The proof of a formula concerning the asymptotic behavior of the reciprocal sum of the square of multiple-angle Fibonacci numbers

Abstract

Let $(F_{n})_{n}$ be the Fibonacci sequence defined by $F_{n+2}=F_{n+1}+F_{n}$ with $F_{0}=0$ and $F_{1}=1$. In this paper, we prove that for any integer $m\geq 1$ there exists a positive constant $C_{m}$ for which

$$ \lim_{n\to \infty } \Biggl\{ \Biggl( \sum_{k=n}^{ \infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} - \bigl(F_{mn}^{2}-F_{m(n-1)}^{2}+(-1)^{mn}C_{m} \bigr) \Biggr\} =0. $$

Furthermore, we show that $C_{m}$ tends to $2/5$ as $m\to \infty$ (indeed, we provide quantitative versions of the previous results as well as an explicit form for $C_{m}$). This confirms some questions proposed by Lee and Park [J. Inequal. Appl. 2020(1):91 2020].

Introduction

It is well known that if a series $\sum_{k\geq 1}a_{k}$ is convergent, then its “tail” $(\sum_{k=n}^{\infty }a_{k})_{n}$ tends to 0 as $n\to \infty$. In particular,

$$ \lim_{n\to \infty } \Biggl(\sum_{k=n}^{\infty }a_{k} \Biggr)^{-1}=\infty . $$

In the past years, many mathematicians have been interested in studying the properties and forms of the reciprocal tails (as above) of the convergent series, where $a_{n}$ is related to some recurrence sequences. Here, we restrict ourselves only to cases in which $a_{n}$ is related to the Fibonacci sequence $(F_{n})_{n\geq 0}$ which is defined by the binary recurrence

$$ F_{n+2}=F_{n+1}+F_{n}, $$

with initial values, $F_{0}=0$ and $F_{1}=1$. In 2008, Ohtsuka and Nakamura [11] studied the partial infinite sums of reciprocal Fibonacci numbers and showed that

$$\begin{aligned}& \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor = \textstyle\begin{cases} F_{n-2}, & \text{if } n \text{ is even}, n\geq 2; \\ F_{n-2}-1, & \text{if } n \text{ is odd}, n\geq 1; \end{cases}\displaystyle \end{aligned}$$

and

$$\begin{aligned}& \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor = \textstyle\begin{cases} F_{n-1}F_{n}-1, & \text{if } n \text{ is even}, n\geq 2; \\ F_{n-1}F_{n}, & \text{if } n \text{ is odd}, n\geq 1. \end{cases}\displaystyle \end{aligned}$$

(1)

In the same year, Choi and Choo [2] provided formulas related to the sums of reciprocals of the products of Fibonacci and Lucas numbers, namely

$$ \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{k} L_{k+m}} \Biggr)^{-1} \Biggr\rfloor \quad \text{and}\quad \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{L_{k} F_{k+m}} \Biggr)^{-1} \Biggr\rfloor $$

(recall that the Lucas sequence $(L_{n})_{n}$ satisfies the same recurrence as Fibonacci numbers, but with initial values $L_{0}=2$ and $L_{1}=1$). For more facts in this topic, we recommend to the reader the papers [1, 3–6, 10, 12, 13].

To study the analytic behavior of these sequences, one introduces another (more qualitative) definition. We then say that $f_{n} \sim g_{n}$ if $f_{n}-g_{n}$ tends to 0 as $n\to \infty$. Very recently, motivated by this definition, Lee and Park [8, 9] proved, among other things, that

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{k}} \Biggr)^{-1} \sim F_{n-2}, \qquad \Biggl( \sum _{k=n}^{\infty } \frac{1}{F_{k}^{2}} \Biggr)^{-1} \sim F_{n}^{2}-F_{n-1}^{2}+ \frac{2}{3}(-1)^{n} $$

and

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{3k}^{2}} \Biggr)^{-1} \sim F_{3n}^{2}-F_{3n-3}^{2}+ \frac{4}{9}(-1)^{n}. $$

Moreover, as formula (5.1) of [8, Sect. 5], they stated (without proof) the following expected general formula:

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \sim F_{mn}^{2}-F_{m(n-1)}^{2}+(-1)^{mn}C_{m}, $$

(2)

which should hold for any positive integer m, where $C_{m}$ is a positive constant. Additionally, they remarked that “it looks not easy to find the explicit values of $C_{m}$ satisfying (2) except for $m = 1$ and 3 (for which $C_{1}=2/3$ and $C_{3}=4/9$). By using computer software programs (Maple 17 and wolframalpha.com), they estimated this constant for $m\in [1,9]$. These computations suggest (as written by them): “We might expect that $C_{m}$ tends to $2/5$ as $m\to \infty$”.

The aim of this paper is to confirm the expectation by proving these facts (indeed, we provide quantitative versions for them as well as a completely explicit formula for $C_{m}$). More precisely, we have the following.

