Symmetric Inequalities for Reciprocal Sums of Fibonacci Numbers

Abstract

This paper establishes symmetric inequalities for the reciprocal sums of Fibonacci numbers. By using elementary methods, recurrence properties, Cassini’s identity, and the Fibonacci–Lucas connection, we deduce the precise bounds for these reciprocal sums and an asymptotic formula for the corresponding floor function. This work is novel in two key aspects: it is the first to combine reciprocals of linear recurrence sequences with the tail of the Riemann zeta function, and the first to derive symmetric inequalities for such reciprocal sums.

1. Introduction and Main Results

For any integer $n\ge 0$, Fibonacci numbers are defined as follows:

$$\begin{array}{c}\hfill F_0=0,F_1=1,F_{n+2}=F_{n+1}+F_n.\end{array}$$

(1)

Despite being a sequence of integers, the Fibonacci numbers can be expressed using powers of irrational numbers, known as Binet’s formula:

$$\begin{array}{c}\hfill F_n={\displaystyle \frac{{\alpha }^n-{\beta }^n}{\sqrt{5}}},\end{array}$$

(2)

where

$$\begin{array}{c}\hfill \alpha ={\displaystyle \frac{1+\sqrt{5}}{2}}\approx 1.618,\beta ={\displaystyle \frac{1-\sqrt{5}}{2}}=-{\displaystyle \frac{1}{\alpha }}\approx -0.618.\end{array}$$

(3)

It is well-known that the Fibonacci sequence possesses a wealth of classic and important properties, spanning diverse fields such as number theory, algebra, combinatorics, and even geometry. Particularly noteworthy are the interesting summation formulas and combinatorial identities related to Fibonacci numbers, which play a vital role in number theory and combinatorial mathematics. These include, for instance, expressions for the partial sums

$$\begin{array}{c}\hfill \sum _{k=0}^n{F}_k=F_{n+2}-1,\sum _{k=1}^n{F}_{2k-1}=F_{2n},\sum _{k=1}^n{F}_{2k}=F_{2n+1}-1,\end{array}$$

the square sum

$$\begin{array}{c}\hfill \sum _{k=0}^n{F}_k^2=F_n{F}_{n+1},\end{array}$$

and the famous Cassini’s identity

$$\begin{array}{c}\hfill F_{n-1}{F}_{n+1}-F_n^2={(-1)}^n.\end{array}$$

(4)

For comprehensive treatments of Fibonacci numbers and their properties, we refer to the classic literature [1,2,3,4,5,6,7,8,9].

1.1. Reciprocal Sums in Linear Recurrence Sequences and Zeta Tails

The study of reciprocal sums of Fibonacci numbers gained significant momentum with the groundbreaking work of Ohtsuka and Nakamura [10], who discovered remarkable properties of infinite reciprocal sums. They established the following exact identities:

$$\begin{array}{cc}\hfill & ⌊{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{F_k}}\right)}^{-1}⌋=\left\{\begin{array}{cc}{\displaystyle F_{n-2},}\hfill & \mathrm{if}n\mathrm{is}\mathrm{even}\mathrm{and}n\ge 2,\hfill \\ \hfill & \hfill \\ {\displaystyle F_{n-2}-1,}\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd}\mathrm{and}n\ge 1,\hfill \end{array}\right.\hfill \\ \hfill \\ \hfill & ⌊{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{F_k^2}}\right)}^{-1}⌋=\left\{\begin{array}{cc}{\displaystyle F_{n-1}{F}_n-1,}\hfill & \mathrm{if}n\mathrm{is}\mathrm{even}\mathrm{and}n\ge 2,\hfill \\ \hfill & \hfill \\ {\displaystyle F_{n-1}{F}_n,}\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd}\mathrm{and}n\ge 1,\hfill \end{array}\right.\hfill \end{array}$$

where $\lfloor x\rfloor$ denotes the largest integer not exceeding x, and $F_{-1}=1$.

Subsequently, considerable mathematical effort has been devoted to studying reciprocal sums of linear recurrence sequences, producing a range of substantial outcomes. For example, Zhang and the second author [11] investigated the reciprocal sum of the Pell numbers and gained

$$\begin{array}{cc}\hfill & ⌊{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{P_k}}\right)}^{-1}⌋=\left\{\begin{array}{cc}{\displaystyle P_{n-1}+P_{n-2},}\hfill & \mathrm{if}n\mathrm{is}\mathrm{even}\mathrm{and}n\ge 2,\hfill \\ \hfill & \hfill \\ {\displaystyle P_{n-1}+P_{n-2}-1,}\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd}\mathrm{and}n\ge 1,\hfill \end{array}\right.\hfill \end{array}$$

where Pell numbers are defined by

$$\begin{array}{c}\hfill P_0=0,P_1=1,P_{n+2}=2P_{n+1}+P_n(n\ge 0),\end{array}$$

and $P_{-1}=P_1=1$. The second author [12] focused on the reciprocal sum of the cube Fibonacci numbers. By guessing terms and comparing coefficients and complicated numerical calculations, she drew the conclusion that

$$\begin{array}{c}\hfill ⌊{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{F_k^3}}\right)}^{-1}⌋=\left\{\begin{array}{cc}{\displaystyle F_n{F}_{n-1}^2+F_{n-2}{F}_n^2+⌊\frac{1}{11}(14F_{n-2}-5F_n)⌋,}\hfill & \mathrm{if}n\mathrm{is}\mathrm{even}\mathrm{and}n\ge 2,\hfill \\ \hfill & \hfill \\ {\displaystyle F_n{F}_{n-1}^2+F_{n-2}{F}_n^2+⌊\frac{1}{11}(5F_n-14F_{n-2})⌋,}\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd}\mathrm{and}n\ge 1,\hfill \end{array}\right.\end{array}$$

where $F_{-1}=F_1=1$. Other relevant results connected with the reciprocal sums of linear recurrence sequences can be found in [13,14,15,16].

A parallel line of research concerns the tail sums of the Riemann zeta function. For a positive integer n and complex number s with $\mathrm{Re}s>1$, define

$$\begin{array}{c}\hfill {\zeta }_n\left(s\right)=\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k^s}}.\end{array}$$

Significant work has been carried out on the reciprocal sums

$$\begin{array}{c}\hfill ⌊{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k^s}}\right)}^{-1}⌋\end{array}$$

for integer values of s. For specific aspects of this topic, one may refer to References [17,18].

1.2. Methodological Challenges and Research Motivation

The existing literature demonstrates that increasing the complexity of the summand—whether through raising the power exponent in linear recurrence sequences or considering higher orders in zeta function tails—significantly increases computational difficulty and produces more intricate results.

This observation naturally leads to our research question: what novel phenomena emerge when combining reciprocals of linear recurrence sequences with the tail sums of the Riemann zeta function? Specifically, for the hybrid reciprocal sums

$$\begin{array}{c}\hfill ⌊{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k^s{F}_k}}\right)}^{-1}⌋,\end{array}$$

(5)

do analogous results exist as the previous reciprocal sums? This is an exceptionally challenging problem, and one that has remained entirely unexplored in previous research.

In fact, the hybrid nature of the problem in (5) introduces substantial complexity beyond previous cases. Our preliminary investigations revealed that established methods from prior works [10,11,12] are inadequate for this problem. The fundamental difficulty arises because the equation constructed from products of Fibonacci numbers and polynomial terms admits no exact solution.

Consequently, we adopt an alternative analytical strategy based on term-by-term approximation and inequality techniques.

