefore, we obtain

\frac{1}{F_{n+1}} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{k}} < \frac{1}{F_{n+1} − 1},

which shows that the statement is true when n is even. We now turn to consider the case where n ≥ 5 is odd. If mn is odd, it is easy to see that

f_{3}(mn) − \frac{1}{F_{mn+2} + 1} > 0.

Lemma 10 tells us that f_{3}(n) + f_{3}(n + 1) > 0. Therefore,

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{k}} = \frac{−1}{F_{n+1} + 1} − ∑_{k=n}^{mn−1} f_{3}(k) − (f_{3}(mn) − \frac{1}{F_{mn+2} + 1}) < \frac{−1}{F_{n+1} + 1}.

If mn is even, employing Lemma 10 again, we can deduce

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{k}} = \frac{−1}{F_{n+1} + 1} − ∑_{k=n+2}^{mn} f_{3}(k) − (f_{3}(n) + f_{3}(n + 1) − \frac{1}{F_{mn+2} + 1})

≤ \frac{−1}{F_{n+1} + 1} − ∑_{k=n+2}^{mn} f_{3}(k) − (f_{3}(n) + f_{3}(n + 1) − \frac{1}{F_{2n+2} + 1})

< \frac{−1}{F_{n+1} + 1}.

Now we can conclude that if n ≥ 5 is odd, then

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{k}} < \frac{−1}{F_{n+1} + 1}.

If mn is even, then Lemma 11 implies that

∑_{k=n}^{mn} f_{4}(k) < 0.

If mn is odd, invoking Lemma 11 and Lemma 12, we can get

∑_{k=n}^{mn} f_{4}(k) = ∑_{k=n+2}^{mn−1} f_{4}(k) + (f_{4}(n) + f_{4}(n + 1) + f_{4}(mn)) < 0.

Thus, we always have

∑_{k=n}^{mn} f_{4}(k) < 0,

from which we obtain

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{k}} = \frac{−1}{F_{n+1}} + \frac{1}{F_{mn+2}} − ∑_{k=n}^{mn} f_{4}(k) > \frac{−1}{F_{n+1}} .

Therefore, we arrive at

\frac{−1}{F_{n+1}} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{k}} < \frac{−1}{F_{n+1} + 1},

which shows that the result holds for odd n.

3 Results for a = 2

We first introduce the following notations

g_{1}(n) = \frac{1}{F_{2n−2} + F_{2n}} − \frac{(−1)^{n}}{F_{2n}} − \frac{1}{F_{2n} + F_{2n+2}} ,
g_{2}(n) = \frac{1}{F_{2n−2} + F_{2n} − 1} − \frac{(−1)^{n}}{F_{2n}} − \frac{1}{F_{2n} + F_{2n+2} − 1},
g_{3}(n) = \frac{1}{F_{2n−2} + F_{2n} + 1} − \frac{(−1)^{n}}{F_{2n}} − \frac{1}{F_{2n} + F_{2n+2} + 1},
g_{4}(n) = \frac{−1}{F_{2n−2} + F_{2n}} − \frac{(−1)^{n}}{F_{2n}} + \frac{1}{F_{2n} + F_{2n+2}} ,
g_{5}(n) = \frac{−1}{F_{2n−2} + F_{2n} + 1} − \frac{(−1)^{n}}{F_{2n}} + \frac{1}{F_{2n} + F_{2n+2} + 1}.

It is routine to check that for 1 ≤ i ≤ 5, g_{i}(n) is positive if n is odd, and negative otherwise. Lemma 14. If n ≥ 1, then g_{1}(n) + g_{1}(n + 1) > 0 and

g_{1}(n) + g_{1}(n + 1) > g_{1}(n + 2) + g_{1}(n + 3).

Proof. If n is odd, we have

g_{1}(n) + g_{1}(n + 1) = (\frac{1}{F_{2n−2} + F_{2n}} − \frac{1}{F_{2n+2} + F_{2n+4}} ) + (\frac{1}{F_{2n}} − \frac{1}{F_{2n+2}} )
= \frac{5F_{2n+1}}{(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})} + \frac{F_{2n+1}}{F_{2n}F_{2n+2}}
> 0.

Applying the easily checked fact

\frac{F_{2n+1}}{F_{2n−2} + F_{2n}} > \frac{F_{2n+5}}{F_{2n+6} + F_{2n+8}} ,
\frac{F_{2n+1}}{F_{2n}F_{2n+2}} > \frac{F_{2n+5}}{F_{2n+4}F_{2n+6}} ,

we can conclude that g_{1}(n) + g_{1}(n + 1) > g_{1}(n + 2) + g_{1}(n + 3).

Now we consider the case where n is even. Doing some elementary manipulations and using Lemma 5, we have

g_{1}(n) + g_{1}(n + 1) = (\frac{1}{F_{2n−2} + F_{2n}} − \frac{1}{F_{2n}} ) + (\frac{1}{F_{2n+2}} − \frac{1}{F_{2n+2} + F_{2n+4}} )

= \frac{F_{2n−2}(F_{2n}F_{2n+4} − F_{2n+2}^{2}) + (F_{2n}^{2} − F_{2n−2}F_{2n+2})F_{2n+4}}{F_{2n}F_{2n+2}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})}
= \frac{F_{2n+4} − F_{2n−2}}{F_{2n}F_{2n+2}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})}
= \frac{4F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})}
> 0.

Applying the above identity, we see that

\frac{g_{1}(n) + g_{1}(n + 1)}{g_{1}(n + 2) + g_{1}(n + 3)} = \frac{F_{2n+1}F_{2n+4}F_{2n+6}}{F_{2n}F_{2n+2}F_{2n+5}} · \frac{F_{2n+6} + F_{2n+8}}{F_{2n−2} + F_{2n}} > 1.

Thus, g_{1}(n) + g_{1}(n + 1) > g_{1}(n + 2) + g_{1}(n + 3) also holds.

Lemma 15. For n ≥ 1, we have

F_{6n+2} > F_{2n}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4}).

Proof. It follows from Lemma 5 that

F_{2n−1}F_{2n+3} − F_{2n−2}F_{2n+4} = 5,
F_{2n−1}F_{2n+1} − F_{2n}^{2} = 1,
F_{2n+1}F_{2n+3} − F_{2n}F_{2n+4} = 2.

Thus, F_{2n−1}F_{2n+3} > F_{2n−2}F_{2n+4}, F_{2n−1}F_{2n+1} > F_{2n}^{2}, and F_{2n+1}F_{2n+3} > F_{2n}F_{2n+4}. Employing Lemma 3 repeatedly and the above three inequalities, we have

F_{6n+2} = F_{2n}F_{4n+1} + F_{2n+1}F_{4n+2}
= F_{2n}(F_{2n−2}F_{2n+2} + F_{2n−1}F_{2n+3}) + F_{2n+1}(F_{2n−1}F_{2n+2} + F_{2n}F_{2n+3})
> F_{2n−2}F_{2n}F_{2n+2} + F_{2n−2}F_{2n+4}F_{2n} + F_{2n}^{2}F_{2n+2} + F_{2n}F_{2n+4}F_{2n}
= F_{2n}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4}),

which completes the proof.

Lemma 16. If n ≥ 1 and m ≥ 3, then

g_{1}(n) + g_{1}(n + 1) + g_{1}(mn) > 0.

Proof. If mn is odd, then the result follows from Lemma 14 and the fact g_{1}(mn) > 0. Thus we focus on the case where mn is even. For k ≥ 1,

\frac{1}{F_{2k−2} + F_{2k}} − \frac{1}{F_{2k}} = −\frac{F_{2k−2}}{(F_{2k−2} + F_{2k})F_{2k}}
= −\frac{F_{2k−2}}{F_{2k−2}F_{2k} + F_{2k}^{2}}
> −\frac{F_{2k−2}}{F_{2k−2}F_{2k+2}}
= −\frac{1}{F_{2k+2}} ,

where the inequality follows from F_{2k}^{2} − F_{2k−2}F_{2k+2} = 1. Since mn is even, employing the above inequality, we have

g_{1}(mn) > −\frac{1}{F_{2mn+2}} − \frac{1}{F_{2mn} + F_{2mn+2}} > −\frac{2}{F_{2mn+2}} ≥ −\frac{2}{F_{6n+2}} .