Theorem 1

For any integer $m\geq 1$, there exists a positive constant $C_{m}$ such that

$$ \Biggl\vert \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} - \bigl(F_{mn}^{2}-F_{m(n-1)}^{2}+(-1)^{mn}C_{m} \bigr) \Biggr\vert < \frac{9.83}{\alpha ^{2m(n-1)}} $$

(3)

for all $n\geq \max \{\frac{4}{m}, 2\}$ (where, as usual, $\alpha =(1+\sqrt{5})/2$ denotes the golden ratio). In particular,

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \sim F_{mn}^{2}-F_{m(n-1)}^{2}+(-1)^{mn}C_{m}. $$

(4)

Moreover, for all $m\geq 1$,

$$ C_{m}= \frac{2}{5} \biggl(1+ \frac{r_{m}+\beta s_{m}}{2-(-1)^{m}L_{4m}} \biggr), $$

(5)

where $\beta =(1-\sqrt{5})/2$,

$$ r_{m} := \bigl(1 - (-1)^{m}\bigr) L_{2 m} + 5 \bigl(1 + (-1)^{m}\bigr) F_{m - 1} F_{m} F_{2 m}, $$

and

$$ s_{m}:=\bigl(1+(-1)^{m}\bigr)F_{2m}(L_{2m}-2). $$

Furthermore, the estimate

$$ \vert C_{m}-2/5 \vert < 1.2/\alpha ^{2m} $$

(6)

holds for all $m\geq 1$, and so $C_{m}$ tends to $2/5$ as $m\to \infty$.

Now, we shall present two interesting consequences of the previous result. First, observe that it is immediate, after a standard calculation, that $C_{m}$ in formula (5) agrees with values $C_{1}=2/3$ and $C_{3}=4/9$ (provided in [8]). Moreover, the first 10 values of $C_{m}$ for m odd are as follows:

$$ \frac{2}{3}, \frac{4}{9}, \frac{50}{123}, \frac{338}{843}, \frac{1156}{2889}, \frac{15{,}842}{39{,}603}, \frac{108{,}578}{271{,}443}, \frac{372{,}100}{930{,}249}, \frac{5{,}100{,}818}{12{,}752{,}043}, \frac{34{,}961{,}522}{87{,}403{,}803}, $$

which are rational numbers. However,

$$\begin{aligned}& C_{2}= \frac{2}{15}(1-2\beta ),\qquad C_{4}= \frac{6}{35}(1-2\beta ),\qquad C_{6}= \frac{8}{45}(1-2\beta ), \\& C_{8}= \frac{42}{235}(1-2\beta ),\qquad C_{10}= \frac{22}{123}(1-2\beta ),\qquad C_{12}= \frac{144}{161}(1-2 \beta ), \end{aligned}$$

which (since $1-2\beta =\sqrt{5}$) is the same list as

$$\begin{aligned}& C_{2}= \frac{2\sqrt{5}}{15},\qquad C_{4}= \frac{6\sqrt{5}}{35},\qquad C_{6}= \frac{8\sqrt{5}}{45}, \\& C_{8}= \frac{42\sqrt{5}}{235},\qquad C_{10}= \frac{22\sqrt{5}}{123},\qquad C_{12}= \frac{144}{805} \sqrt{5} \end{aligned}$$

are irrational numbers.

This suggests that $C_{2m-1}\in \mathbb{Q}_{>0}$ and $C_{2m}\in \sqrt{5}\cdot \mathbb{Q}_{>0}$ for all $m\geq 1$. Indeed, the next result confirms this fact by providing a cleaner formula for $C_{m}$ depending on the parity of m. More precisely,

Corollary 1

Let m be a positive integer. We have

• (i) If m is even, then

  $$ C_{m} = \frac{2(L_{2m}-2)}{25F_{2m}}\sqrt{5}. $$

• (ii) If m is odd, then

  $$ C_{m} = \frac{2(L_{2m}+2)}{5L_{2m}}. $$

In particular, $C_{m}$ is a rational number if and only if m is odd.

The previous explicit formulas for $C_{m}$ provide better bounds to $C_{m}-2/5$ which together with Theorem 1 allows to prove the following.

Corollary 2

We have that

• (i) Let m be an even positive integer. Then

  $$ \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \Biggr\rfloor = F_{mn}^{2}-F_{m(n-1)}^{2} $$

  holds for all $n\geq 3$.

• (ii) Let m be an odd positive integer. Then

  $$\begin{aligned} \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \Biggr\rfloor =& \textstyle\begin{cases} F_{mn}^{2}-F_{m(n-1)}^{2}, & \textit{if } n \textit{ is even}; \\ F_{mn}^{2}-F_{m(n-1)}^{2}-1, & \textit{if } n \textit{ is odd}, \end{cases}\displaystyle \end{aligned}$$

  holds for all $n\geq 3$.

The proofs of these results combine several estimates, properties of Fibonacci and Lucas numbers as well as some facts about the convergence of series. The computations in this work were performed with Mathematica software.

Auxiliary results

In this section, we present a few auxiliary facts which will be very useful in all proofs.

Lemma 1

Let $(F_{n})_{n}$ and $(L_{n})_{n}$ be the Fibonacci and Lucas sequences, respectively, and $\alpha =(1+\sqrt{5})/2$ and $\beta =(1-\sqrt{5})/2$. We have

• (i) (Binet’s formula for $F_{n}$) formula

  $$ F_{n}= \frac{\alpha ^{n}-\beta ^{n}}{\sqrt{5}} $$

  holds for all $n\geq 1$.

• (ii) (Binet’s formula for $L_{n}$) formula

  $$ L_{n}=\alpha ^{n}+\beta ^{n} $$

  holds for all $n\geq 1$.

The next two lemmas follow from the previous one.

Lemma 2

We have that

• (i) $F_{2n}=F_{n}L_{n}$.

• (ii) $L_{n}=F_{n+1}+F_{n-1}$.