1.3. Our Approach and Main Results

The main purpose of this paper is to analyze the reciprocal sum derived from the Fibonacci numbers:

$$\begin{array}{c}\hfill {\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}.\end{array}$$

(6)

Our methodology proceeds according to the following steps:

(1)

For the sum in (6), we first classify the initial term n according to its parity (even or odd cases).

(2)

For the first two terms in the summation, we employ the method of undetermined coefficients to construct upper and lower bounds, which are expressed in a difference form involving consecutive terms of k. The specific constructions are detailed in formulae (29), (33), (38) and (42).

(3)

We then sum the inequalities deduced from the previous step from the initial term n to infinity, thereby obtaining the leading terms of the upper and lower bounds for (6).

(4)

By iterating the above process, we subsequently derive the remaining terms of the bounds.

The execution of Steps (2)–(4) relies critically on the repeated application of Cassini’s identity (4), the recurrence properties of Fibonacci numbers, and the interconnections between Fibonacci and Lucas numbers. By employing these tools, we establish a series of symmetric inequalities for (6). As a corollary, we also obtain an asymptotic formula for its integer part. We now present our main results.

Theorem 1.

If n is even and $n\ge 6$, then we have

$$\begin{array}{c}\hfill (n+1)F_{n-2}+{\displaystyle \frac{\sqrt{5}}{2}}{F}_{n-4}-{\displaystyle \frac{1}{2}}{F}_{n-4}<{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}<(n+1)F_{n-2}+{\displaystyle \frac{\sqrt{5}}{2}}{F}_{n-4}+{\displaystyle \frac{1}{2}}{F}_{n-4}.\end{array}$$

Theorem 2.

If n is odd and $n\ge 5$, then we have

$$\begin{array}{c}\hfill (n+1)F_{n-2}+{\displaystyle \frac{\sqrt{5}}{2}}{F}_{n-4}-{\displaystyle \frac{1}{2}}{F}_{n-4}-1<{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}<(n+1)F_{n-2}+{\displaystyle \frac{\sqrt{5}}{2}}{F}_{n-4}+{\displaystyle \frac{1}{2}}{F}_{n-4}.\end{array}$$

From the above theorems, we can immediately deduce the following conclusion:

Corollary 1.

For any integer $n\ge 4$, we have

$$\begin{array}{c}\hfill ⌊{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}⌋=n{F}_{n-2}+O\left(F_{n-2}\right).\end{array}$$

Remark 1.

The inequalities in Theorem 1 exhibit elegant symmetry: both bounds share the core structure $(n+1)F_{n-2}+{\displaystyle \frac{\sqrt{5}}{2}}{F}_{n-4}$, differing only in the sign of ${\displaystyle \frac{1}{2}}{F}_{n-4}$. This “mirror symmetry” maintains consistency in the main expression while using simple sign alternation to bound the reciprocal sum precisely.

Theorem 2 demonstrates a “perturbation symmetry,” where the lower bound incorporates a minor “$-1$” term. Relative to the order of $F_{n-4}$, this perturbation is negligible, reflecting the sophisticated construction of symmetric relations in mathematics through minimal adjustment.

Remark 2.

This paper is structured as follows. Section 2 establishes the necessary technical lemmas. Building on these foundations, Section 3 presents the proofs of Theorems 1 and 2. Finally, the proof of Corollary 1, which is a direct consequence of these theorems, is provided in Section 4.

2. Several Preliminary Lemmas

For any integer $n\ge 0$, the Lucas numbers are defined by

$$\begin{array}{c}\hfill L_0=2,L_1=1,L_{n+2}=L_{n+1}+L_n.\end{array}$$

(7)

Like the Fibonacci sequence, the Lucas sequence is a second-order linear recurrence sequence, and Binet’s formula is

$$\begin{array}{c}\hfill L_n={\alpha }^n+{\beta }^n,\end{array}$$

(8)

where $\alpha$ and $\beta$ are defined by (3). In addition, it is easy to derive from (2), (4) and (8) that

$$\begin{array}{cc}\hfill 5F_s{F}_t& =L_{s+t}-{(-1)}^t{L}_{s-t},s\ge t,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill L_s{F}_t& =F_{s+t}-{(-1)}^t{F}_{s-t},s\ge t,.\hfill \end{array}$$

(10)

where s and t are non-negative integers.

We display four lemmas in this section, which will be indispensable in the proofs of Theorems 1 and 2. These four lemmas involve inequalities of Fibonacci and Lucas numbers with differing orders (e.g., $F_{6k-1}$, $F_{2k+2}$ and $L_{4k+3}$ in Lemma 3), which necessitates not only diverse proof techniques but also a case-by-case analysis of the parameter k.

Lemma 1.

Let

$$\begin{array}{cc}\hfill H_1\left(k\right)=& \left[-4k^2+(4-4\beta -2{\beta }^2)k+(2{\beta }^2-4\beta +3)\right]·F_{6k-1}\hfill \\ \hfill & +\left[-4\beta k^2+(4{\beta }^2-2\beta )k-(3{\beta }^2-4\beta )\right]·F_{6k-2}\hfill \\ \hfill & -\left[80k^3+(-160\beta +148)k^2+(74{\beta }^2-192\beta +72)k+(21{\beta }^2-32\beta +9)\right]·F_{2k+2}\hfill \\ \hfill & +\left[80k^3+(-184\beta +152)k^2+(110{\beta }^2-210\beta +68)k+(31{\beta }^2-29\beta +6)\right]·F_{2k+1},\hfill \end{array}$$

then, $H_1\left(k\right)<0$ holds for any integer $k\ge 1$.

Proof. 

Firstly, we prove that for $k\ge 4$, one has

$$\begin{array}{cc}\hfill & \left[-4\beta k^2+(4{\beta }^2-2\beta )k-(3{\beta }^2-4\beta )\right]·F_{6k-2}\hfill \\ \hfill <& \left[4k^2+(2{\beta }^2+4\beta -4)k-(2{\beta }^2-4\beta +3)\right]·F_{6k-1}.\hfill \end{array}$$

(11)

Recall that

$$\begin{array}{c}\hfill F_{6k-1}=F_{6k-2}+F_{6k-3}.\end{array}$$

(12)

Substituting (12) into (11), one can get that

$$\begin{array}{cc}\hfill & \left[(-4\beta -4)k^2+(2{\beta }^2-6\beta +4)k-({\beta }^2-3)\right]·F_{6k-2}\hfill \\ \hfill <& \left[4k^2+(2{\beta }^2+4\beta -4)k-(2{\beta }^2-4\beta +3)\right]·F_{6k-3}.\hfill \end{array}$$

(13)

Let

$$\begin{array}{cc}\hfill & f_1\left(k\right)=(-4\beta -4)k^2+(2{\beta }^2-6\beta +4)k-({\beta }^2-3),\hfill \\ \hfill & g_1\left(k\right)=4k^2+(2{\beta }^2+4\beta -4)k-(2{\beta }^2-4\beta +3).\hfill \end{array}$$

Noting that $\beta$ is a constant, $f_1\left(k\right)$ and $g_1\left(k\right)$ are quadratic functions in k. Then, it follows from the fundamental properties of quadratic functions that $f_1\left(k\right)<0$ holds for $k\ge 6$; meanwhile, $g_1\left(k\right)>0$ is always true for $k\ge 3$.

Furthermore, since $F_{6k-2}$ and $F_{6k-3}$ are positive for all integers $k\ge 1$, inequality (13) holds for $k\ge 6$. Direct verification for $k=4$ and $k=5$ confirms that (13) also holds in these cases. We therefore conclude that inequality (11) is valid for all integers $k\ge 4$.