From the proof of Lemma 14 we know that whether n is even or odd, we always have

g_{1}(n) + g_{1}(n + 1) ≥ \frac{4F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})}.

Therefore,

g_{1}(n) + g_{1}(n + 1) + g_{1}(mn) > \frac{4F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})} − \frac{2}{F_{6n+2}}
> \frac{2}{F_{2n}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})} − \frac{2}{F_{6n+2}}
> 0,

where the last inequality follows from Lemma 15.

Lemma 17. If n > 0, then

2F_{4n}(F_{4n} + F_{4n+2}) > F_{2n+2}F_{4n+3}(F_{2n−2} + F_{2n}).

Proof. It suffices to show that 2F_{4n}^{2} > F_{2n−2}F_{2n+2}F_{4n+3} and 2F_{4n}F_{4n+2} > F_{2n}F_{2n+2}F_{4n+3}. These two inequalities can be proved using similar arguments, so we only prove the first one.

Applying Lemma 5 repeatedly and Lemma 3, we can obtain

2F_{4n}^{2} = 2F_{4n−3}F_{4n+3} − 8
= 2(F_{2n−2}^{2} + F_{2n−1}^{2})F_{4n+3} − 8
> (F_{2n−2}F_{2n−1} + 2F_{2n−1}^{2})F_{4n+3} − 8
= F_{2n−1}F_{2n+1}F_{4n+3} − 8
= (F_{2n−2}F_{2n+2} + 2)F_{4n+3} − 8
> F_{2n−2}F_{2n+2}F_{4n+3}.

The proof is completed.

Lemma 18. For all n ≥ 2, we have

g_{2}(n) + g_{2}(n + 1) + g_{2}(2n) > 0.

Proof. It is straightforward to verify that F_{2n−2} + F_{2n} + F_{2n+2} + F_{2n+4} = 3(F_{2n} + F_{2n+2}). Applying Lemma 5 repeatedly, we get

(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4}) = F_{2n−2}F_{2n+2} + F_{2n−2}F_{2n+4} + F_{2n}F_{2n+2} + F_{2n}F_{2n+4}
= F_{2n−2}F_{2n+2} + (F_{2n}F_{2n+2} − 3) + F_{2n}F_{2n+2}
+ F_{2n}(2F_{2n+2} + F_{2n+1})
= (F_{2n−2}F_{2n+2} − F_{2n}^{2}) + (F_{2n}^{2} + F_{2n}F_{2n+1})
+ 4F_{2n}F_{2n+2} − 3
= 5F_{2n}F_{2n+2} − 4.

It follows from the definition of g_{2}(n) and the above two equations that

g_{2}(n) + g_{2}(n + 1) ≥ (\frac{1}{F_{2n−2} + F_{2n} − 1} − \frac{1}{F_{2n+2} + F_{2n+4} − 1}) − (\frac{1}{F_{2n}} − \frac{1}{F_{2n+2}} )
= \frac{5F_{2n+1}}{(F_{2n−2} + F_{2n} − 1)(F_{2n+2} + F_{2n+4} − 1)} − \frac{F_{2n+1}}{F_{2n}F_{2n+2}}
= \frac{3(F_{2n} + F_{2n+2} + 1)F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n−2} + F_{2n} − 1)(F_{2n+2} + F_{2n+4} − 1)}
> \frac{1}{F_{2n+2}(F_{2n−2} + F_{2n} − 1)},

where the last inequality follows from 3F_{n} > F_{n+2}. It is routine to show

2(F_{4n+2} − F_{4n−2}) = 2(2F_{4n} + F_{4n−1} − F_{4n−2})
= 3F_{4n} + F_{4n} + 2F_{4n−3}
> 3F_{4n} + (2F_{4n−2} + F_{4n−3}) + F_{4n−2}
> 3(F_{4n−2} + F_{4n}),

which means

F_{4n+2} − F_{4n−2} > \frac{3}{2}(F_{4n−2} + F_{4n}).

Employing the above inequality, we can deduce that

g_{2}(2n) = \frac{F_{4n+2} − F_{4n−2}}{(F_{4n−2} + F_{4n} − 1)(F_{4n} + F_{4n+2} − 1)} − \frac{1}{F_{4n}}
> \frac{3}{2(F_{4n} + F_{4n+2} − 1)} − \frac{1}{F_{4n}}
= \frac{−F_{4n+3} + 2}{2F_{4n}(F_{4n} + F_{4n+2} − 1)}
> −\frac{F_{4n+3}}{2F_{4n}(F_{4n} + F_{4n+2} − 1)}.

Now we conclude that

g_{2}(n) + g_{2}(n + 1) + g_{2}(2n) > \frac{1}{F_{2n+2}(F_{2n−2} + F_{2n} − 1)} − \frac{F_{4n+3}}{2F_{4n}(F_{4n} + F_{4n+2} − 1)} > 0,

where the last inequality follows from Lemma 17.

Applying the argument in the proof of Lemma 18, it can be readily seen the following property of g_{3}(n), whose proof is omitted here.

Lemma 19. If n ≥ 2 is even, we have

g_{3}(n) + g_{3}(n + 1) < 0.

Imitating the proof of Lemma 14 and Lemma 16 respectively, we can easily get the following results on g_{4}(n).

Lemma 20. For n ≥ 1, we have

g_{4}(n) + g_{4}(n + 1) < 0.

Lemma 21. If n ≥ 1 and m ≥ 2, then

g_{4}(n) + g_{4}(n + 1) + g_{4}(mn) < 0.

Lemma 22. If n ≥ 1 is odd, we have

g_{5}(n) + g_{5}(n + 1) > \frac{1}{F_{4n} + F_{4n+2} + 1}.

Proof. It is easy to see that the result is true for n = 1, thus we assume that n ≥ 3. From the proof of Lemma 18, we can easily obtain that if n ≥ 3 is odd, then

g_{5}(n) + g_{5}(n + 1) = \frac{3(F_{2n} + F_{2n+2} − 1)F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n−2} + F_{2n} + 1)(F_{2n+2} + F_{2n+4} + 1)}
> \frac{1}{F_{2n+2}(F_{2n−2} + F_{2n} + 1)}.

Employing Lemma 3 repeatedly, it is easy to see that

F_{2n+2}(F_{2n−2} + F_{2n} + 1) < F_{2n−2}F_{2n+3} + F_{2n}F_{2n+3} + F_{2n+2}
= F_{4n} − F_{2n−3}F_{2n+2} + F_{4n+2} − F_{2n−1}F_{2n+2} + F_{2n+2}
< F_{4n} + F_{4n+2}.

Combining the above two inequalities yields the desired result. Lemma 23. For n ≥ 2, we have

F_{4n−2}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4}) > F_{4n}(F_{4n−2} + F_{4n}).

Proof. We first consider the right-hand side. Applying F_{4n}^{2} − F_{4n−1}F_{4n+1} = −1, we have F_{4n}(F_{4n−2} + F_{4n}) = F_{4n−2}F_{4n} + F_{4n}^{2} = F_{4n−2}F_{4n} + F_{4n−1}F_{4n+1} − 1 = F_{8n−1} − 1. For the left-hand side, we have that if n ≥ 2, then

(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4}) = F_{2n−2}F_{2n+2} + F_{2n}F_{2n+2} + F_{2n−2}F_{2n+4} + F_{2n}F_{2n+4}
> (F_{2n−2}F_{2n+1} + F_{2n−1}F_{2n+2}) + (F_{2n−2}F_{2n+3}
+ F_{2n−1}F_{2n+4}) + F_{2n−2}F_{2n+4}
> F_{4n} + F_{4n+2} + 2.