• (iii) $L_{2n}=5F_{n}^{2}+2(-1)^{n}$.

• (iv) $L_{2n}=L_{n}^{2}-2(-1)^{n}$.

• (v) (D’Ocagne’s identity) $(-1)^{n}F_{m-n}=F_{m}F_{n+1}-F_{m+1}F_{n}$.

We know that $\mathbb{Q}(\beta )=\mathbb{Q}(\sqrt{5})$ is a quadratic field extension of $\mathbb{Q}$ with $\mathbb{Q}$-basis $\{1,\beta \}$. The next result asserts the exact coefficients of the $\mathbb{Q}$-linear combinations for powers of β, namely,

Lemma 3

For any $n\geq 1$, one has that

$$ \beta ^{n}=\beta F_{n}+F_{n-1}. $$

The last ingredients are some known lower and upper bounds for $F_{n}$, that is,

Lemma 4

The inequalities

$$ \alpha ^{n-2}\leq F_{n} \leq \alpha ^{n-1} $$

hold for all $n\geq 1$.

We refer the reader to [7] for the proofs of these results as well as for the history, properties, and rich applications of the Fibonacci sequence and some of its generalizations.

With these tools in hand, we are now in a position to prove our results.

The proof of Theorem 1

By Lemma 1(i), we have that

$$ F_{mk}= \frac{\alpha ^{mk}}{\sqrt{5}} \biggl(1- \biggl( \frac{\beta }{\alpha } \biggr)^{mk} \biggr), $$

hence

$$ \frac{1}{F_{mk}^{2}}= \frac{5}{\alpha ^{2mk}(1-(\beta /\alpha )^{mk})^{2}}. $$

However, $1/(1-x)^{2}=1+2x+3x^{2}+\cdots$ holds for $|x|<1$. Thus, since $|\beta /\alpha |<1$ and $\alpha \beta =-1$, we have

$$\begin{aligned} \frac{1}{F_{mk}^{2}} = & \frac{5}{\alpha ^{2mk}} \biggl(1+2 \biggl( \frac{\beta }{\alpha } \biggr)^{mk}+3 \biggl( \frac{\beta }{\alpha } \biggr)^{2mk} + \cdots \biggr) \\ = & \frac{5}{\alpha ^{2mk}} + 5 \sum_{i=2}^{ \infty } i (-1)^{mk} \biggl( \frac{\beta }{\alpha } \biggr)^{imk}. \end{aligned}$$

By summing up from $k=n$ to infinity and after a straightforward calculation, we arrive at

$$\begin{aligned} \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} = & \frac{5}{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}+5 \sum_{i=2}^{\infty } i\sum_{k=n}^{\infty } (-1)^{mk} \biggl( \frac{\beta }{\alpha } \biggr)^{imk} \\ = & \frac{5}{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)} + 5 \sum_{i=2}^{\infty } i (-1)^{mn} \biggl( \frac{\beta }{\alpha } \biggr)^{imn} \frac{1}{1-(-1)^{m}(\beta /\alpha )^{im}} \\ = & \frac{5}{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}\bigl(1+f(m,n)\bigr), \end{aligned}$$

(7)

where $f(m,n)$ denotes the following summatory

$$ f(m,n):= \biggl( \frac{\alpha ^{2m}-1}{\alpha ^{2m}} \biggr) \sum_{i=2}^{\infty } i \biggl( \frac{\beta }{\alpha } \biggr)^{mn(i-1)} \frac{1}{1-(-1)^{m}(\beta /\alpha )^{im}} . $$

Note that, by the reverse triangle inequality (and $\beta /\alpha =-1/\alpha ^{2}$, which follows from $\beta =-1/\alpha$), one has

$$ \bigl\vert 1-(-1)^{m}(\beta /\alpha )^{im} \bigr\vert \geq 1- \biggl\vert \frac{\beta }{\alpha } \biggr\vert ^{2m}>1- \frac{1}{\alpha ^{4}}> 0.85 . $$

Thus, we deduce the following upper bound for $|f(m,n)|$:

$$\begin{aligned} \bigl\vert f(m,n) \bigr\vert \leq & \frac{1}{0.85} \frac{\alpha ^{2m}-1}{\alpha ^{2m}} \sum_{i=2}^{\infty } i \biggl( \frac{1}{\alpha ^{2}} \biggr)^{mn(i-1)} \\ \leq & \frac{1.18}{\alpha ^{2mn}} \sum_{i=2}^{ \infty } i \biggl( \frac{1}{\alpha ^{2mn}} \biggr)^{i-2} \\ = & \frac{1.18}{\alpha ^{2mn}} \Biggl(2 + \frac{3}{\alpha ^{2mn}} + \sum _{i=4}^{\infty } \frac{i}{\alpha ^{2mn(i-2)}} \Biggr) \\ = & \frac{1.18}{\alpha ^{2mn}} \Biggl(2 + \frac{3}{\alpha ^{2mn}} + \sum _{i=3}^{\infty } \frac{i+1}{\alpha ^{mn(2i-2)}} \Biggr) \\ = & \frac{1.18}{\alpha ^{2mn}} \Biggl(2 + \frac{3}{\alpha ^{2mn}} + \sum _{i=3}^{\infty } \biggl( \frac{i+1}{\alpha ^{imn}} \cdot \frac{1}{\alpha ^{(i-2)mn}} \biggr) \Biggr) \\ \leq & \frac{1.18}{\alpha ^{2mn}} \Biggl(2.065 + \frac{1}{60} \sum _{i=1}^{\infty } \frac{1}{\alpha ^{imn}} \Biggr) \\ \leq & \frac{1.18}{\alpha ^{2mn}} \biggl(2.065 + \frac{0.171}{60} \biggr) \\ < & \frac{2.45}{\alpha ^{2mn}}, \end{aligned}$$