Secondly, we prove that for $k\ge 1$, the following inequality holds:

$$\begin{array}{cc}\hfill & \left[80k^3+(-184\beta +152)k^2+(110{\beta }^2-210\beta +68)k+(31{\beta }^2-29\beta +6)\right]·F_{2k+1}\hfill \\ \hfill <& \left[80k^3+(-160\beta +148)k^2+(74{\beta }^2-192\beta +72)k+(21{\beta }^2-32\beta +9)\right]·F_{2k+2}.\hfill \end{array}$$

(14)

Noting that

$$\begin{array}{c}\hfill F_{2k+2}=2F_{2k}+F_{2k-1},F_{2k+1}=F_{2k}+F_{2k-1},\end{array}$$

so, (14) is equivalent to

$$\begin{array}{cc}\hfill & \left[80k^3+(-136\beta +144)k^2+(38{\beta }^2-174\beta +76)k+(31{\beta }^2-35\beta +12)\right]·F_{2k}\hfill \\ \hfill >& \left[(-24\beta +4)k^2+(36{\beta }^2-18\beta -4)k-(10{\beta }^2-3\beta +3)\right]·F_{2k-1}.\hfill \end{array}$$

(15)

It follows from the value of $\beta$ defined by (3) that inequality (15) holds for all $k\ge 1$, and consequently, inequality (14) holds as well.

The combination of (11) and (14) shows $H_1\left(k\right)<0$ for $k\ge 4$. A direct check confirms the cases $k=1,2,3$. □

Lemma 2.

Let

$$\begin{array}{cc}\hfill H_2\left(k\right)=& \left[-4k^2+(4+4\alpha -2{\alpha }^2)k+(2{\alpha }^2+4\alpha +3)\right]·F_{6k-1}\hfill \\ \hfill & +\left[4\alpha k^2+(4{\alpha }^2+2\alpha )k-(3{\alpha }^2+4\alpha )\right]·F_{6k-2}\hfill \\ \hfill & -\left[80k^3+(160\alpha +148)k^2+(74{\alpha }^2+192\alpha +72)k+(21{\alpha }^2+32\alpha +9)\right]·F_{2k+2}\hfill \\ \hfill & +\left[80k^3+(184\alpha +152)k^2+(110{\alpha }^2+210\alpha +68)k+(31{\alpha }^2+29\alpha +6)\right]·F_{2k+1},\hfill \end{array}$$

then, $H_2\left(k\right)>0$ holds for any integer $k\ge 3$.

Proof. 

It is easy to verify that $H_2\left(3\right)>0$, so let us focus on the case $k\ge 4$. By (10), we have

$$\begin{array}{cc}\hfill & F_{6k-1}=F_{2k+2}·L_{4k-3}+F_{2k-5},\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill & F_{6k-2}=F_{2k+1}·L_{4k-3}-F_{2k-4}.\hfill \end{array}$$

(17)

On the other hand, from the recurrence of Fibonacci numbers, one has

$$\begin{array}{cc}\hfill & F_{2k-4}=5F_{2k+2}-8F_{2k+1},\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill & F_{2k-5}=-8F_{2k+2}+13F_{2k+1}.\hfill \end{array}$$

(19)

Substituting (16)–(19) into $H_2\left(k\right)$, we get

$$\begin{array}{c}\hfill H_2\left(k\right)=-4k^2·L_{4k-3}{F}_{2k+2}+4\alpha k^2·L_{4k-3}{F}_{2k+1}+g_2\left(k\right)·F_{2k+2}+f_2\left(k\right)·F_{2k+1},\end{array}$$

(20)

where

$$\begin{array}{cc}\hfill g_2\left(k\right)=& \left[(4+4\alpha -2{\alpha }^2)k+(2{\alpha }^2+4\alpha +3)\right]·L_{4k-3}\hfill \\ \hfill & -\left[80k^3+(180\alpha +116)k^2+(78{\alpha }^2+234\alpha +104)k+(22{\alpha }^2+44\alpha +33)\right],\hfill \\ \hfill f_2\left(k\right)=& \left[(4{\alpha }^2+2\alpha )k-(3{\alpha }^2+4\alpha )\right]·L_{4k-3}\hfill \\ \hfill & +\left[80k^3+(216\alpha +100)k^2+(116{\alpha }^2+278\alpha +120)k+(33{\alpha }^2+49\alpha +45)\right].\hfill \end{array}$$

Note that

$$\begin{array}{cc}\hfill & -4k^2·L_{4k-3}{F}_{2k+2}+4\alpha k^2·L_{4k-3}{F}_{2k+1}\hfill \\ \hfill =& 4k^2·L_{4k-3}·(-F_{2k+2}+\alpha F_{2k+1})\hfill \\ \hfill =& 4k^2·L_{4k-3}·\left[(\alpha -1)F_{2k+1}-F_{2k}\right].\hfill \end{array}$$

By induction, we establish that $(\alpha -1)F_{2k+1}-F_{2k}>0$ for all integers $k\ge 4$. Furthermore, from the value of $\alpha$ defined by (3) and the properties of Lucas numbers (8), it is straightforward to verify that both $g_2\left(k\right)>0$ and $f_2\left(k\right)>0$ hold for any $k\ge 1$.

Consequently, we prove that $H_2\left(k\right)>0$ for all $k\ge 4$. □

Lemma 3.

Let

$$\begin{array}{cc}\hfill H_3\left(k\right)=& -\left[4k^2+(4\beta +2{\beta }^2)k-({\beta }^2-6\beta +4)\right]·F_{6k+2}\hfill \\ \hfill & -\left[4\beta k^2-(4{\beta }^2-6\beta )k+({\beta }^2-2\beta )\right]·F_{6k+1}\hfill \\ \hfill & -\left[4k^2+(14+2\beta )k+(10-2\beta )\right]·L_{4k+3}-\left[8k^2+(14-6\beta )k+(4+\beta )\right]·L_{4k+2}\hfill \\ \hfill & +\left[80k^3+(-160\beta +268)k^2+(74{\beta }^2-352\beta +290)k+(58{\beta }^2-168\beta +97)\right]·F_{2k+3}\hfill \\ \hfill & -\left[80k^3+(-184\beta +272)k^2+(110{\beta }^2-394\beta +280)k+(86{\beta }^2-180\beta +83)\right]·F_{2k+2}\hfill \\ \hfill & -\left[40k^2+(82-50\beta )k+(38-40\beta )\right],\hfill \end{array}$$

then $H_3\left(k\right)<0$ holds for any integer $k\ge 2$.

Proof. 

Firstly, we prove that for $k\ge 3$, the following inequality holds:

$$\begin{array}{c}\hfill \left[-4\beta k^2+(4{\beta }^2-6\beta )k-({\beta }^2-2\beta )\right]F_{6k+1}<\left[4k^2+(2{\beta }^2+4\beta )k-({\beta }^2-6\beta +4)\right]F_{6k+2}.\end{array}$$

(21)

Let

$$\begin{array}{cc}\hfill & f_3\left(k\right)=-4\beta k^2+(4{\beta }^2-6\beta )k-({\beta }^2-2\beta ),\hfill \\ \hfill & g_3\left(k\right)=4k^2+(2{\beta }^2+4\beta )k-({\beta }^2-6\beta +4),\hfill \end{array}$$

then we can derive that

$$\begin{array}{c}\hfill g_3\left(k\right)-f_3\left(k\right)=(4+4\beta )k^2+(-2{\beta }^2+10\beta )k+(4\beta -4)>0\end{array}$$

holds for $k\ge 6$, which implies that (21) is valid in this range. Furthermore, direct verification confirms that (21) also holds for the remaining cases $k=3,4,5$.