Therefore, using the fact F_{4n−2}F_{4n+2} − F_{4n−1}F_{4n+1} = −2, we have

F_{4n−2}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4}) > F_{4n−2}F_{4n} + F_{4n−2}F_{4n+2} + 2
= F_{4n−2}F_{4n} + F_{4n−1}F_{4n+1}
= F_{8n−1}.

Thus the left-hand side is greater than the right-hand side.

Theorem 24. If n ≥ 2 is even and m ≥ 2, then

⌊(∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k}} )^{−1}⌋ = \{ \begin{matrix}F_{2n−2} + F_{2n} − 1, & if m = 2; \\ F_{2n−2} + F_{2n}, & if m > 2. \end{matrix}

Proof. We first consider the case where m = 2. From Lemma 14 we know that

∑_{k=n}^{2n−1} g_{1}(k) < \frac{4F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})} · \frac{n}{2} < \frac{1}{(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})}.

In addition,

g_{1}(2n) + \frac{1}{F_{4n} + F_{4n+2}} = \frac{1}{F_{4n−2} + F_{4n}} − \frac{1}{F_{4n}} = \frac{−F_{4n−2}}{F_{4n}(F_{4n−2} + F_{4n})}.

Therefore, invoking Lemma 23, we have

∑_{k=n}^{2n} g_{1}(k) + \frac{1}{F_{4n} + F_{4n+2}} < \frac{1}{(F_{2n−2} + F_{2n})(F_{2n+2} + F_{2n+4})} − \frac{F_{4n−2}}{F_{4n}(F_{4n−2} + F_{4n})} < 0.

Now with the help of g_{1}(n), we can obtain

∑_{k=n}^{2n} \frac{(−1)^{k}}{F_{2k}} = \frac{1}{F_{2n−2} + F_{2n}} − \frac{1}{F_{4n} + F_{4n+2}} − ∑_{k=n}^{2n} g_{1}(k) > \frac{1}{F_{2n−2} + F_{2n}} .

From the proof of Lemma 18, we know that g_{2}(n) + g_{2}(n + 1) > 0. Moreover, applying Lemma 18, we can deduce

∑_{k=n}^{2n} g_{2}(k) = g_{2}(n) + g_{2}(n + 1) + g_{2}(2n) + ∑_{k=n+2}^{2n−1} g_{2}(k) > 0.

Therefore,

∑_{k=n}^{2n} \frac{(−1)^{k}}{F_{2k}} = \frac{1}{F_{2n−2} + F_{2n} − 1} − \frac{1}{F_{4n} + F_{4n+2} − 1} − ∑_{k=n}^{2n} g_{2}(k) < \frac{1}{F_{2n−2} + F_{2n} − 1}.

We now conclude that

\frac{1}{F_{2n−2} + F_{2n}} < ∑_{k=n}^{2n} \frac{(−1)^{k}}{F_{2k}} < \frac{1}{F_{2n−2} + F_{2n} − 1},

which shows that the statement for m = 2 is true.

Next we turn to consider the case where m > 2. First, employing Lemma 14 and Lemma 16, we see that

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k}} < \frac{1}{F_{2n−2} + F_{2n}} − (g_{1}(n) + g_{1}(n + 1) + g_{1}(mn)) − ∑_{k=n+2}^{mn−1} g_{1}(k) < \frac{1}{F_{2n−2} + F_{2n}} .

We write the sum in terms of g_{3}(n) as

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k}} = \frac{1}{F_{2n−2} + F_{2n} + 1} − ∑_{k=n}^{mn−1} g_{3}(k) − (g_{3}(mn) + \frac{1}{F_{2mn} + F_{2mn+2} + 1})

= \frac{1}{F_{2n−2} + F_{2n} + 1} − ∑_{k=n}^{mn−1} g_{3}(k) − (\frac{1}{F_{2mn−2} + F_{2mn} + 1} − \frac{1}{F_{2mn}} )

> \frac{1}{F_{2n−2} + F_{2n} + 1},

where the last inequality follows from Lemma 19. Now we get

\frac{1}{F_{2n−2} + F_{2n} + 1} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k}} < \frac{1}{F_{2n−2} + F_{2n}} ,

which yields the desired identity.

Theorem 25. If n ≥ 1 is odd and m ≥ 2, then

⌊(∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k}} )^{−1}⌋ = −F_{2n−2} − F_{2n} − 1.

Proof. If mn is even, it follows from Lemma 20 that

∑_{k=n}^{mn} g_{4}(k) < 0.

If mn is odd, then Lemma 20 and Lemma 21 ensure that

∑_{k=n}^{mn} g_{4}(k) = ∑_{k=n+2}^{mn−1} g_{4}(k) + (g_{4}(n) + g_{4}(n + 1) + g_{4}(mn)) < 0.

Therefore, we always have

∑_{k=n}^{mn} g_{4}(k) < 0.

With the help of g_{4}(n), we have

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k}} = \frac{−1}{F_{2n−2} + F_{2n}} + \frac{1}{F_{2mn−2} + F_{2mn}} − ∑_{k=n}^{mn} g_{4}(k) > \frac{−1}{F_{2n−2} + F_{2n}} .

From Lemma 22 we know that if n is odd, then g_{5}(n) + g_{5}(n + 1) > 0. Now we claim that

∑_{k=n}^{mn} g_{5}(k) > \frac{1}{F_{2mn} + F_{2mn+2} + 1}.

If mn is even, employing Lemma 22, we obtain

∑_{k=n}^{mn} g_{5}(k) − \frac{1}{F_{2mn} + F_{2mn+2} + 1} ≥ ∑_{k=n}^{mn} g_{5}(k) − \frac{1}{F_{4n} + F_{4n+2} + 1}
≥ g_{5}(n) + g_{5}(n + 1) − \frac{1}{F_{4n} + F_{4n+2} + 1}
> 0.

If mn is odd, then

∑_{k=n}^{mn} g_{5}(k) − \frac{1}{F_{2mn} + F_{2mn+2} + 1} = ∑_{k=n}^{mn−1} g_{5}(k) + (g_{5}(mn) − \frac{1}{F_{2mn} + F_{2mn+2} + 1})
> −\frac{1}{F_{2mn−2} + F_{2mn} + 1} + \frac{1}{F_{2mn}}
> 0.

Therefore, we have

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k}} = \frac{−1}{F_{2n−2} + F_{2n} + 1} + \frac{1}{F_{2mn−2} + F_{2mn} + 1} − ∑_{k=n}^{mn} g_{5}(k) < \frac{−1}{F_{2n−2} + F_{2n} + 1}.

Now we can conclude that

\frac{−1}{F_{2n−2} + F_{2n}} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k}} < \frac{−1}{F_{2n−2} + F_{2n} + 1},

from which the desired result follows.

Similarly, we can prove the following results. Theorem 26. If n ≥ 4 is even and m ≥ 2, then

⌊(∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k−1}} )^{−1}⌋ = F_{2n−3} + F_{2n−1} − 1.

Theorem 27. If n ≥ 3 is odd and m ≥ 2, then

⌊(∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{2k−1}} )^{−1}⌋ = \{ \begin{matrix}−F_{2n−3} − F_{2n−1} − 1, & if m = 2; \\ −F_{2n−3} − F_{2n−1}, & if m > 2. \end{matrix}

4 Results for a = 3

We first introduce the following notations:

s_{1}(n) = \frac{1}{2F_{3n−1}} − \frac{(−1)^{n}}{F_{3n}} − \frac{1}{2F_{3n+2}} ,
s_{2}(n) = \frac{1}{2F_{3n−1} − 1} − \frac{(−1)^{n}}{F_{3n}} − \frac{1}{2F_{3n+2} − 1},
s_{3}(n) = \frac{−1}{2F_{3n−1}} − \frac{(−1)^{n}}{F_{3n}} + \frac{1}{2F_{3n+2}} ,
s_{4}(n) = \frac{−1}{2F_{3n−1} + 1} − \frac{(−1)^{n}}{F_{3n}} + \frac{1}{2F_{3n+2} + 1}.