where we used that

$$ \alpha ^{imn} > 1.6^{imn}\geq 6.55^{i} > 60(i+1) $$

for all $i\geq 3$ (since $mn\geq 4$) together with

$$ \frac{1}{\alpha ^{mn}}+ \frac{1}{\alpha ^{2mn}}+ \frac{1}{\alpha ^{3mn}}+\cdots < \frac{1}{\alpha ^{4}}+ \frac{1}{\alpha ^{8}}+ \frac{1}{\alpha ^{12}}+\cdots = \frac{1}{\alpha ^{4}-1}< 0.171 , $$

because $\alpha ^{mn}\geq \alpha ^{4}$. Thus $f(m,n)$ tends to 0 as $\min \{m,n\}\to \infty$. Furthermore, $|f(m,n)|<0.06$ for all integers m and n with $mn\geq 4$.

Turning back to (7), we have that

$$ \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}}= \frac{5}{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}\cdot \bigl(1+f(m,n)\bigr), $$

and hence

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} = \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5} \biggl( \frac{1}{1+f(m,n)} \biggr). $$

(8)

Since $|f(m,n)|<1$, for $mn\geq 4$, then

$$ \frac{1}{1+f(m,n)}=1-f(m,n)+\bigl(f(m,n)\bigr)^{2}-\cdots =1-f(m,n)+R_{m,n}, $$

where $R_{m,n}:=(f(m,n))^{2}-(f(m,n))^{3}+(f(m,n))^{4}-\cdots$ . For our purposes, we need to find an upper bound for $|R_{m,n}|$. For that, one has

$$\begin{aligned} \vert R_{m,n} \vert \leq & \bigl\vert \bigl(f(m,n)\bigr)^{2}\bigl(1-f(m,n)+\bigl(f(m,n)\bigr)^{2}- \cdots \bigr) \bigr\vert \\ = & \biggl\vert \frac{(f(m,n))^{2}}{1+f(m,n)} \biggr\vert \\ \leq & \frac{(f(m,n))^{2}}{1- \vert f(m,n) \vert } \\ \leq & 1.07 \bigl(f(m,n)\bigr)^{2} < \frac{6.4}{\alpha ^{4mn}}, \end{aligned}$$

(9)

where we used that

$$ \bigl\vert 1+f(m,n) \bigr\vert \geq 1- \bigl\vert f(m,n) \bigr\vert >1- \frac{2.37}{\alpha ^{2mn}}>1-0.06=0.94. $$

Turning back to (8), we have

$$\begin{aligned} \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} = & \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}\bigl(1-f(m,n)+R_{m,n} \bigr) \\ = & \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}f(m,n) \\ &{} + \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5} R_{m,n} \\ = & \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}f(m,n) \\ & + T_{m,n} , \end{aligned}$$

where

$$ T_{m,n}:= \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5} \cdot R_{m,n} $$

satisfies (by (9))

$$ \vert T_{m,n} \vert = \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1) \vert R_{m,n} \vert }{5}< \frac{\alpha ^{2mn}}{5}\cdot \frac{6.4}{\alpha ^{4mn}}< \frac{1.3}{\alpha ^{2mn}}. $$

(10)

Hence

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1}= \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}f(m,n)+T_{m,n}. $$

(11)

Now, let us work with the second term of the right-hand side of (11). By the definition of $f(m,n)$, we can write

$$\begin{aligned}& \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}f(m,n) \\& \quad =\frac{2\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} \biggl( \frac{\beta }{\alpha } \biggr)^{mn} \frac{1}{1-(-1)^{m} (\beta /\alpha )^{2m}} + D_{m,n}, \end{aligned}$$

where

$$ D_{m,n}:= \frac{\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} \sum_{i=3}^{\infty } i \biggl( \frac{\beta }{\alpha } \biggr)^{mn(i-1)} \frac{1}{1-(-1)^{m} (\beta /\alpha )^{im}} . $$

Since $\beta /\alpha =-1/\alpha ^{2}$, we deduce that

$$\begin{aligned}& \frac{2\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} \biggl( \frac{\beta }{\alpha } \biggr)^{mn} \frac{1}{1-(-1)^{m}(\beta /\alpha )^{2m}} \\& \quad = \frac{2\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} \biggl( \frac{-1}{\alpha ^{2}} \biggr)^{mn} \frac{1}{1-(-1/\alpha ^{4})^{m}} \\& \quad = \frac{2(-1)^{mn}}{5} \cdot \frac{\alpha ^{4m} - 2\alpha ^{2m} + 1}{\alpha ^{4m}} \Biggl(1+ \sum _{i=1}^{ \infty } \biggl( \frac{-1}{\alpha ^{4}} \biggr)^{mi} \Biggr) \\& \quad = \frac{2(-1)^{mn}}{5}(1-S_{m}) (1+E_{m}) \\& \quad = \frac{2(-1)^{mn}}{5}+(-1)^{mn}G_{m}, \end{aligned}$$

where

$$ G_{m}:= \frac{2}{5}(E_{m}-S_{m}-E_{m}S_{m}) $$

(12)

and $E_{m}$ and $S_{m}$ are defined as

$$ S_{m}:= \frac{2}{\alpha ^{2m}}- \frac{1}{\alpha ^{4m}}\quad \text{and}\quad E_{m}:=\sum_{i=1}^{\infty } \biggl( \frac{-1}{\alpha ^{4}} \biggr)^{mi} = \frac{(-1)^{m}}{\alpha ^{4m}-(-1)^{m}}. $$

