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The proof of a formula concerning the asymptotic behavior of the reciprocal sum of the square of multipleangle Fibonacci numbers
Journal of Inequalities and Applications volume 2022, Article number: 21 (2022)
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
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,
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
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
and
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
(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
and
Moreover, as formula (5.1) of [8, Sect. 5], they stated (without proof) the following expected general formula:
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
for all \(n\geq \max \{\frac{4}{m}, 2\}\) (where, as usual, \(\alpha =(1+\sqrt{5})/2\) denotes the golden ratio). In particular,
Moreover, for all \(m\geq 1\),
where \(\beta =(1\sqrt{5})/2\),
and
Furthermore, the estimate
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:
which are rational numbers. However,
which (since \(12\beta =\sqrt{5}\)) is the same list as
are irrational numbers.
This suggests that \(C_{2m1}\in \mathbb{Q}_{>0}\) 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
$$ \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{1} \Biggr\rfloor = F_{mn}^{2}F_{m(n1)}^{2} $$holds for all \(n\geq 3\).

(ii)
Let m be an odd positive integer. Then
$$\begin{aligned} \Biggl\lfloor \Biggl( \sum_{k=n}^{\infty } \frac{1}{F_{mk}^{2}} \Biggr)^{1} \Biggr\rfloor =& \textstyle\begin{cases} F_{mn}^{2}F_{m(n1)}^{2}, & \textit{if } n \textit{ is even}; \\ F_{mn}^{2}F_{m(n1)}^{2}1, & \textit{if } n \textit{ is odd}, \end{cases}\displaystyle \end{aligned}$$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)
\(L_{n}=F_{n+1}+F_{n1}\).

(iii)
\(L_{2n}=5F_{n}^{2}+2(1)^{n}\).

(iv)
\(L_{2n}=L_{n}^{2}2(1)^{n}\).

(v)
(D’Ocagne’s identity) \((1)^{n}F_{mn}=F_{m}F_{n+1}F_{m+1}F_{n}\).
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
The last ingredients are some known lower and upper bounds for \(F_{n}\), that is,
Lemma 4
The inequalities
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
hence
However, \(1/(1x)^{2}=1+2x+3x^{2}+\cdots \) holds for \(x<1\). Thus, since \(\beta /\alpha <1\) and \(\alpha \beta =1\), we have
By summing up from \(k=n\) to infinity and after a straightforward calculation, we arrive at
where \(f(m,n)\) denotes the following summatory
Note that, by the reverse triangle inequality (and \(\beta /\alpha =1/\alpha ^{2}\), which follows from \(\beta =1/\alpha \)), one has
Thus, we deduce the following upper bound for \(f(m,n)\):
where we used that
for all \(i\geq 3\) (since \(mn\geq 4\)) together with
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
and hence
Since \(f(m,n)<1\), for \(mn\geq 4\), then
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
where we used that
Turning back to (8), we have
where
satisfies (by (9))
Hence
Now, let us work with the second term of the righthand side of (11). By the definition of \(f(m,n)\), we can write
where
Since \(\beta /\alpha =1/\alpha ^{2}\), we deduce that
where
and \(E_{m}\) and \(S_{m}\) are defined as
Now, we use that \(S_{m}<2/\alpha ^{2m}\) and
(since \(x\mapsto x/(x1)\) 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
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
since \((\alpha ^{2m}1)^{2}<\alpha ^{4m}\). Therefore, by using again \(\beta /\alpha =1/\alpha ^{2}\), we get
where we used that \(i<2^{i}<\alpha ^{2i}\) holds for every integer \(i\geq 3\). Furthermore, since \(mn\geq 4\), we have for \(i\geq 2\)
Thus,
By combining (15) and (16), we infer that
In particular,
Summarizing, we have
On the other hand, we can apply Binet’s formula again to obtain
where we used that \(\alpha \beta =1\). Now, we can combine the previous relation with the formula in (18) to write
where
Now, we use the formula for \(A_{m,n}\), estimates (17) and (10) to obtain
which implies in (3) (we used that \(0.76/\alpha ^{2mn4}=0.76\alpha ^{4}/\alpha ^{2mn}\)). Moreover, in particular, \(A_{m,n}\) tends to 0 as \(n\to \infty \), then (4) holds, i.e.,
Moreover, \(C_{m}\) is positive, because by (14) one has
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
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
Now, let us work with the expression
By rationalizing, one has
and finally
By putting all this information together and after a straightforward computation, we deduce that
where
We then use \(\beta ^{s}=\beta F_{s}+F_{s1}\) (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 \(F_{4m1}1 = L_{2m1}F_{2m}\), see [7]), we arrive at
and
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
where
and
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
Furthermore, it holds
For \(r_{m}\), we infer that
Thus,
where we used Lemma 2(i) and that \(12\beta =\sqrt{5}\).
The proof of item (ii)
When m is odd, we note that, by Lemma 2(iv), one has
Clearly, we get by (22) and (23) that
Thus,
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
However, since
we can use Lemma 2(v) to deduce that
and
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
In the case in which m is even, one has
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
Thus, inequality (3) yields
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 \(C_{m}\in (0.28, 2/3)\), \(2m(n1)\geq 8\), and so (24) becomes
Thus, \(\lfloor X_{m,n} \rfloor =0\) and so
Hence, if mn is an even integer, we have
The case mn odd
In this case, m and n are odd integers. Thus \(C_{m}\in (2/5, 2/3)\) and (24) implies
where we used that \(2m(n1)\geq 12\) (since \(m\geq 3\), because ≥2 is odd). Thus, \(\lfloor X_{m,n} \rfloor =1\), and so
Therefore, if mn is an odd integer, we have
This finishes the proof.
Conclusions
In this paper, for any \(m\geq 1\), we provide an explicit constant \(C_{m}>0\) for which
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
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.
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Data sharing is not applicable to this paper as no datasets were generated or analyzed during the current study.
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Acknowledgements
The first author is grateful to CNPqBrazil for financial support. The second author thanks University of Hradec Kralove for support.
Funding
The second author was supported by Project of Excellence of Faculty of Science No. 2209/20222023, University of Hradec Kralove, Czech Republic.
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MD dealt with the conceptualization, supervision, methodology, investigation, and writing—original draft preparation. PT made the formal analysis, writing—review and editing, project administration, and funding acquisition. Both authors read and approved the final manuscript.
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Marques, D., Trojovský, P. The proof of a formula concerning the asymptotic behavior of the reciprocal sum of the square of multipleangle Fibonacci numbers. J Inequal Appl 2022, 21 (2022). https://doi.org/10.1186/s13660022027557
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DOI: https://doi.org/10.1186/s13660022027557
MSC
 11B39
 11B05
Keywords
 Fibonacci
 Series
 Upper bounds
 Inequalities
 Asymptotic
 Recurrence sequences