Next, we prove that for $k\ge 3$, one has

$$\begin{array}{cc}\hfill & \left[4k^2+(14+2\beta )k+(10-2\beta )\right]·L_{4k+3}+\left[8k^2+(14-6\beta )k+(4+\beta )\right]·L_{4k+2}\hfill \\ \hfill & +\left[80k^3+(-184\beta +272)k^2+(110{\beta }^2-394\beta +280)k+(86{\beta }^2-180\beta +83)\right]·F_{2k+2}\hfill \\ \hfill >& \left[80k^3+(-160\beta +268)k^2+(74{\beta }^2-352\beta +290)k+(58{\beta }^2-168\beta +97)\right]·F_{2k+3}.\hfill \end{array}$$

(22)

Equation (9) yields that for all integers $k\ge 3$,

$$\begin{array}{cc}\hfill & L_{4k+3}=5F_{2k}{F}_{2k+3}+4>5·2k·F_{2k+3},\hfill \\ \hfill & L_{4k+2}=5F_{2k}{F}_{2k+2}+3>5·2k·F_{2k+2}.\hfill \end{array}$$

Hence, it suffices to show

$$\begin{array}{cc}\hfill & \left[4k^2+(14+2\beta )k+(10-2\beta )\right]·10k{F}_{2k+3}+\left[8k^2+(14-6\beta )k+(4+\beta )\right]·10k{F}_{2k+2}\hfill \\ \hfill & +\left[80k^3+(-184\beta +272)k^2+(110{\beta }^2-394\beta +280)k+(86{\beta }^2-180\beta +83)\right]·F_{2k+2}\hfill \\ \hfill >& \left[80k^3+(-160\beta +268)k^2+(74{\beta }^2-352\beta +290)k+(58{\beta }^2-168\beta +97)\right]·F_{2k+3},\hfill \end{array}$$

(23)

that is,

$$\begin{array}{cc}\hfill & \left[120k^3+(-64\beta +284)k^2+(36{\beta }^2-52\beta +130)k+(28{\beta }^2-12\beta -14)\right]·F_{2k+2}\hfill \\ \hfill >& \left[40k^3+(-180\beta +128)k^2+(74{\beta }^2-332\beta +190)k+(58{\beta }^2-168\beta +97)\right]·F_{2k+1}.\hfill \end{array}$$

(24)

Given that $\beta <0$ and $F_{2k+2}>F_{2k+1}$ for any positive integer k, inequality (24) clearly holds for all $k\ge 3$. It immediately follows that inequalities (22) and (23) must also hold.

Furthermore, from the value of $\beta$ defined by (3) and the properties of quadratic functions, we can derive that

$$\begin{array}{c}\hfill 40k^2+(82-50\beta )k+(38-40\beta )>0\end{array}$$

(25)

holds for all $k\ge 1$.

It follows from inequalities (21), (22) and (25) that $H_3\left(k\right)<0$ for all $k\ge 3$. Moreover, a direct substitution of $k=2$ readily verifies that $H_3\left(2\right)<0$ also holds. □

Lemma 4.

Let

$$\begin{array}{cc}\hfill H_4\left(k\right)=& \left[-4k^2+(4\alpha -2{\alpha }^2)k+({\alpha }^2+6\alpha +4)\right]·F_{6k+2}\hfill \\ \hfill & +\left[4\alpha k^2+(4{\alpha }^2+6\alpha )k-({\alpha }^2+2\alpha )\right]·F_{6k+1}\hfill \\ \hfill & +\left[80k^3+(160\alpha +268)k^2+(74{\alpha }^2+352\alpha +280)k+(58{\alpha }^2+168\alpha +92)\right]·F_{2k+3}\hfill \\ \hfill & -\left[80k^3+(184\alpha +272)k^2+(110{\alpha }^2+394\alpha +280)k+(86{\alpha }^2+180\alpha +88)\right]·F_{2k+2},\hfill \end{array}$$

then $H_4\left(k\right)>0$ holds for any integer $k\ge 1$.

Proof. 

First of all, we prove that for $k\ge 1$, we have

$$\begin{array}{c}\hfill \left[-4k^2+(4\alpha -2{\alpha }^2)k+({\alpha }^2+6\alpha +4)\right]F_{6k+2}+\left[4\alpha k^2+(4{\alpha }^2+6\alpha )k-({\alpha }^2+2\alpha )\right]F_{6k+1}>0.\end{array}$$

(26)

Let

$$\begin{array}{cc}\hfill & f_4\left(k\right)=-4k^2·F_{6k+2}+4\alpha k^2·F_{6k+1}=4k^2·\left[(\alpha -1)F_{6k+1}-F_{6k}\right],\hfill \\ \hfill & g_4\left(k\right)=\left[(4\alpha -2{\alpha }^2)k+({\alpha }^2+6\alpha +4)\right]F_{6k+2}+\left[(4{\alpha }^2+6\alpha )k-({\alpha }^2+2\alpha )\right]F_{6k+1}.\hfill \end{array}$$

By induction, we establish that $(\alpha -1)F_{6k+1}-F_{6k}>0$ for all $k\ge 1$, which implies $f_4\left(k\right)>0$. Moreover, from the value of $\alpha$ defined by (3) we can show $g_4\left(k\right)>0$. Hence, inequality (26) holds for every integer $k\ge 1$.

From the recurrence of the Fibonacci sequence, we can deduce that for $k\ge 1$,

$$\begin{array}{cc}\hfill & \left[80k^3+(160\alpha +268)k^2+(74{\alpha }^2+352\alpha +280)k+(58{\alpha }^2+168\alpha +92)\right]·F_{2k+3}\hfill \\ \hfill & -\left[80k^3+(184\alpha +272)k^2+(110{\alpha }^2+394\alpha +280)k+(86{\alpha }^2+180\alpha +88)\right]·F_{2k+2}\hfill \\ \hfill =& \left[80k^3+(136\alpha +264)k^2+(38{\alpha }^2+310\alpha +280)k+(30{\alpha }^2+156\alpha +96)\right]·F_{2k+1}\hfill \\ \hfill & -\left[(24\alpha +4)k^2+(36{\alpha }^2+42\alpha )k+(28{\alpha }^2+12\alpha -4)\right]·F_{2k}>0.\hfill \end{array}$$

(27)

Combining (26) and (27), we complete the proof of Lemma 4. □

3. Proofs of the Theorems

3.1. Proof of Theorem 1

If n is an even number, let $n=2m$ $(m\ge 3)$; then, Theorem 1 is reduced to

$$\begin{array}{c}\hfill (2m+1)F_{2m-2}-\beta F_{2m-4}<{\left(\sum _{k=2m}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}<(2m+1)F_{2m-2}+\alpha F_{2m-4},\end{array}$$

which is equivalent to

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{(2m+1)F_{2m-2}+\alpha F_{2m-4}}}<\sum _{k=2m}^{\infty }{\displaystyle \frac{1}{k{F}_k}}<{\displaystyle \frac{1}{(2m+1)F_{2m-2}-\beta F_{2m-4}}}.\end{array}$$

(28)

3.1.1. Proof of the Right-Hand-Side Inequality of (28)

Now, we prove that for all positive integers $k\ge 3$, the following inequality holds:

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2k{F}_{2k}}}+{\displaystyle \frac{1}{(2k+1)F_{2k+1}}}<{\displaystyle \frac{1}{(2k+1)F_{2k-2}-\beta F_{2k-4}}}-{\displaystyle \frac{1}{(2k+3)F_{2k}-\beta F_{2k-2}}}.\end{array}$$

(29)