(13)

Now, we use that $S_{m}<2/\alpha ^{2m}$ and

$$ \vert E_{m} \vert \leq \sum_{i=1}^{\infty } \biggl\vert \frac{-1}{\alpha ^{4}} \biggr\vert ^{mi} = \frac{1}{\alpha ^{4m}-1} = \frac{1}{\alpha ^{4m}} \frac{\alpha ^{4m}}{\alpha ^{4m}-1} \leq \frac{1}{\alpha ^{4m}} \frac{\alpha ^{4}}{\alpha ^{4}-1} < \frac{1.2}{\alpha ^{4m}} $$

(since $x\mapsto x/(x-1)$ is a decreasing function for $x>1$, and so the maximum of $\alpha ^{4m}/(\alpha ^{4m}-1)$ is attained at $m=1$) to infer that

$$ \vert G_{m} \vert \leq \frac{2}{5} \bigl( \vert E_{m} \vert + \vert S_{m} \vert + \vert E_{m}S_{m} \vert \bigr)\leq \frac{2}{5} \biggl( \frac{1.2}{\alpha ^{4m}}+ \frac{2}{\alpha ^{2m}}+ \frac{2.4}{\alpha ^{6m}} \biggr)\leq \frac{1.2}{\alpha ^{2m}}, $$

(14)

which proves (6). Observe that, in particular, $G_{m}$ tends to 0 as $m\to \infty$.

For the remaining terms, i.e., $D_{m,n}$, first we can realize that

$$ \frac{\alpha ^{2mn}(\alpha ^{2m}-1)^{2}}{5\alpha ^{4m}} < \frac{\alpha ^{2mn}}{5} , $$

(15)

since $(\alpha ^{2m}-1)^{2}<\alpha ^{4m}$. Therefore, by using again $\beta /\alpha =-1/\alpha ^{2}$, we get

$$\begin{aligned} \Biggl\vert \sum_{i=3}^{\infty } i \frac{ (\beta /\alpha ) ^{mn(i-1)}}{1-(-1)^{m} (\beta /\alpha )^{im}} \Biggr\vert = & \Biggl\vert \sum _{i=3}^{\infty } i \biggl( \frac{-1}{\alpha ^{2}} \biggr)^{mn(i-1)} \frac{1}{1-(-1)^{m}(\beta /\alpha )^{im}} \Biggr\vert \\ < & 1.32 \cdot \sum_{i=3}^{\infty } i \frac{1}{\alpha ^{2mn(i-1)}} \\ < & 1.32 \cdot \sum_{i=3}^{\infty } \frac{1}{\alpha ^{2mn(i-1)-2i}}, \end{aligned}$$

where we used that $i<2^{i}<\alpha ^{2i}$ holds for every integer $i\geq 3$. Furthermore, since $mn\geq 4$, we have for $i\geq 2$

$$ a^{2mn(i-1)-2i} \geq \bigl(\alpha ^{2mn (\frac{i-1}{i} )-2} \bigr)^{i} \geq \bigl(\alpha ^{ \frac{2mn}{2}-2} \bigr)^{i} \geq \bigl(\alpha ^{2} \bigr)^{i}> 2.6^{i} . $$

Thus,

$$\begin{aligned} &\Biggl\vert \sum_{i=3}^{\infty } i \frac{ (\beta /\alpha ) ^{mn(i-1)}}{1-(-1)^{m} (\beta /\alpha )^{im}} \Biggr\vert \\ &\quad \leq 1.32\cdot \sum_{i=3}^{\infty } \frac{1}{\alpha ^{2mn(i-1)-2i}} = \frac{1.32}{\alpha ^{4mn-4}} \sum_{i=3}^{\infty } \frac{1}{\alpha ^{2mn(i-3)-2(i-2)}} \\ &\quad = \frac{1.32}{\alpha ^{4mn-4}} \sum_{i=1}^{\infty } \frac{1}{\alpha ^{2mn(i-1)-2i}} = \frac{1.32}{\alpha ^{4mn-4}} \Biggl(\frac{1}{\alpha ^{-2}} + \sum _{i=2}^{ \infty } \frac{1}{\alpha ^{2mn(i-1)-2i}} \Biggr) \\ &\quad < \frac{1.32}{\alpha ^{4mn-4}} \Biggl( \frac{1}{\alpha ^{-2}} + \sum _{i=2}^{\infty } \frac{1}{2.6^{i}} \Biggr) \\ &\quad < \frac{3.78}{\alpha ^{4mn-4}}. \end{aligned}$$

(16)

By combining (15) and (16), we infer that

$$ \vert D_{m,n} \vert < \frac{0.76}{\alpha ^{2mn-4}}. $$

(17)

In particular,

$$ \lim_{\min \{m,n\}\to \infty }D_{m,n}=0. $$

Summarizing, we have

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1}= \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{2(-1)^{mn}}{5}-(-1)^{mn}G_{m}+D_{m,n}+T_{m,n}. $$

(18)