It is easy to see that for each i, s_{i}(n) is positive if n is odd, and negative otherwise. Lemma 28. If n ≥ 2 is even, then

s_{1}(n) + s_{1}(n + 1) < 0.

Proof. Since n is even, applying Lemma 5 twice, we have

s_{1}(n) + s_{1}(n + 1) = (\frac{1}{2F_{3n−1}} − \frac{1}{2F_{3n+5}} ) − (\frac{1}{F_{3n}} − \frac{1}{F_{3n+3}} )
= \frac{2F_{3n+2}}{F_{3n−1}F_{3n+5}} − \frac{2F_{3n+1}}{F_{3n}F_{3n+3}}
= 2 · \frac{F_{3n}F_{3n+2}F_{3n+3} − F_{3n−1}F_{3n+1}F_{3n+5}}{F_{3n−1}F_{3n}F_{3n+3}F_{3n+5}}
= 2 · \frac{F_{3n}F_{3n+2}F_{3n+3} − (F_{3n}^{2} + 1)F_{3n+5}}{F_{3n−1}F_{3n}F_{3n+3}F_{3n+5}}
= 2 · \frac{F_{3n}(F_{3n+2}F_{3n+3} − F_{3n}F_{3n+5}) − F_{3n+5}}{F_{3n−1}F_{3n}F_{3n+3}F_{3n+5}}
= 2 · \frac{2F_{3n} − F_{3n+5}}{F_{3n−1}F_{3n}F_{3n+3}F_{3n+5}}
< 0,

which completes the proof.

Lemma 29. For all n ≥ 1, we have

s_{2}(n) + s_{2}(n + 1) > 0.

Proof. It is clear that the result holds if n is odd. In the rest, we assume that n is even. Applying the analysis in the proof of Lemma 28, we can easily obtain

s_{2}(n) + s_{2}(n + 1) = (\frac{1}{2F_{3n−1} − 1} − \frac{1}{2F_{3n+5} − 1}) − (\frac{1}{F_{3n}} − \frac{1}{F_{3n+3}} )

= \frac{8F_{3n+2}}{(2F_{3n−1} − 1)(2F_{3n+5} − 1)} − \frac{2F_{3n+1}}{F_{3n}F_{3n+3}}

= \frac{8(F_{3n}F_{3n+2}F_{3n+3} − F_{3n−1}F_{3n+1}F_{3n+5}) + 2F_{3n+1}(2F_{3n−1} + 2F_{3n+5} − 1)}{(2F_{3n−1} − 1)(2F_{3n+5} − 1)F_{3n}F_{3n+3}}

= \frac{16F_{3n} − 8F_{3n+5} + 2F_{3n+1}(2F_{3n−1} + 2F_{3n+5} − 1)}{(2F_{3n−1} − 1)(2F_{3n+5} − 1)F_{3n}F_{3n+3}}

> \frac{4F_{3n+1}F_{3n+5} − 8F_{3n+5}}{(2F_{3n−1} − 1)(2F_{3n+5} − 1)F_{3n}F_{3n+3}}

> 0.

The proof is completed.

Lemma 30. If n ≥ 1 and m ≥ 2, then

s_{2}(n) + s_{2}(n + 1) + s_{2}(mn) > 0.

Proof. If mn is odd, then the result follows from Lemma 29 and the fact s_{2}(mn) > 0. So we assume that mn is even. Now it is clear that

s_{2}(mn) = \frac{1}{2F_{3mn−1} − 1} − \frac{1}{F_{3mn}} − \frac{1}{2F_{3mn+2} − 1} > −\frac{1}{F_{3mn}} ≥ −\frac{1}{F_{6n}} .

If n is odd, we have

s_{2}(n) + s_{2}(n + 1) > \frac{1}{F_{3n}} − \frac{1}{F_{3n+3}} = \frac{2F_{3n+1}}{F_{3n}F_{3n+3}} > \frac{2}{F_{3n}F_{3n+3}} .

If n is even, then from Lemma 29 we know that

s_{2}(n) + s_{2}(n + 1) > \frac{4F_{3n+1}F_{3n+5} − 8F_{3n+5}}{(2F_{3n−1} − 1)(2F_{3n+5} − 1)F_{3n}F_{3n+3}}
= \frac{4F_{3n+5}(2F_{3n−1} + F_{3n−2} − 2)}{(2F_{3n−1} − 1)(2F_{3n+5} − 1)F_{3n}F_{3n+3}}
> \frac{2}{F_{3n}F_{3n+3}} .

Now we can derive the conclusion that

s_{2}(n) + s_{2}(n + 1) + s_{2}(mn) > \frac{2}{F_{3n}F_{3n+3}} − \frac{1}{F_{6n}} ≥ 0,

where the last inequality follows from

2F_{6n} = F_{3n}(2F_{3n−1} + 2F_{3n+1}) > F_{3n}(F_{3n} + 2F_{3n+1}) = F_{3n}F_{3n+3}.

This completes the proof. Lemma 31. For all n ≥ 1,

s_{3}(n) + s_{3}(n + 1) < 0.

Proof. The result clearly holds when n is even. If n is odd, applying similar analysis in the proof of Lemma 28, we can easily derive

s_{3}(n) + s_{3}(n + 1) = 2 · \frac{2F_{3n} − F_{3n+5}}{F_{3n−1}F_{3n}F_{3n+3}F_{3n+5}} < 0,

which completes the proof. Lemma 32. If n ≥ 1 and m ≥ 2, then

s_{3}(n) + s_{3}(n + 1) + s_{3}(mn) < 0.

Proof. If mn is even, then the result follows from Lemma 31 and the fact s_{3}(mn) < 0. Now we assume that mn is odd, which implies that n is odd and m ≥ 3. First we have

s_{3}(mn) = \frac{−1}{2F_{3mn−1}} + \frac{1}{F_{3mn}} + \frac{1}{2F_{3mn+2}} < \frac{1}{F_{3mn}} ≤ \frac{1}{F_{9n}} .

Moreover, from the proof of Lemma 31 we know

s_{3}(n) + s_{3}(n + 1) = −\frac{2(F_{3n+5} − 2F_{3n})}{F_{3n−1}F_{3n}F_{3n+3}F_{3n+5}} < −\frac{1}{F_{3n−1}F_{3n}F_{3n+3}} .

Now we arrive at

s_{3}(n) + s_{3}(n + 1) + s_{3}(mn) < −\frac{1}{F_{3n−1}F_{3n}F_{3n+3}} + \frac{1}{F_{9n}} < 0,

where the last inequality follows from

F_{9n} = F_{3n−2}F_{6n+1} + F_{3n−1}F_{6n+2} > F_{3n−1}(F_{3n−1}F_{3n+2} + F_{3n}F_{3n+3}) > F_{3n−1}F_{3n}F_{3n+3}. The proof is completed.

Lemma 33. If n ≥ 1 is odd, then

s_{4}(n) + s_{4}(n + 1) > \frac{1}{2F_{6n+2} + 1}.

Proof. It is easy to check that the result holds for n = 1, so we assume that n ≥ 3. Applying the similar analysis in the proof of Lemma 28, we have that, for n ≥ 3,

s_{4}(n) + s_{4}(n + 1) = − (\frac{1}{2F_{3n−1} + 1} − \frac{1}{2F_{3n+5} + 1}) + (\frac{1}{F_{3n}} − \frac{1}{F_{3n+3}} )
= −\frac{2F_{3n+5} − 2F_{3n−1}}{(2F_{3n−1} + 1)(2F_{3n+5} + 1)} + \frac{2F_{3n+1}}{F_{3n}F_{3n+3}}
> −\frac{F_{3n+5} − F_{3n−1}}{(2F_{3n−1} + 1)F_{3n+5}} + \frac{2F_{3n+1}}{F_{3n}F_{3n+3}}
= −\frac{4F_{3n+2}}{(2F_{3n−1} + 1)F_{3n+5}} + \frac{2F_{3n+1}}{F_{3n}F_{3n+3}}
= \frac{4(F_{3n−1}F_{3n+1}F_{3n+5} − F_{3n}F_{3n+2}F_{3n+3}) + 2F_{3n+1}F_{3n+5}}{(2F_{3n−1} + 1)F_{3n}F_{3n+3}F_{3n+5}}
= \frac{4(2F_{3n} − F_{3n+5}) + 2F_{3n+1}F_{3n+5}}{(2F_{3n−1} + 1)F_{3n}F_{3n+3}F_{3n+5}}
> \frac{(2F_{3n+1} − 4)F_{3n+5}}{(2F_{3n−1} + 1)F_{3n}F_{3n+3}F_{3n+5}}
> \frac{1}{F_{3n}F_{3n+3}} .