For convenience, we let

$$\begin{array}{ccc}\hfill & {\displaystyle \frac{1}{2k{F}_{2k}}}+{\displaystyle \frac{1}{(2k+1)F_{2k+1}}}={\displaystyle \frac{A_1}{B_1}},\hfill & \hfill {\displaystyle \frac{1}{(2k+1)F_{2k-2}-\beta F_{2k-4}}}-{\displaystyle \frac{1}{(2k+3)F_{2k}-\beta F_{2k-2}}}={\displaystyle \frac{C_1}{D_1}}.\end{array}$$

Next, we prove

$$\begin{array}{c}\hfill {\displaystyle \frac{A_1}{B_1}}<{\displaystyle \frac{C_1}{D_1}},\end{array}$$

(30)

where $A_1$, $B_1$, $C_1$, $D_1$ can be deduced from (1), (7) and (9) as

$$\begin{array}{cc}\hfill A_1=& 2k{F}_{2k+2}+F_{2k+1},\hfill \\ \hfill B_1=& 2k(2k+1)(L_{4k+1}-1)/5,\hfill \\ \hfill C_1=& (2k+3)F_{2k-1}+2F_{2k-2}-\beta F_{2k-3},\hfill \\ \hfill D_1=& (2k+1)(2k+3)(L_{4k-2}-3)/5-(2k+1)\beta (L_{4k-4}-2)/5\hfill \\ \hfill & -(2k+3)\beta (L_{4k-4}-7)/5+{\beta }^2(L_{4k-6}-3)/5.\hfill \end{array}$$

Noting that for $k\ge 3$, $A_1$, $B_1$, $C_1$, $D_1$ are positive, so (30) can be reduced to

$$\begin{array}{c}\hfill A_1{D}_1<B_1{C}_1.\end{array}$$

(31)

From (9), (10) and the recurrence of the Fibonacci sequence, we can get that

$$\begin{array}{cc}\hfill 5A_1{D}_1=& 2k(2k+1)(2k+3)(F_{6k}-F_{2k-4})+(2k+1)(2k+3)(F_{6k-1}+F_{2k-3})\hfill \\ \hfill & -2k(4k+4)\beta (F_{6k-2}-F_{2k-6})-(4k+4)\beta (F_{6k-3}+F_{2k-5})\hfill \\ \hfill & +2k{\beta }^2·(F_{6k-4}-F_{2k-8})+{\beta }^2·(F_{6k-5}+F_{2k-7})\hfill \\ \hfill & -\left[12k^2+(24-18\beta )k+(3{\beta }^2-23\beta +9)\right]·(2k{F}_{2k+2}+F_{2k+1})\hfill \\ \hfill =& 2k(2k+1)(2k+3)F_{6k}+\left[4k^2+(8-4\beta -2{\beta }^2)k+(2{\beta }^2-4\beta +3)\right]F_{6k-1}\hfill \\ \hfill & +\left[-8\beta k^2+(4{\beta }^2-4\beta )k+(-3{\beta }^2+4\beta )\right]F_{6k-2}\hfill \\ \hfill & -\left[64k^3+(-140\beta +140)k^2+(74{\beta }^2-182\beta +72)k+(21{\beta }^2-32\beta +9)\right]F_{2k+2}\hfill \\ \hfill & +\left[64k^3+(-168\beta +136)k^2+(110{\beta }^2-202\beta +64)k+(31{\beta }^2-29\beta +6)\right]F_{2k+1}.\hfill \end{array}$$

Moreover,

$$\begin{array}{cc}\hfill 5B_1{C}_1=& 2k(2k+1)(2k+3)(F_{6k}+F_{2k+2})+4k(2k+1)(F_{6k-1}-F_{2k+3})\hfill \\ \hfill & -2k(2k+1)\beta (F_{6k-2}+F_{2k+4})-2k(2k+1)(2k+3)F_{2k-1}\hfill \\ \hfill & -4k(2k+1)F_{2k-2}+2k(2k+1)\beta F_{2k-3}\hfill \\ \hfill =& 2k(2k+1)(2k+3)F_{6k}+4k(2k+1)F_{6k-1}-2k(2k+1)\beta ·F_{6k-2}\hfill \\ \hfill & +\left[16k^3+(-20\beta +8)k^2-10\beta k\right]F_{2k+2}-\left[16k^3+(-16\beta +16)k^2+(-8\beta +4)k\right]F_{2k+1}.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill & 5(A_1{D}_1-B_1{C}_1)\hfill \\ \hfill =& \left[-4k^2+(4-4\beta -2{\beta }^2)k+2{\beta }^2-4\beta +3\right]F_{6k-1}+\left[-4\beta k^2+(4{\beta }^2-2\beta )k-3{\beta }^2+4\beta \right]F_{6k-2}\hfill \\ \hfill & -\left[80k^3+(-160\beta +148)k^2+(74{\beta }^2-192\beta +72)k+(21{\beta }^2-32\beta +9)\right]F_{2k+2}\hfill \\ \hfill & +\left[80k^3+(-184\beta +152)k^2+(110{\beta }^2-210\beta +68)k+(31{\beta }^2-29\beta +6)\right]F_{2k+1}.\hfill \end{array}$$

It follows directly from Lemma 1 that

$$\begin{array}{c}\hfill 5(A_1{D}_1-B_1{C}_1)=H_1\left(k\right)<0\end{array}$$

holds for all $k\ge 3$. Consequently, inequalities (29)–(31) are established.

Now applying (29) repeatedly, we get

$$\begin{array}{c}\hfill \sum _{k=m}^{\infty }\left({\displaystyle \frac{1}{2k{F}_{2k}}}+{\displaystyle \frac{1}{(2k+1)F_{2k+1}}}\right)<\sum _{k=m}^{\infty }\left({\displaystyle \frac{1}{(2k+1)F_{2k-2}-\beta F_{2k-4}}}-{\displaystyle \frac{1}{(2k+3)F_{2k}-\beta F_{2k-2}}}\right),\end{array}$$

that is,

$$\begin{array}{c}\hfill \sum _{k=2m}^{\infty }{\displaystyle \frac{1}{k{F}_k}}<{\displaystyle \frac{1}{(2m+1)F_{2m-2}-\beta F_{2m-4}}}.\end{array}$$

(32)

□

3.1.2. Proof of the Left-Hand-Side Inequality of (28)

On the other hand, we prove that for any integer $k\ge 3$, the following inequality holds:

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2k{F}_{2k}}}+{\displaystyle \frac{1}{(2k+1)F_{2k+1}}}>{\displaystyle \frac{1}{(2k+1)F_{2k-2}+\alpha F_{2k-4}}}-{\displaystyle \frac{1}{(2k+3)F_{2k}+\alpha F_{2k-2}}}.\end{array}$$

(33)

Let

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{(2k+1)F_{2k-2}+\alpha F_{2k-4}}}-{\displaystyle \frac{1}{(2k+3)F_{2k}+\alpha F_{2k-2}}}={\displaystyle \frac{C_2}{D_2}},\end{array}$$

then we will prove

$$\begin{array}{c}\hfill {\displaystyle \frac{A_1}{B_1}}>{\displaystyle \frac{C_2}{D_2}},\end{array}$$

(34)

where

$$\begin{array}{cc}\hfill C_2=& (2k+3)F_{2k-1}+2F_{2k-2}+\alpha F_{2k-3},\hfill \\ \hfill D_2=& (2k+1)(2k+3)(L_{4k-2}-3)/5+(2k+1)\alpha ·(L_{4k-4}-2)/5\hfill \\ \hfill & +(2k+3)\alpha ·(L_{4k-4}-7)/5+{\alpha }^2·(L_{4k-6}-3)/5\hfill \\ \hfill =& (2k+1)(2k+3)L_{4k-2}/5+(4k+4)\alpha ·L_{4k-4}/5+{\alpha }^2·L_{4k-6}/5\hfill \\ \hfill & -\left[12k^2+(24+18\alpha )k+(3{\alpha }^2+23\alpha +9)\right]/5.\hfill \end{array}$$