On the other hand, we can apply Binet’s formula again to obtain

$$\begin{aligned} F_{mn}^{2}- F_{m(n-1)}^{2}& = \biggl( \frac{\alpha ^{mn}-\beta ^{mn}}{\sqrt{5}} \biggr)^{2} + \biggl( \frac{\alpha ^{m(n-1)}-\beta ^{m(n-1)}}{\sqrt{5}} \biggr)^{2} \\ & = \frac{\alpha ^{2mn}-2(-1)^{mn}+\beta ^{2mn}-\alpha ^{2m(n-1)}+2(-1)^{m(n-1)}-\beta ^{2m(n-1)}}{5} \\ & = \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{2(-1)^{mn}}{5}\bigl(1+(-1)^{m}\bigr) + \biggl( \frac{\beta ^{2mn}-\beta ^{2m(n-1)}}{5} \biggr), \\ & = \frac{\alpha ^{2m(n-1)}(\alpha ^{2m}-1)}{5}- \frac{2(-1)^{mn}}{5}- \frac{2(-1)^{m(n-1)}}{5} - \biggl( \frac{\alpha ^{2m}-1}{5\alpha ^{2mn}} \biggr), \end{aligned}$$

where we used that $\alpha \beta =-1$. Now, we can combine the previous relation with the formula in (18) to write

$$ A_{m,n}= \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1}- \bigl(F_{mn}^{2}-F_{m(n-1)}^{2}+(-1)^{mn}C_{m} \bigr), $$

(19)

where

$$ C_{m}:= \frac{2}{5}+(-1)^{m}G_{m}\quad \text{and}\quad A_{m,n}:= \frac{1-\alpha ^{2m}}{5\alpha ^{2mn}} + D_{m,n}+T_{m,n} . $$

Now, we use the formula for $A_{m,n}$, estimates (17) and (10) to obtain

$$ \vert A_{m,n} \vert < \frac{0.2}{\alpha ^{2m(n-1)}}+ \frac{5.3}{\alpha ^{2mn}}+ \frac{1.3}{\alpha ^{2mn}}= \frac{1}{\alpha ^{2m(n-1)}} \biggl(0.2+ \frac{5.3}{\alpha ^{2m}}+ \frac{1.3}{\alpha ^{2m}} \biggr)< \frac{9.83}{\alpha ^{2m(n-1)}}, $$

which implies in (3) (we used that $0.76/\alpha ^{2mn-4}=0.76\alpha ^{4}/\alpha ^{2mn}$). Moreover, in particular, $A_{m,n}$ tends to 0 as $n\to \infty$, then (4) holds, i.e.,

$$ \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \sim F_{mn}^{2}-F_{m(n-1)}^{2}+(-1)^{mn}C_{m}. $$

Moreover, $C_{m}$ is positive, because by (14) one has

$$ C_{m}:= \frac{2}{5}+(-1)^{m}G_{m}\geq \frac{2}{5}- \vert G_{m} \vert > \frac{2}{5}- \frac{1.2}{\alpha ^{2m}}>0, $$

where we used that $1.2/\alpha ^{2m}\leq 1.2/\alpha ^{4}<0.18$ for all $m\geq 2$. Additionally, we combine (12) and (13) to obtain

$$\begin{aligned} C_{m} =& \frac{2}{5} \biggl( 1 + \frac{1}{\alpha ^{4m}-(-1)^{m}}- \frac{2(-1)^{m}}{\alpha ^{2m}}+ \frac{(-1)^{m}}{\alpha ^{4m}} \\ &{}- \frac{2}{\alpha ^{2m}(\alpha ^{4m}-(-1)^{m})}+ \frac{1}{\alpha ^{4m}(\alpha ^{4m}-(-1)^{m})} \biggr). \end{aligned}$$

(20)

To obtain the formula in (5), we shall rationalize every fraction in (20). First, we observe that $(\alpha \beta )^{im}=1$ for $i\in \{2,4\}$ and

$$ \bigl(\alpha ^{4m}-(-1)^{m}\bigr) \bigl(\beta ^{4m}-(-1)^{m}\bigr)=(\alpha \beta )^{4m}-(-1)^{m} \bigl( \alpha ^{4m}+\beta ^{4m}\bigr)+1=2-(-1)^{m}L_{4m}. $$

Now, let us work with the expression

$$ \frac{1}{\alpha ^{4m}-(-1)^{m}}- \frac{2(-1)^{m}}{\alpha ^{2m}}+ \frac{(-1)^{m}}{\alpha ^{4m}}- \frac{2}{\alpha ^{2m}(\alpha ^{4m}-(-1)^{m})}+ \frac{1}{\alpha ^{4m}(\alpha ^{4m}-(-1)^{m})}. $$

(21)

By rationalizing, one has

$$\begin{aligned}& \frac{1}{\alpha ^{4m}-(-1)^{m}}= \frac{\beta ^{4m}-(-1)^{m}}{2-(-1)^{m}L_{m}} \quad \text{and}\quad \frac{2(-1)^{m}}{\alpha ^{2m}}=2(-1)^{m} \beta ^{2m}, \\& \frac{(-1)^{m}}{\alpha ^{4m}}=(-1)^{m}\beta ^{4m}\quad \text{and}\quad \frac{2}{\alpha ^{2m}(\alpha ^{4m}-(-1)^{m})}= \frac{2\beta ^{2m}(\alpha ^{4m}-(-1)^{m})}{2-(-1)^{m}L_{4m}}, \end{aligned}$$

and finally

$$ \frac{1}{\alpha ^{4m}(\alpha ^{4m}-(-1)^{m})}= \frac{\beta ^{4m}(\beta ^{4m}-(-1)^{m})}{2-(-1)^{m}L_{4m}}. $$