In addition, we have

F_{2n+2} = F_{n−1}F_{n+2} + F_{n}F_{n+3} > F_{n}F_{n+3}.

Combining the above two inequalities together yields the desired result.

Theorem 34. If n ≥ 1 and m ≥ 2, then

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

Proof. We first consider the case where n is even. With the help of s_{1}(n), we have

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k}} = \frac{1}{2F_{3n−1}} − ∑_{k=n}^{mn−1} s_{1}(k) − (s_{1}(mn) + \frac{1}{2F_{3mn+2}} )

= \frac{1}{2F_{3n−1}} − ∑_{k=n}^{mn−1} s_{1}(k) − (\frac{1}{2F_{3mn+2}} − \frac{1}{F_{3mn}} )

> \frac{1}{2F_{3n−1}} − ∑_{k=n}^{mn−1} s_{1}(k)

> \frac{1}{2F_{3n−1}} ,

where the last inequality follows from Lemma 28.

Employing Lemma 29 and Lemma 30, we can deduce that

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k}} = \frac{1}{2F_{3n−1} − 1} − \frac{1}{2F_{3mn+2} − 1} − ∑_{k=n+2}^{mn−1} s_{2}(k) − (s_{2}(n) + s_{2}(n + 1) + s_{2}(mn))

< \frac{1}{2F_{3n−1} − 1}.

Therefore, we obtain

\frac{1}{2F_{3n−1}} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k}} < \frac{1}{2F_{3n−1} − 1},

which shows that the statement is true when n is even.

We now turn to consider the case where n is odd. If m is even, applying Lemma 31 and Lemma 33, we can deduce that

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k}} = \frac{−1}{2F_{3n−1}} + \frac{1}{2F_{3mn+2}} − ∑_{k=n}^{mn} s_{3}(k) > \frac{−1}{2F_{3n−1}} ,

and

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k}} = \frac{−1}{2F_{3n−1} + 1} − ∑_{k=n+2}^{mn} s_{4}(k) − (s_{4}(n) + s_{4}(n + 1) − \frac{1}{2F_{3mn+2} + 1})

≤ \frac{−1}{2F_{3n−1} + 1} − ∑_{k=n+2}^{mn} s_{4}(k) − (s_{4}(n) + s_{4}(n + 1) − \frac{1}{2F_{6n+2} + 1})

< \frac{−1}{2F_{3n−1} + 1}.

Thus, if n is odd and m is even, we have

\frac{−1}{2F_{3n−1}} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k}} < \frac{−1}{2F_{3n−1} + 1}.

If m is odd, then Lemma 31 and Lemma 32 implies that

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k}} = \frac{−1}{2F_{3n−1}} + \frac{1}{2F_{3mn+2}} − ∑_{k=n+2}^{mn−1} s_{3}(k) − (s_{3}(n) + s_{3}(n + 1) + s_{3}(mn)) > \frac{−1}{2F_{3n−1}} .

And it follows from Lemma 33 that

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k}} = \frac{−1}{2F_{3n−1} + 1} − ∑_{k=n}^{mn−1} s_{4}(k) − (s_{4}(mn) − \frac{1}{2F_{3mn+2} + 1})
= \frac{−1}{2F_{3n−1} + 1} − ∑_{k=n}^{mn−1} s_{4}(k) − (\frac{1}{F_{3mn}} − \frac{1}{2F_{3mn−1} + 1})
< \frac{−1}{2F_{3n−1} + 1}.

Thus, if n and m are both odd, then

\frac{−1}{2F_{3n−1}} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k}} < \frac{−1}{2F_{3n−1} + 1}

also holds. Hence, the statement is true when n is odd. Theorem 35. If n ≥ 2, then

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

Theorem 36. If n ≥ 1 and m ≥ 3, then

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

Theorem 37. If n ≥ 1 and m ≥ 2, then

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

Remark 38. We will prove Theorem 35 and Theorem 36 in detail in the next section. The proof of Theorem 37 is very similar to that of Theorem 34, thus omitted here.

5 Proof of Theorem 35 and Theorem 36

We begin with introducing the following auxiliary functions:

t_{1}(n) = \frac{1}{2F_{3n}} − \frac{(−1)^{n}}{F_{3n+1}} − \frac{1}{2F_{3n+3}} ,
t_{2}(n) = \frac{1}{2F_{3n} − 1} − \frac{(−1)^{n}}{F_{3n+1}} − \frac{1}{2F_{3n+3} − 1},
t_{3}(n) = \frac{1}{2F_{3n} + 1} − \frac{(−1)^{n}}{F_{3n+1}} − \frac{1}{2F_{3n+3} + 1},
t_{4}(n) = \frac{−1}{2F_{3n}} − \frac{(−1)^{n}}{F_{3n+1}} + \frac{1}{2F_{3n+3}} ,
t_{5}(n) = \frac{−1}{2F_{3n} + 1} − \frac{(−1)^{n}}{F_{3n+1}} + \frac{1}{2F_{3n+3} + 1},
t_{6}(n) = \frac{−1}{2F_{3n} − 1} − \frac{(−1)^{n}}{F_{3n+1}} + \frac{1}{2F_{3n+3} − 1}.

It is straightforward to check that each t_{i}(n) is positive if n is odd, and negative otherwise. Lemma 39. For all n ≥ 1, we have t_{1}(n) + t_{1}(n + 1) > 0 and

t_{1}(n) + t_{1}(n + 1) > t_{1}(n + 2) + t_{1}(n + 3).

Proof. If n is odd, we have

t_{1}(n) + t_{1}(n + 1) = (\frac{1}{2F_{3n}} − \frac{1}{2F_{3n+6}} ) + (\frac{1}{F_{3n+1}} − \frac{1}{F_{3n+4}} ) = \frac{2F_{3n+3}}{F_{3n}F_{3n+6}} + \frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}} > 0.

Since

\frac{F_{3n+3}}{F_{3n}F_{3n+6}} > \frac{F_{3n+9}}{F_{3n+6}F_{3n+12}} ,
\frac{F_{3n+2}}{F_{3n+1}F_{3n+4}} > \frac{F_{3n+8}}{F_{3n+7}F_{3n+10}} ,

we can conclude that t_{1}(n) + t_{1}(n + 1) > t_{1}(n + 2) + t_{1}(n + 3).

Now we consider the case where n is even. Applying Lemma 5 repeatedly, we have

t_{1}(n) + t_{1}(n + 1) = (\frac{1}{2F_{3n}} − \frac{1}{2F_{3n+6}} ) − (\frac{1}{F_{3n+1}} − \frac{1}{F_{3n+4}} )
= \frac{2F_{3n+3}}{F_{3n}F_{3n+6}} − \frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}
= 2 · \frac{F_{3n+1}F_{3n+3}F_{3n+4} − F_{3n}F_{3n+2}F_{3n+6}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}
= 2 · \frac{F_{3n+1}F_{3n+2}F_{3n+3} + F_{3n+1}F_{3n+3}^{2}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}

− 2 · \frac{F_{3n}F_{3n+2}F_{3n+4} + F_{3n}F_{3n+2}F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}

= 2 · \frac{F_{3n+2}(F_{3n+1}F_{3n+3} − F_{3n}F_{3n+4}) + F_{3n+1}(F_{3n+1}F_{3n+5} − 1)}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}

− 2 · \frac{F_{3n}F_{3n+2}F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}

= 2 · \frac{2F_{3n+2} + F_{3n+5}(F_{3n+1}^{2} − F_{3n}F_{3n+2}) − F_{3n+1}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}
= 2 · \frac{2F_{3n+2} + F_{3n+5} − F_{3n+1}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}
= 2 · \frac{F_{3n} + F_{3n+2} + F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}
> 0.