Recall that for integer $k\ge 3$, $B_1$ and $D_2$ are both positive, so inequality (34) is equivalent to

$$\begin{array}{c}\hfill A_1{D}_2>B_1{C}_2.\end{array}$$

(35)

From (1), (7), (9) and (10), we can deduce

$$\begin{array}{cc}\hfill 5A_1{D}_2=& 2k·(2k+1)(2k+3)F_{6k}+(2k+1)(2k+3)F_{6k-1}+2k·(4k+4)\alpha ·F_{6k-2}\hfill \\ \hfill & +(4k+4)\alpha ·F_{6k-3}+2k{\alpha }^2·F_{6k-4}+{\alpha }^2·F_{6k-5}-2k·(2k+1)(2k+3)F_{2k-4}\hfill \\ \hfill & +(2k+1)(2k+3)F_{2k-3}-2k\alpha (4k+4)F_{2k-6}+(4k+4)\alpha F_{2k-5}-2k{\alpha }^2{F}_{2k-8}\hfill \\ \hfill & +{\alpha }^2{F}_{2k-7}-\left[12k^2+(24+18\alpha )k+(3{\alpha }^2+23\alpha +9)\right]·(2k{F}_{2k+2}+F_{2k+1}),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill 5B_1{C}_2=& 2k·(2k+1)(2k+3)(F_{6k}+F_{2k+2})+4k·(2k+1)(F_{6k-1}-F_{2k+3})\hfill \\ \hfill & +2k\alpha ·(2k+1)(F_{6k-2}+F_{2k+4})-2k·(2k+1)(2k+3)F_{2k-1}\hfill \\ \hfill & -4k·(2k+1)F_{2k-2}-2k\alpha ·(2k+1)F_{2k-3}.\hfill \end{array}$$

From the recurrence of the Fibonacci sequence, we have

$$\begin{array}{cc}\hfill & 5(A_1{D}_2-B_1{C}_2)\hfill \\ \hfill =& \left[-4k^2+(4+4\alpha -2{\alpha }^2)k+2{\alpha }^2+4\alpha +3\right]F_{6k-1}+\left[4\alpha k^2+(4{\alpha }^2+2\alpha )k-3{\alpha }^2-4\alpha \right]F_{6k-2}\hfill \\ \hfill & -\left[80k^3+(160\alpha +148)k^2+(74{\alpha }^2+192\alpha +72)k+(21{\alpha }^2+32\alpha +9)\right]·F_{2k+2}\hfill \\ \hfill & +\left[80k^3+(184\alpha +152)k^2+(110{\alpha }^2+210\alpha +68)k+(31{\alpha }^2+29\alpha +6)\right]·F_{2k+1}.\hfill \end{array}$$

An application of Lemma 2 shows that

$$\begin{array}{c}\hfill 5(A_1{D}_2-B_1{C}_2)=H_2\left(k\right)>0\end{array}$$

for any integer $k\ge 3$. Therefore, the inequalities (33)–(35) hold.

Applying (33) repeatedly, one has

$$\begin{array}{c}\hfill \sum _{k=m}^{\infty }\left({\displaystyle \frac{1}{2k{F}_{2k}}}+{\displaystyle \frac{1}{(2k+1)F_{2k+1}}}\right)>\sum _{k=m}^{\infty }\left({\displaystyle \frac{1}{(2k+1)F_{2k-2}+\alpha F_{2k-4}}}-{\displaystyle \frac{1}{(2k+3)F_{2k}+\alpha F_{2k-2}}}\right),\end{array}$$

that is

$$\begin{array}{c}\hfill \sum _{k=2m}^{\infty }{\displaystyle \frac{1}{k{F}_k}}>{\displaystyle \frac{1}{(2m+1)F_{2m-2}+\alpha F_{2m-4}}}.\end{array}$$

(36)

□

Then, (28) follows from (32) and (36), which completes the proof of Theorem 1. □

3.2. Proof of Theorem 2

Let $n=2m+1$ $(m\ge 2)$; then, Theorem 2 is equivalent to

$$\begin{array}{c}\hfill (2m+2)F_{2m-1}-\beta F_{2m-3}-1<{\left(\sum _{k=2m+1}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}<(2m+2)F_{2m-1}+\alpha F_{2m-3},\end{array}$$

or

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{(2m+2)F_{2m-1}+\alpha F_{2m-3}}}<\sum _{k=2m+1}^{\infty }{\displaystyle \frac{1}{k{F}_k}}<{\displaystyle \frac{1}{(2m+2)F_{2m-1}-\beta F_{2m-3}-1}}.\end{array}$$

(37)

3.2.1. Proof of the Right-Hand-Side Inequality of (37)

Next, we prove that for any integer $k\ge 2$, the following inequality holds:

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{(2k+1)F_{2k+1}}}+{\displaystyle \frac{1}{(2k+2)F_{2k+2}}}<{\displaystyle \frac{1}{(2k+2)F_{2k-1}-\beta F_{2k-3}-1}}-{\displaystyle \frac{1}{(2k+4)F_{2k+1}-\beta F_{2k-1}-1}}.\end{array}$$

(38)

For convenience, we let

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{(2k+1)F_{2k+1}}}+{\displaystyle \frac{1}{(2k+2)F_{2k+2}}}={\displaystyle \frac{A_2}{B_2}},\hfill \\ \hfill & {\displaystyle \frac{1}{(2k+2)F_{2k-1}-\beta F_{2k-3}-1}}-{\displaystyle \frac{1}{(2k+4)F_{2k+1}-\beta F_{2k-1}-1}}={\displaystyle \frac{C_3}{D_3}}.\hfill \end{array}$$

Then, we can prove

$$\begin{array}{c}\hfill {\displaystyle \frac{A_2}{B_2}}<{\displaystyle \frac{C_3}{D_3}},\end{array}$$

(39)

where $A_2$, $B_2$, $C_3$, $D_3$ can be calculated from (1), (7) and (9) as

$$\begin{array}{cc}\hfill A_2=& (2k+1)F_{2k+3}+F_{2k+2},\hfill \\ \hfill B_2=& (2k+1)(2k+2)(L_{4k+3}+1)/5,\hfill \\ \hfill C_3=& (2k+4)F_{2k}+2F_{2k-1}-\beta F_{2k-2},\hfill \\ \hfill D_3=& (2k+2)(2k+4)(L_{4k}+3)/5-(2k+2)\beta (L_{4k-2}+2)/5\hfill \\ \hfill & -(2k+2)F_{2k-1}-(2k+4)\beta (L_{4k-2}+7)/5+{\beta }^2·(L_{4k-4}+3)/5\hfill \\ \hfill & +\beta F_{2k-3}-(2k+4))F_{2k+1}+\beta F_{2k-1}+1\hfill \\ \hfill =& (2k+2)(2k+4)L_{4k}/5-(4k+6)\beta L_{4k-2}/5+{\beta }^2·L_{4k-4}/5-(2k+2)F_{2k-1}\hfill \\ \hfill & -(2k+4)F_{2k+1}+\beta F_{2k-3}+\beta F_{2k-1}+\left[12k^2+(36-18\beta )k+(3{\beta }^2-32\beta +29)\right]/5.\hfill \end{array}$$

For $k\ge 2$, we notice that $B_2$ and $D_3$ are both positive, so inequality (39) can be reduced to

$$\begin{array}{c}\hfill A_2{D}_3<B_2{C}_3.\end{array}$$

(40)