By putting all this information together and after a straightforward computation, we deduce that

$$ C_{m}= \frac{2}{5} \biggl(1 + \frac{B_{m}}{(2-(-1)^{m}L_{4m})} \biggr), $$

where

$$\begin{aligned} B_{m} := & \beta ^{4m}-(-1)^{m}-4(-1)^{m} \beta ^{2m}+ 2\beta ^{2m}L_{4m}+2(-1)^{m} \beta ^{4m} \\ & {}-\beta ^{4m}L_{4m}-2\beta ^{6m}+2(-1)^{m} \beta ^{2m}+\beta ^{8m}-(-1)^{m} \beta ^{4m}. \end{aligned}$$

We then use $\beta ^{s}=\beta F_{s}+F_{s-1}$ (Lemma 3) to write $B_{m}$ in the form $r_{m}+\beta s_{m}$ (where $r_{m}$ and $s_{m}$ belong to $\mathbb{Q}$ and they are called the rational and irrational parts of $B_{m}$, respectively). After some manipulations (by using $F_{2\ell m}=F_{\ell m}L_{\ell m}$, where ℓ is a positive integer, and identity $F_{4m-1}-1 = L_{2m-1}F_{2m}$, see [7]), we arrive at

$$ r_{m} := \bigl(1 - (-1)^{m}\bigr) L_{2 m} + 5 \bigl(1 + (-1)^{m}\bigr) F_{m - 1} F_{m} F_{2 m} $$

(22)

and

$$ s_{m}:=\bigl(1+(-1)^{m}\bigr)F_{2m}(L_{2m}-2) $$

(23)

as desired. The proof is then complete.

The proof of Corollary 1

Now, our goal is to provide a simpler characterization of $C_{m}$ depending on the parity of m (and to show how this affects its arithmetic nature). By Theorem 1, we have that

$$ C_{m}= \frac{2}{5} \biggl(1+ \frac{r_{m}+\beta s_{m}}{2-(-1)^{m}L_{4m}} \biggr), $$

where

$$ r_{m} := \bigl(1 - (-1)^{m}\bigr) L_{2 m} + 5 \bigl(1 + (-1)^{m}\bigr) F_{m - 1} F_{m} F_{2 m} $$

and

$$ s_{m}:=\bigl(1+(-1)^{m}\bigr)F_{2m}(L_{2m}-2). $$

Thus, we only need to work with $r_{m}$ and $s_{m}$ for the case in which m is odd and m is even. Therefore, the proof conveniently splits into two cases as follows.

The proof of item (i)

When m is even, we start by noting that, by Lemma 2(iii), one has

$$ 2-(-1)^{m}L_{4m}=2-(-1)^{m}\bigl(5F_{2m}^{2}+2(-1)^{2m} \bigr) = 2 - \bigl(5F_{2m}^{2}+2\bigr) = -5 F_{2m}^{2}. $$

Furthermore, it holds

$$ s_{m} = 2 F_{2m}(L_{2m}-2) > 0 . $$

For $r_{m}$, we infer that

$$ r_{m} = 10 F_{m - 1} F_{m} F_{2 m} = 2(F_{4m-1}+F_{2m}-1). $$

Thus,

$$\begin{aligned} C_{m} = & \frac{2}{5} \biggl(1+ \frac{10 F_{m - 1} F_{m} F_{2 m}+2\beta F_{2m}(L_{2m}-2)}{-5F_{2m}^{2}} \biggr) \\ = & \frac{2}{5} \frac{5(F_{2m}-2F_{m-1}F_{m})-2(L_{2m}-2)\beta }{5F_{2m}} \\ = & \frac{2}{5} \frac{(L_{2m}-2)-2(L_{2m}-2)\beta }{5F_{2m}} \\ = & \frac{2(L_{2m}-2)}{25F_{2m}} (1-2\beta ) = \frac{2(L_{2m}-2)}{25F_{2m}}\sqrt{5} , \end{aligned}$$

where we used Lemma 2(i) and that $1-2\beta =\sqrt{5}$.

The proof of item (ii)

When m is odd, we note that, by Lemma 2(iv), one has

$$ 2-(-1)^{m}L_{4m}=2-(-1)^{m}\bigl(L_{2m}^{2}-2(-1)^{2m} \bigr)=L_{2m}^{2}. $$

Clearly, we get by (22) and (23) that

$$ s_{m} = 0 \quad \text{and}\quad r_{m} = 2 L_{2m} . $$

Thus,

$$ C_{m}= \frac{2}{5} \biggl(1+ \frac{2L_{2m}}{L_{2m}^{2}} \biggr)= \frac{2}{5} \biggl(1+ \frac{2}{L_{2m}} \biggr)= \frac{2(L_{2m}+2)}{5L_{2m}} $$

for any odd $m\geq 1$. The proof is then complete.