In addition, it is easy to see that F_{3n} + F_{3n+2} + F_{3n+5} = 3F_{3n} + 3F_{3n+1} + F_{3n+4}, thus

t_{1}(n) + t_{1}(n + 1) = \frac{6}{F_{3n+1}F_{3n+4}F_{3n+6}} + \frac{6}{F_{3n}F_{3n+4}F_{3n+6}} + \frac{2}{F_{3n}F_{3n+1}F_{3n+6}} ,

which decreases as n grows. Lemma 40. For all n ≥ 1, we have

2F_{3n+3} > F_{n}F_{n+1}F_{n+6}.

Proof. Applying Lemma 3 repeatedly, we obtain

F_{3n+3} = F_{n}F_{2n+2} + F_{n+1}F_{2n+3}
= F_{n}(F_{n}F_{n+1} + F_{n+1}F_{n+2}) + F_{n+1}(F_{n}F_{n+2} + F_{n+1}F_{n+3})
= F_{n}F_{n+1}(F_{n} + 2F_{n+2}) + F_{n+1}^{2}(F_{n} + 2F_{n+1})
= F_{n}F_{n+1}(F_{n} + F_{n+1} + 2F_{n+2}) + 2F_{n+1}^{3}
> F_{n}F_{n+1}(3F_{n+2} + 2F_{n+1})
= F_{n}F_{n+1}F_{n+5}.

Therefore,

2F_{3n+3} − F_{n}F_{n+1}F_{n+6} > 2F_{n}F_{n+1}F_{n+5} − F_{n}F_{n+1}F_{n+6} = F_{n}F_{n+1}(2F_{n+5} − F_{n+6}) > 0, which completes the proof. Lemma 41. If n ≥ 1 and m ≥ 3, then

t_{1}(n) + t_{1}(n + 1) + t_{1}(mn) > 0.

Proof. If mn is odd, then the result follows from Lemma 39 and the fact t_{1}(mn) > 0. Now we assume that mn is even. It follows from Lemma 5 that F_{3mn}F_{3mn+1} = F_{3mn−2}F_{3mn+3} + 2, from which we get

t_{1}(mn) = \frac{1}{2F_{3mn}} − \frac{1}{F_{3mn+1}} − \frac{1}{2F_{3mn+3}}
= −\frac{F_{3mn−2}}{2(F_{3mn−2}F_{3mn+3} + 2)} − \frac{1}{2F_{3mn+3}}
> −\frac{F_{3mn−2}}{2F_{3mn−2}F_{3mn+3}} − \frac{1}{2F_{3mn+3}}
= −\frac{1}{F_{3mn+3}}
≥ −\frac{1}{F_{9n+3}} .

On the other hand, it follows from the proof of Lemma 39 that

t_{1}(n) + t_{1}(n + 1) > \frac{2}{F_{3n}F_{3n+1}F_{3n+6}} .

Now we arrive at

t_{1}(n) + t_{1}(n + 1) + t_{1}(mn) > \frac{2}{F_{3n}F_{3n+1}F_{3n+6}} − \frac{1}{F_{9n+3}} > 0,

where the last inequality follows from Lemma 40.

Lemma 42. For all n ≥ 2, we have

F_{2n}F_{2n+1} − F_{n+1}F_{n+4}F_{2n−2} < 0.

Proof. It follows from Lemma 4 and Lemma 5 respectively that

F_{n+2}F_{n+3} − F_{n}F_{n+1} = F_{2n+3},
F_{n+1}F_{n+4} − F_{n+2}F_{n+3} = (−1)^{n},

from which we can deduce that

F_{n+1}F_{n+4} = F_{n}F_{n+1} + F_{2n+3} + (−1)^{n} > F_{2n+3} + 2.

Therefore,

F_{2n}F_{2n+1} − F_{n+1}F_{n+4}F_{2n−2} < F_{2n}F_{2n+1} − (F_{2n+3} + 2)F_{2n−2}
= (F_{2n}F_{2n+1} − F_{2n−2}F_{2n+3}) − 2F_{2n−2}
= 2 − 2F_{2n−2}
≤ 0,

where the last equality follows from Lemma 5. Lemma 43. If n ≥ 2 is even, then

∑_{k=n}^{2n} t_{1}(k) + \frac{1}{2F_{6n+3}} < 0.

Proof. From the proof of Lemma 39 we know that if n is even, then

t_{1}(n) + t_{1}(n + 1) = 2 · \frac{F_{3n} + F_{3n+2} + F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}} < \frac{2F_{3n+6}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}} = \frac{2}{F_{3n}F_{3n+1}F_{3n+4}} .

Applying Lemma 39 again and the above inequality, we have

∑_{k=n}^{2n} t_{1}(k) + \frac{1}{2F_{6n+3}} = ∑_{k=n}^{2n−1} t_{1}(k) + (t_{1}(2n) + \frac{1}{2F_{6n+3}} )
< \frac{2}{F_{3n}F_{3n+1}F_{3n+4}} · \frac{n}{2} + (\frac{1}{2F_{6n}} − \frac{1}{F_{6n+1}} )
= \frac{n}{F_{3n}F_{3n+1}F_{3n+4}} − \frac{F_{6n−2}}{2F_{6n}F_{6n+1}}
< \frac{1}{2F_{3n+1}F_{3n+4}} − \frac{F_{6n−2}}{2F_{6n}F_{6n+1}}
< 0,

where the last inequality follows from Lemma 42.

t_{2}(n) + t_{2}(n + 1) > 0.

Proof. It is easy to see that the result is true when n is odd. So we assume that n is even. It follows from the definition of t_{2}(n) that

t_{2}(n) + t_{2}(n + 1) = (\frac{1}{2F_{3n} − 1} − \frac{1}{2F_{3n+6} − 1}) − \frac{1}{F_{3n+1}} + \frac{1}{F_{3n+4}}
= \frac{2F_{3n+6} − 2F_{3n}}{(2F_{3n} − 1)(2F_{3n+6} − 1)} − \frac{1}{F_{3n+1}} + \frac{1}{F_{3n+4}}
> \frac{F_{3n+6} − F_{3n}}{2F_{3n}F_{3n+6}} − \frac{1}{F_{3n+1}} + \frac{1}{F_{3n+4}}
= \frac{1}{2F_{3n}} − \frac{1}{F_{3n+1}} + \frac{1}{F_{3n+4}} − \frac{1}{2F_{3n+6}}
= t_{1}(n) + t_{1}(n + 1)
> 0,

where the last inequality follows from the proof of Lemma 39.

Lemma 45. If n ≥ 1 and m ≥ 2, then

t_{2}(n) + t_{2}(n + 1) + t_{2}(mn) > 0.

Proof. If mn is odd, then the result follows from Lemma 44 and the fact t_{2}(mn) > 0. Thus we assume that mn is even in the rest. Applying the argument in the proof of Lemma 44 and Lemma 41, we can easily obtain

t_{2}(mn) > t_{1}(mn) > −\frac{1}{F_{3mn+3}} ≥ −\frac{1}{F_{6n+3}} .

If n is odd, we have

t_{2}(n) + t_{2}(n + 1) > \frac{1}{F_{3n+1}} − \frac{1}{F_{3n+4}} = \frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}} > \frac{2}{(2F_{3n} − 1)F_{3n+4}} .