By applying (1), (7), (9) and (10), one can obtain that

$$\begin{array}{cc}\hfill 5A_2{D}_3=& (2k+1)(2k+2)(2k+4)(F_{6k+3}+F_{2k-3})+(2k+2)(2k+4)(F_{6k+2}-F_{2k-2})\hfill \\ \hfill & -(2k+1)(4k+6)·\beta ·(F_{6k+1}+F_{2k-5})-(4k+6)·\beta ·(F_{6k}-F_{2k-4})\hfill \\ \hfill & +{\beta }^2·(2k+1)(F_{6k-1}+F_{2k-7})+{\beta }^2·(F_{6k-2}-F_{2k-6})-(2k+2)(L_{4k+1}+4)\hfill \\ \hfill & -(2k+1)(2k+2)(L_{4k+2}+7)-(2k+4)(L_{4k+3}+1)-(2k+1)(2k+4)(L_{4k+4}+3)\hfill \\ \hfill & +(2k+1)·\beta ·(L_{4k}+18)+\beta ·(L_{4k-1}+11)+(2k+1)·\beta ·(L_{4k+2}+7)\hfill \\ \hfill & +\beta (L_{4k+1}+4)+\left[12k^2+(36-18\beta )k+3{\beta }^2-32\beta +29\right]·\left[(2k+1)F_{2k+3}+F_{2k+2}\right]\hfill \\ \hfill =& (2k+1)(2k+2)(2k+4)F_{6k+3}+\left[4k^2+(12-4\beta -2{\beta }^2)k+({\beta }^2-6\beta +8)\right]F_{6k+2}\hfill \\ \hfill & -\left[8\beta k^2-(4{\beta }^2-12\beta )k+{\beta }^2\right]F_{6k+1}-\left[40k^2+(82-50\beta )k+(38-40\beta )\right]\hfill \\ \hfill & -\left[4k^2+(14+2\beta )k+(10-2\beta )\right]L_{4k+3}-\left[8\beta k^2+(14-6\beta )k+(4+\beta )\right]L_{4k+2}\hfill \\ \hfill & +\left[64k^3+(236-140\beta )k^2+(74{\beta }^2-322\beta +270)k+(58{\beta }^2-158\beta +93)\right]F_{2k+3}\hfill \\ \hfill & -\left[64k^3+(232-168\beta )k^2+(110{\beta }^2-370\beta +248)k+(86{\beta }^2-172\beta +75)\right]F_{2k+2}.\hfill \end{array}$$

Furthermore,

$$\begin{array}{cc}\hfill 5B_2{C}_3=& (2k+1)(2k+2)(2k+4)(F_{6k+3}-F_{2k+3})+2·(2k+1)(2k+2)(F_{6k+2}+F_{2k+4})\hfill \\ \hfill & -(2k+1)(2k+2)\beta ·(F_{6k+1}-F_{2k+5})+(2k+1)(2k+2)(2k+4)F_{2k}\hfill \\ \hfill & +2·(2k+1)(2k+2)F_{2k-1}-(2k+1)(2k+2)\beta ·F_{2k-2}\hfill \\ \hfill =& (2k+1)(2k+2)(2k+4)F_{6k+3}+2·(2k+1)(2k+2)F_{6k+2}-(2k+1)(2k+2)\beta F_{6k+1}\hfill \\ \hfill & -\left[16k^3-(20\beta -32)k^2-(30\beta -20)k-(10\beta -4)\right]·F_{2k+3}\hfill \\ \hfill & +\left[16k^3+(40-16\beta )k^2+(32-24\beta )k+(8-8\beta )\right]·F_{2k+2}.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill & 5(A_2{D}_3-B_2{C}_3)\hfill \\ \hfill =& -\left[4k^2+(2{\beta }^2+4\beta )k-({\beta }^2-6\beta +4)\right]F_{6k+2}-\left[4\beta k^2-(4{\beta }^2-6\beta )k+({\beta }^2-2\beta )\right]F_{6k+1}\hfill \\ \hfill & -\left[4k^2+(14+2\beta )k+(10-2\beta )\right]·L_{4k+3}-\left[8k^2+(14-6\beta )k+(4+\beta )\right]·L_{4k+2}\hfill \\ \hfill & +\left[80k^3+(-160\beta +268)k^2+(74{\beta }^2-352\beta +290)k+(58{\beta }^2-168\beta +97)\right]·F_{2k+3}\hfill \\ \hfill & -\left[80k^3+(-184\beta +272)k^2+(110{\beta }^2-394\beta +280)k+(86{\beta }^2-180\beta +83)\right]·F_{2k+2}\hfill \\ \hfill & -\left[40k^2+(82-50\beta )k+(38-40\beta )\right].\hfill \end{array}$$

From Lemma 3, we can easily find that

$$\begin{array}{c}\hfill 5(A_2{D}_3-B_2{C}_3)=H_3\left(k\right)<0\end{array}$$

holds for any integer $k\ge 2$; hence, inequalities (38)–(40) hold.

Applying (38) repeatedly, we get

$$\begin{array}{cc}\hfill \sum _{k=2m+1}^{\infty }{\displaystyle \frac{1}{k{F}_k}}=& \sum _{k=m}^{\infty }\left({\displaystyle \frac{1}{(2k+1)F_{2k+1}}}+{\displaystyle \frac{1}{(2k+2)F_{2k+2}}}\right)\hfill \\ \hfill <& \sum _{k=m}^{\infty }\left({\displaystyle \frac{1}{(2k+2)F_{2k-1}-\beta F_{2k-3}-1}}-{\displaystyle \frac{1}{(2k+4)F_{2k+1}-\beta F_{2k-1}-1}}\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{(2m+2)F_{2m-1}-\beta F_{2m-3}-1}}.\hfill \end{array}$$

(41)

□

3.2.2. Proof of the Left-Hand-Side Inequality of (37)

Finally, we prove that for any integer $k\ge 2$, the following inequality holds:

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{(2k+1)F_{2k+1}}}+{\displaystyle \frac{1}{(2k+2)F_{2k+2}}}>{\displaystyle \frac{1}{(2k+2)F_{2k-1}+\alpha F_{2k-3}}}-{\displaystyle \frac{1}{(2k+4)F_{2k+1}+\alpha F_{2k-1}}}.\end{array}$$

(42)

As usual, let

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{(2k+2)F_{2k-1}+\alpha F_{2k-3}}}-{\displaystyle \frac{1}{(2k+4)F_{2k+1}+\alpha F_{2k-1}}}={\displaystyle \frac{C_4}{D_4}},\end{array}$$

then we can prove

$$\begin{array}{c}\hfill {\displaystyle \frac{A_2}{B_2}}>{\displaystyle \frac{C_4}{D_4}},\end{array}$$

(43)

where

$$\begin{array}{cc}\hfill C_4=& (2k+4)F_{2k}+2F_{2k-1}+\alpha F_{2k-2},\hfill \\ \hfill D_4=& (2k+2)(2k+4)(L_{4k}+3)/5+(2k+2)\alpha (L_{4k-2}+2)/5\hfill \\ \hfill & +(2k+4)\alpha (L_{4k-2}+7)/5+{\alpha }^2·(L_{4k-4}+3)/5.\hfill \end{array}$$

Note that for integer $k\ge 3$, $B_2$ and $D_4$ are both positive, so inequality (43) is equivalent to

$$\begin{array}{c}\hfill A_2{D}_4>B_2{C}_4.\end{array}$$

(44)