The proof of Corollary 2

First, for $m=1$, we observe from (1) that

$$\begin{aligned} \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor =&\textstyle\begin{cases} F_{n-1}F_{n}-1, & \text{if } n \text{ is even}, n\geq 2; \\ F_{n-1}F_{n}, & \text{if } n \text{ is odd}, n\geq 1. \end{cases}\displaystyle \end{aligned}$$

However, since

$$ F_{n}^{2}-F_{n-1}^{2}=(F_{n}-F_{n-1}) (F_{n}+F_{n-1})=F_{n-2}F_{n+1}, $$

we can use Lemma 2(v) to deduce that

$$ F_{n-2}F_{n+1}-1=F_{n-1}F_{n}, \quad \text{if } n \text{ is odd} $$

and

$$ F_{n-2}F_{n+1}=F_{n-1}F_{n}-1,\quad \text{if } n \text{ is even.} $$

Thus, the formula in Corollary 2(ii) holds for $m=1$.

To deal with $m\geq 2$, we use Corollary 1 to obtain better bounds for $C_{m}$. If m is odd, then

$$ \frac{2}{5}< C_{m}< \frac{2}{5} \biggl(1+ \frac{2}{L_{2}} \biggr)= \frac{2}{3}. $$

In the case in which m is even, one has

$$ 0.28< \frac{2\sqrt{5}}{25} \biggl( \frac{L_{2m}}{F_{2m}} - \frac{2}{F_{2m}} \biggr)< C_{m} < \frac{2}{5}, $$

where we combine items (iii) and (iv) of Lemma 2 to get $L_{2m}/F_{2m}>\sqrt{5}$.

To simplify our notation, we shall denote $X_{m,n}$ as

$$ X_{m,n}:= \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} - \bigl(F_{mn}^{2}-F_{m(n-1)}^{2} \bigr). $$

Thus, inequality (3) yields

$$ (-1)^{mn}C_{m}- \frac{9.83}{\alpha ^{2m(n-1)}}< X_{m,n}< (-1)^{mn}C_{m}+ \frac{9.83}{\alpha ^{2m(n-1)}}. $$

(24)

To prove items (i) and (ii), we may split the proof into two cases as follows.

The case mn even

In this case, we have $C_{m}\in (0.28, 2/3)$, $2m(n-1)\geq 8$, and so (24) becomes

$$ 0.07< 0.28- \frac{9.83}{\alpha ^{8}}< C_{m}+ \frac{9.83}{\alpha ^{2m(n-1})}< X_{m,n}< C_{m}+ \frac{9.83}{\alpha ^{8}}< \frac{2}{3}+ \frac{9.83}{\alpha ^{8}}< 0.88. $$

Thus, $\lfloor X_{m,n} \rfloor =0$ and so

$$ 0 = \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} - \bigl(F_{mn}^{2}-F_{m(n-1)}^{2} \bigr) \Biggr\rfloor = \Biggl\lfloor \Biggl( \sum_{k=n}^{ \infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \Biggr\rfloor - \bigl(F_{mn}^{2}-F_{m(n-1)}^{2} \bigr). $$

Hence, if mn is an even integer, we have

$$ \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \Biggr\rfloor = F_{mn}^{2}-F_{m(n-1)}^{2}. $$

The case mn odd

In this case, m and n are odd integers. Thus $C_{m}\in (2/5, 2/3)$ and (24) implies

$$\begin{aligned} -0.7 < &- \frac{2}{3}- \frac{9.83}{\alpha ^{12}}< -C_{m}- \frac{9.83}{\alpha ^{2m(n-1)}}< X_{m,n} \\ < &-C_{m}+ \frac{9.83}{\alpha ^{2m(n-1)}}< - \frac{2}{5}+ \frac{9.83}{\alpha ^{12}}< -0.36, \end{aligned}$$

where we used that $2m(n-1)\geq 12$ (since $m\geq 3$, because ≥2 is odd). Thus, $\lfloor X_{m,n} \rfloor =-1$, and so

$$ -1 = \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} - \bigl(F_{mn}^{2}-F_{m(n-1)}^{2} \bigr) \Biggr\rfloor = \Biggl\lfloor \Biggl( \sum_{k=n}^{ \infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \Biggr\rfloor - \bigl(F_{mn}^{2}-F_{m(n-1)}^{2} \bigr). $$

Therefore, if mn is an odd integer, we have

$$ \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \Biggr\rfloor = F_{mn}^{2}-F_{m(n-1)}^{2}-1. $$

This finishes the proof.

Conclusions

In this paper, for any $m\geq 1$, we provide an explicit constant $C_{m}>0$ for which

$$ \Biggl\vert \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} - \bigl(F_{mn}^{2}-F_{m(n-1)}^{2}+(-1)^{mn}C_{m} \bigr) \Biggr\vert < \frac{9.83}{\alpha ^{2m(n-1)}}, $$

where $(F_{n})_{n}$ is the Fibonacci sequence and $\alpha =(1+\sqrt{5})/2$ is the golden number. Moreover, we show that the estimate $|C_{m}-2/5|<1.2/\alpha ^{2m}$ holds for all $m\geq 1$. These results solve effectively (and quantitatively) some questions proposed by Lee and Park [8]. As an application of the previous facts, we find the closed formula

$$\begin{aligned} \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{-1} \Biggr\rfloor =& \textstyle\begin{cases} F_{mn}^{2}-F_{m(n-1)}^{2}, & \text{if } mn \text{ is even}; \\ F_{mn}^{2}-F_{m(n-1)}^{2}-1, & \text{if } mn \text{ is odd.} \end{cases}\displaystyle \end{aligned}$$

The proof combines several estimates, inequalities, properties of Fibonacci and Lucas numbers as well as some facts about the convergence of series. The computations in this work were performed with Mathematica software.