If n is even, then from the proof of Lemma 44 and Lemma 39 we know that

t_{2}(n) + t_{2}(n + 1) > \frac{F_{3n+6} − F_{3n}}{(2F_{3n} − 1)F_{3n+6}} − \frac{1}{F_{3n+1}} + \frac{1}{F_{3n+4}}
= \frac{4F_{3n+3}}{(2F_{3n} − 1)F_{3n+6}} − \frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}
= \frac{4(F_{3n+1}F_{3n+3}F_{3n+4} − F_{3n}F_{3n+2}F_{3n+6}) + 2F_{3n+2}F_{3n+6}}{(2F_{3n} − 1)F_{3n+1}F_{3n+4}F_{3n+6}}
> \frac{2F_{3n+2}F_{3n+6}}{(2F_{3n} − 1)F_{3n+1}F_{3n+4}F_{3n+6}}
> \frac{2}{(2F_{3n} − 1)F_{3n+4}} .

Therefore, we always have

t_{2}(n) + t_{2}(n + 1) > \frac{2}{(2F_{3n} − 1)F_{3n+4}} ,

from which we get

t_{2}(n) + t_{2}(n + 1) + t_{2}(mn) > \frac{2}{(2F_{3n} − 1)F_{3n+4}} − \frac{1}{F_{6n+3}} > 0,

where the last inequality follows from the fact F_{6n+3} = F_{3n−1}F_{3n+3} + F_{3n}F_{3n+4}. Lemma 46. If n ≥ 2 is even, then

t_{3}(n) + t_{3}(n + 1) < 0.

Proof. Applying the analysis in the proof of Lemma 39, we can deduce that

t_{3}(n) + t_{3}(n + 1) = (\frac{1}{2F_{3n} + 1} − \frac{1}{2F_{3n+6} + 1}) − (\frac{1}{F_{3n+1}} − \frac{1}{F_{3n+4}} )
= \frac{2F_{3n+6} − 2F_{3n}}{(2F_{3n} + 1)(2F_{3n+6} + 1)} − \frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}
< \frac{F_{3n+6} − F_{3n}}{(2F_{3n} + 1)F_{3n+6}} − \frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}
= \frac{4F_{3n+3}}{(2F_{3n} + 1)F_{3n+6}} − \frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}
= \frac{4(F_{3n+1}F_{3n+3}F_{3n+4} − F_{3n}F_{3n+2}F_{3n+6}) − 2F_{3n+2}F_{3n+6}}{(2F_{3n} + 1)F_{3n+1}F_{3n+4}F_{3n+6}}
= \frac{4(F_{3n} + F_{3n+2} + F_{3n+5}) − 2F_{3n+2}F_{3n+6}}{(2F_{3n} + 1)F_{3n+1}F_{3n+4}F_{3n+6}}
< \frac{4F_{3n+6} − 2F_{3n+2}F_{3n+6}}{(2F_{3n} + 1)F_{3n+1}F_{3n+4}F_{3n+6}}
< 0.

The proof is completed. Lemma 47. If n ≥ 1 is odd, then

t_{4}(n) + t_{4}(n + 1) > \frac{1}{2F_{9n+3}} .

Proof. Applying similar arguments in the proof of Lemma 39, we obtain that if n is odd,

t_{4}(n) + t_{4}(n + 1) = 2 · \frac{F_{3n} + F_{3n+2} + F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}} > \frac{2}{F_{3n}F_{3n+1}F_{3n+6}} .

It follows from Lemma 40 that

\frac{1}{F_{3n}F_{3n+1}F_{3n+6}} > \frac{1}{2F_{9n+3}} .

Combining the above two inequalities yields the desired result. Lemma 48. For all n ≥ 2, we have

∑_{k=n}^{2n} t_{4}(k) < \frac{1}{2F_{6n+3}} .

Proof. If n is even, it is easy to see that t_{4}(n) + t_{4}(n + 1) < 0. Thus,

∑_{k=n}^{2n} t_{4}(k) = ∑_{k=n}^{2n−1} t_{4}(k) + t_{4}(2n) < 0 < \frac{1}{2F_{6n+3}} .

If n is odd,

t_{4}(n) + t_{4}(n + 1) = 2 · \frac{F_{3n} + F_{3n+2} + F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}} < \frac{2}{F_{3n}F_{3n+1}F_{3n+4}} ,

which implies that

∑_{k=n}^{2n} t_{4}(k) < \frac{2}{F_{3n}F_{3n+1}F_{3n+4}} · \frac{n}{2} < \frac{1}{2F_{3n+1}F_{3n+4}} < \frac{1}{2F_{6n+3}} ,

where the last inequality follows from that for n ≥ 1,

F_{3n+1}F_{3n+4} = F_{3n−1}F_{3n+4} + F_{3n}F_{3n+4} > F_{3n−1}F_{3n+3} + F_{3n}F_{3n+4} = F_{6n+3}.

This completes the proof. Lemma 49. If n ≥ 1 is odd, then

t_{5}(n) + t_{5}(n + 1) > \frac{1}{2F_{6n+3} + 1}.

Proof. Imitating the proof of Lemma 46, we can easily obtain

t_{5}(n) + t_{5}(n + 1) > \frac{4(F_{3n} + F_{3n+2} + F_{3n+5}) + 2F_{3n+2}F_{3n+6}}{(2F_{3n} + 1)F_{3n+1}F_{3n+4}F_{3n+6}}
> \frac{2F_{3n+2}F_{3n+6}}{(2F_{3n} + 1)F_{3n+1}F_{3n+4}F_{3n+6}}
> \frac{1}{F_{3n+1}F_{3n+4}} .

In addition, we have

2F_{2n+3} = 2F_{n−1}F_{n+3} + 2F_{n}F_{n+4} > F_{n+1}F_{n+4}.

Combining the above two inequalities together yields the desired result. Lemma 50. For all n ≥ 1, we have

t_{6}(n) + t_{6}(n + 1) < 0.

Proof. It is clear that the result holds if n is even. Now we assume that n is odd. Applying the telescoping technique in the proof of Lemma 45 and the similar analysis in the proof of Lemma 39, we obtain

t_{6}(n) + t_{6}(n + 1) < − \{\frac{4F_{3n+3}}{(2F_{3n} − 1)F_{3n+6}} − \frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}} \}
= −\frac{4(F_{3n+1}F_{3n+3}F_{3n+4} − F_{3n}F_{3n+2}F_{3n+6}) + 2F_{3n+2}F_{3n+6}}{(2F_{3n} − 1)F_{3n+1}F_{3n+4}F_{3n+6}}
= −\frac{−4(F_{3n} + F_{3n+2} + F_{3n+5}) + 2F_{3n+2}F_{3n+6}}{(2F_{3n} − 1)F_{3n+1}F_{3n+4}F_{3n+6}}
= \frac{4(F_{3n} + F_{3n+2} + F_{3n+5}) − 2F_{3n+2}F_{3n+6}}{(2F_{3n} − 1)F_{3n+1}F_{3n+4}F_{3n+6}}
< \frac{4F_{3n+6} − 2F_{3n+2}F_{3n+6}}{(2F_{3n} − 1)F_{3n+1}F_{3n+4}F_{3n+6}}
< 0,

which completes the proof.

Lemma 51. If n ≥ 1 and m ≥ 2, we have

t_{6}(n) + t_{6}(n + 1) + t_{6}(mn) < 0.

Proof. If mn is even, then the result follows from Lemma 50 and the fact t_{6}(mn) < 0, so we assume that mn is odd in the rest. Now we have

t_{6}(mn) = \frac{−1}{2F_{3mn} − 1} + \frac{1}{F_{3mn+1}} + \frac{1}{2F_{3mn+3} − 1}
= \frac{−4F_{3mn+1}}{(2F_{3mn} − 1)(2F_{3mn+3} − 1)} + \frac{1}{F_{3mn+1}}
< \frac{−2F_{3mn+1}}{(2F_{3mn} − 1)F_{3mn+3}} + \frac{1}{F_{3mn+1}}
= \frac{−2F_{3mn+1}^{2} + 2F_{3mn}F_{3mn+3} − F_{3mn+3}}{(2F_{3mn} − 1)F_{3mn+1}F_{3mn+3}}
= \frac{−2(F_{3mn+1}^{2} − F_{3mn}F_{3mn+2}) + 2F_{3mn}F_{3mn+1} − F_{3mn+1} − F_{3mn+2}}{(2F_{3mn} − 1)F_{3mn+1}F_{3mn+3}}
= \frac{2 + (2F_{3mn} − 1)F_{3mn+1} − F_{3mn+2}}{(2F_{3mn} − 1)F_{3mn+1}F_{3mn+3}}
< \frac{(2F_{3mn} − 1)F_{3mn+1}}{(2F_{3mn} − 1)F_{3mn+1}F_{3mn+3}}
= \frac{1}{F_{3mn+3}} .