Then we can derive that

$$\begin{array}{cc}\hfill 5A_2{D}_4=& (2k+1)(2k+2)(2k+4)F_{6k+3}+\left[4k^2+(4\alpha -2{\alpha }^2+12)k+({\alpha }^2+6\alpha +8)\right]F_{6k+2}\hfill \\ \hfill & +\left[8\alpha k^2+(12\alpha +4{\alpha }^2)k-{\alpha }^2\right]F_{6k+1}\hfill \\ \hfill & +\left[64k^3+(140\alpha +236)k^2+(74{\alpha }^2+322\alpha +260)k+(58{\alpha }^2+158\alpha +88)\right]F_{2k+3}\hfill \\ \hfill & -\left[64k^3+(168\alpha +232)k^2+(110{\alpha }^2+370\alpha +248)k+(86{\alpha }^2+172\alpha +80)\right]F_{2k+2}.\hfill \end{array}$$

Meanwhile,

$$\begin{array}{cc}\hfill 5B_2{C}_4=& (2k+1)(2k+2)(2k+4)(F_{6k+3}-F_{2k+3})+2·(2k+1)(2k+2)(F_{6k+2}+F_{2k+4})\hfill \\ \hfill & +(2k+1)(2k+2)\alpha ·(F_{6k+1}-F_{2k+5})+(2k+1)(2k+2)(2k+4)F_{2k}\hfill \\ \hfill & +2·(2k+1)(2k+2)F_{2k-1}+(2k+1)(2k+2)\alpha ·F_{2k-2}.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill & 5(A_2{D}_4-B_2{C}_4)\hfill \\ \hfill =& \left[-4k^2+(4\alpha -2{\alpha }^2)k+({\alpha }^2+6\alpha +4)\right]F_{6k+2}+\left[4\alpha k^2+(4{\alpha }^2+6\alpha )k-({\alpha }^2+2\alpha )\right]F_{6k+1}\hfill \\ \hfill & +\left[80k^3+(160\alpha +268)k^2+(74{\alpha }^2+352\alpha +280)k+(58{\alpha }^2+168\alpha +92)\right]·F_{2k+3}\hfill \\ \hfill & -\left[80k^3+(184\alpha +272)k^2+(110{\alpha }^2+394\alpha +280)k+(86{\alpha }^2+180\alpha +88)\right]·F_{2k+2},\hfill \end{array}$$

From Lemma 4, we obtain

$$\begin{array}{c}\hfill 5(A_2{D}_4-B_2{C}_4)=H_4\left(k\right)>0,\mathrm{for}k\ge 2,\end{array}$$

which implies the validity of (42)–(44).

Applying (42) repeatedly, one can obtain

$$\begin{array}{cc}\hfill \sum _{k=2m+1}^{\infty }{\displaystyle \frac{1}{k{F}_k}}=& \sum _{k=m}^{\infty }\left({\displaystyle \frac{1}{(2k+1)F_{2k+1}}}+{\displaystyle \frac{1}{(2k+2)F_{2k+2}}}\right)\hfill \\ \hfill >& \sum _{k=m}^{\infty }\left({\displaystyle \frac{1}{(2k+2)F_{2k-1}+\alpha F_{2k-3}}}-{\displaystyle \frac{1}{(2k+4)F_{2k+1}+\alpha F_{2k-1}}}\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{(2m+2)F_{2m-1}+\alpha F_{2m-3}}}.\hfill \end{array}$$

(45)

□

Therefore, (37) follows from (41) and (45), which completes the proof of Theorem 2. □

4. Proof of Corollary 1

For the case $n=4$, with the help of Mathematica, we can get

$$\begin{array}{c}\hfill \sum _{k=4}^{\infty }{\displaystyle \frac{1}{k{F}_k}}>\sum _{k=4}^{11}{\displaystyle \frac{1}{k{F}_k}}={\displaystyle \frac{911701073}{5452246800}}>{\displaystyle \frac{1}{6}}.\end{array}$$

(46)

On the other hand,

$$\begin{array}{c}\hfill \sum _{k=4}^{\infty }{\displaystyle \frac{1}{k{F}_k}}<\sum _{k=4}^{N-1}{\displaystyle \frac{1}{k{F}_k}}+\sum _{k\ge N}{\displaystyle \frac{1}{k^2}}<\sum _{k=4}^{N-1}{\displaystyle \frac{1}{k{F}_k}}+\sum _{k\ge N}{\displaystyle \frac{1}{k(k-1)}}=\sum _{k=4}^{N-1}{\displaystyle \frac{1}{k{F}_k}}+{\displaystyle \frac{1}{N-1}}.\end{array}$$

Taking $N=33$, we can obtain

$$\begin{array}{c}\hfill \sum _{k=4}^{\infty }{\displaystyle \frac{1}{k{F}_k}}<\sum _{k=4}^{32}{\displaystyle \frac{1}{k{F}_k}}+{\displaystyle \frac{1}{32}}<{\displaystyle \frac{1}{5}}.\end{array}$$

(47)

It is easy to deduce from (46) and (47) that

$$\begin{array}{c}\hfill 5<{\left(\sum _{k=4}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}<6,\end{array}$$

and hence

$$\begin{array}{c}\hfill ⌊{\left(\sum _{k=4}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}⌋=5=5F_2.\end{array}$$

(48)

For any integer $n\ge 5$, by Theorems 1 and 2, one can easily conclude that

$$\begin{array}{c}\hfill {\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}=(n+1)F_{n-2}+O\left(F_{n-4}\right).\end{array}$$

However, from the definition of the Fibonacci numbers, we know that $F_{n-2}$ and $F_{n-4}$ are of the same order. Hence, we have

$$\begin{array}{c}\hfill {\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}=n{F}_{n-2}+O\left(F_{n-2}\right).\end{array}$$

(49)

Furthermore, for any real number x, we have

$$\begin{array}{c}\hfill ⌊x⌋=x+O\left(1\right).\end{array}$$

(50)

Then, it follows from (49) and (50) that

$$\begin{array}{c}\hfill ⌊{\left(\sum _{k=n}^{\infty }{\displaystyle \frac{1}{k{F}_k}}\right)}^{-1}⌋=n{F}_{n-2}+O\left(F_{n-2}\right).\end{array}$$

(51)

Thus, we complete the proof of Corollary 1 from (48) and (51). □

5. Conclusions

Our study introduces a novel approach by combining reciprocals of linear recurrence sequences with the tail of the Riemann zeta function, and simultaneously establishes symmetric inequalities for these sums, both of which are unprecedented in the literature.

The methodology developed in this paper is, in principle, applicable to other linear recurrence sequences such as Pell numbers, Lucas numbers, and generalized k—Fibonacci sequences. However, it is important to note that different sequences are defined by distinct characteristic equations. Consequently, while the methods can be extended, the elegant symmetric inequalities established here for Fibonacci numbers are not guaranteed to hold for other sequences. We suggest this investigation of symmetry in other sequences as a promising direction for future research.

While our current work remains theoretical in mathematics, these connections suggest promising avenues for practical applications in computer science, combinatorial analysis, and beyond. For example:

• The precise inequalities and asymptotic formulas derived for Fibonacci reciprocal sums can provide tight error bounds in the analysis of algorithms utilizing Fibonacci heaps or golden-ratio-based search techniques.
• In computational number theory, these results offer new tools for studying Diophantine approximations involving the golden ratio and for analyzing the convergence properties of related infinite series.
• The methodology may extend to combinatorial optimization problems where Fibonacci sequences naturally arise, such as in the analysis of certain graph structures and coding schemes.
• In mathematical physics, similar reciprocal sum estimates have proven useful in the spectral analysis of one-dimensional quasicrystals and quantum walk models.