Since mn is odd, we must have that n is odd and m ≥ 3. Therefore,

t_{6}(mn) < \frac{1}{F_{9n+3}} .

It follows from the proof of Lemma 50 that if n is odd,

t_{6}(n) + t_{6}(n + 1) < \frac{4 − 2F_{3n+2}}{(2F_{3n} − 1)F_{3n+1}F_{3n+4}} < \frac{2 − F_{3n+2}}{F_{3n}F_{3n+1}F_{3n+4}} < −\frac{1}{F_{3n}F_{3n+1}F_{3n+4}} .

Now we arrive at

t_{6}(n) + t_{6}(n + 1) + t_{6}(mn) < \frac{1}{F_{9n+3}} − \frac{1}{F_{3n}F_{3n+1}F_{3n+4}} .

Employing Lemma 3, we easily see that F_{3n+3} > F_{2n}F_{n+4} and F_{2n} > F_{n}F_{n+1}, which implies

F_{9n+3} > F_{3n}F_{3n+1}F_{3n+4}.

Therefore,

t_{6}(n) + t_{6}(n + 1) + t_{6}(mn) < \frac{1}{F_{9n+3}} − \frac{1}{F_{3n}F_{3n+1}F_{3n+4}} < 0.

The proof is completed.

Proof of Theorem 35. We first consider the case where n is even. Applying Lemma 43, we have

∑_{k=n}^{2n} \frac{(−1)^{k}}{F_{3k+1}} = \frac{1}{2F_{3n}} − \frac{1}{2F_{6n+3}} − ∑_{k=n}^{2n} t_{1}(k) > \frac{1}{2F_{3n}} .

It follows from Lemma 44 and Lemma 45 that

∑_{k=n}^{2n} \frac{(−1)^{k}}{F_{3k+1}} = \frac{1}{2F_{3n} − 1} − \frac{1}{2F_{6n+3} − 1} − (t_{2}(n) + t_{2}(n + 1) + t_{2}(2n)) − ∑_{k=n+2}^{2n−1} t_{2}(k) < \frac{1}{2F_{3n} − 1}.

Therefore, we have

\frac{1}{2F_{3n}} < ∑_{k=n}^{2n} \frac{(−1)^{k}}{F_{3k+1}} < \frac{1}{2F_{3n} − 1},

which means that the result holds when n is even.

We now turn to consider the case where n is odd. From Lemma 48, we know that

∑_{k=n}^{2n} \frac{(−1)^{k}}{F_{3k+1}} = \frac{−1}{2F_{3n}} + \frac{1}{2F_{6n+3}} − ∑_{k=n}^{2n} t_{4}(k) > \frac{−1}{2F_{3n}} .

With the help of Lemma 49, it is easy to see that

∑_{k=n}^{2n} \frac{(−1)^{k}}{F_{3k+1}} = \frac{−1}{2F_{3n} + 1} − (t_{5}(n) + t_{5}(n + 1) − \frac{1}{2F_{6n+3} + 1}) − ∑_{k=n+2}^{2n} t_{5}(k) < \frac{−1}{2F_{3n} + 1}.

Thus, we obtain

\frac{−1}{2F_{3n}} < ∑_{k=n}^{2n} \frac{(−1)^{k}}{F_{3k+1}} < \frac{−1}{2F_{3n} + 1},

which yields the desired identity. Proof of Theorem 36. We first consider the case where n is even. Applying Lemma 39 and Lemma 41, we see

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k+1}} = \frac{1}{2F_{3n}} − \frac{1}{2F_{3mn+3}} − ∑_{k=n+2}^{mn−1} t_{1}(k) − (t_{1}(n) + t_{1}(n + 1) + t_{1}(mn)) < \frac{1}{2F_{3n}} .

It follows from Lemma 46 that

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k+1}} = \frac{1}{2F_{3n} + 1} − ∑_{k=n}^{mn−1} t_{3}(k) − (t_{3}(mn) + \frac{1}{2F_{3mn+3} + 1})
= \frac{1}{2F_{3n} + 1} − ∑_{k=n}^{mn−1} t_{3}(k) − (\frac{1}{2F_{3mn} + 1} − \frac{1}{F_{3mn+1}} )
> \frac{1}{2F_{3n} + 1}.

Therefore, we obtain

\frac{1}{2F_{3n} + 1} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k+1}} < \frac{1}{2F_{3n}} ,

which shows that the statement is true when n is even.

Now we turn to consider the case where n is odd. Lemma 47 tells us that

t_{4}(n) + t_{4}(n + 1) > 0.

Hence if mn is odd,

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k+1}} = \frac{−1}{2F_{3n}} − ∑_{k=n}^{mn−1} t_{4}(k) − (t_{4}(mn) − \frac{1}{2F_{3mn+3}} )
= \frac{−1}{2F_{3n}} − ∑_{k=n}^{mn−1} t_{4}(k) − (\frac{1}{F_{3mn+1}} − \frac{1}{2F_{3mn}} )
< \frac{−1}{2F_{3n}} .

And it follows from Lemma 50 and Lemma 51 that

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k+1}} = \frac{−1}{2F_{3n} − 1} + \frac{1}{2F_{3mn+3} − 1} − ∑_{k=n+2}^{mn−1} t_{6}(k) − (t_{6}(n) + t_{6}(n + 1) + t_{6}(mn))

> \frac{−1}{2F_{3n} − 1}.

Therefore, we have

\frac{−1}{2F_{3n} − 1} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k+1}} < \frac{−1}{2F_{3n}} .

If mn is even, then Lemma 47 implies that

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k+1}} = \frac{−1}{2F_{3n}} − ∑_{k=n+2}^{mn} t_{4}(k) − (t_{4}(n) + t_{4}(n + 1) − \frac{1}{2F_{3mn+3}} )

< \frac{−1}{2F_{3n}} − ∑_{k=n+2}^{mn} t_{4}(k) − (t_{4}(n) + t_{4}(n + 1) − \frac{1}{2F_{9n+3}} )

< \frac{−1}{2F_{3n}} ,

and from Lemma 50 we obtain

∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k+1}} = \frac{−1}{2F_{3n} − 1} + \frac{1}{2F_{3mn+3} − 1} − ∑_{k=n}^{mn} t_{6}(k) > \frac{−1}{2F_{3n} − 1}.

Hence, we also have

\frac{−1}{2F_{3n} − 1} < ∑_{k=n}^{mn} \frac{(−1)^{k}}{F_{3k+1}} < \frac{−1}{2F_{3n}} ,

which yields the desired identity.

6 Acknowledgments

The authors would like to thank the anonymous reviewers for their helpful comments. This work was supported by the Teaching Reform Research Project of University of Electronic Science and Technology of China (No. 2016XJYYB039).

References

[1] H. Ohtsuka and S. Nakamura, On the sum of reciprocal Fibonacci numbers, Fibonacci Quart. 46/47 (2008/2009), 153–159.

[2] R. R. Stone, General identities for Fibonacci and Lucas numbers with polynomial subscripts in several variables, Fibonacci Quart. 13 (1975), 289–294.

[3] N. N. Vorobiev, Fibonacci Numbers, Springer, 2002.

[4] A. Y. Z. Wang and P. B. Wen, On the partial finite sums of the reciprocals of the Fibonacci numbers, J. Inequal. Appl. 2015 (2015), Article 73.

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

(Concerned with sequence \underline{A000108}.)

Received July 9 2016; revised version received December 24 2016. Published in Journal of Integer Sequences, December 26 2016.

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