Abstract
Let \(\{f_{k} \} _{k=1}^{\infty}\) be a Fibonacci sequence with \(f_{1}=f_{2}=1\). In this paper, we find a simple form \(g_{n}\) such that
where \(a_{k}=\frac{1}{f_{k}^{2}}\), \(\frac{1}{f_{k}f_{k+m}}\), or \(\frac{1}{f_{3k}^{2}}\). For example, we show that
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1 Introduction
Last decade many mathematicians were interested in finding the formula for the integer part of the reciprocal tails of the convergent series. Precisely, one can see the explicit value of \(\lfloor (\sum^{\infty}_{k=n}a_{k} )^{-1} \rfloor\) when \(\sum^{\infty}_{k=1}a_{k}\) converges. This problem starts from the reciprocal sum of Fibonacci numbers. Let \(f_{0}=0\), \(f_{1}=f_{2}=1\), and \(f_{n+2}=f_{n}+f_{n+1}\) for any \(n\in \mathbb{N}\). In [6], Ohtsuka and Nakamura proved
and
where \(\lfloor x \rfloor\) is the floor function. See more results for subsequences of Fibonacci numbers in [11, 12], Pell numbers in [1, 13], and Mathieu series in [4]. Also, recently many interesting results on special numbers have been obtained in [7–9]. The following natural question on the asymptotic behavior can be raised.
Question
Let \(\sum^{\infty}_{k=1}a_{k}\) be a convergent series. Can we find a suitable function \(g_{n}\) such that
Here the notation \(A_{n}\sim B_{n}\) means that \(\lim_{n\rightarrow\infty}(A_{n}-B_{n})=0\).
In [3], we proved that
for any \(m\in\mathbb{N}\) and \(0\leq\ell\leq m-1\). In fact, we proved
In the special case when \(m=1\) and \(\ell=0\), the above equation is reduced to
In [3], we also proved the generalization of (1.1) as the following formula:
for any \(m\in\mathbb{N}\) and \(0\leq\ell\leq m-1\). One can see the results for the product of two Fibonacci numbers in [5].
In this paper, we study the asymptotic behavior of the reciprocal sum of
for \(k,m\in\mathbb{N}\). Precisely, we prove that
where
For the proof of our main theorem, we prove the following inequalities:
- (i)
\(g_{n,0}< ( \sum^{\infty}_{k=n}\frac {1}{f_{k}^{2}} )^{-1}<g_{n,0}+c_{n}\) for all \(n\geq1\),
- (ii)
\(g_{n,\ell}-c_{n}< ( \sum^{\infty}_{k=n}\frac {1}{f_{k}f_{k+2\ell}} )^{-1}\leq g_{n,\ell}\) for all \(\ell\geq1\) and \(n\geq2\ell-1\),
- (iii)
\(h_{n,\ell}-c_{n}< ( \sum^{\infty}_{k=n}\frac {1}{f_{k}f_{k+2\ell-1}} )^{-1}<h_{n,\ell}\) for all \(\ell\geq1\) and \(n\geq2\ell-2\).
Here \(c_{n}=1/f_{n}\). We believe that the conditions \(n\geq2\ell-1\) and \(n\geq2\ell-2\) can be removed. However, it is enough to prove the inequalities for sufficiently largen for the study of asymptotic behavior as \(n\rightarrow\infty\). As an application of the above results, we can obtain
for all \(m\in\mathbb{N}\). For example, our formulas imply that
and
In the final section, we discuss the reciprocal sum of \(f^{2}_{mk}\) for \(m\geq2\). If \(m=3\), then we prove that
where \(\widetilde{g}_{n}=f_{3n}^{2}-f_{3n-3}^{2}+\frac{4}{9}(-1)^{n}\).
As in [3, 6], the following lemma plays an important role in proving the essential inequalities.
Lemma 1.1
([6])
Let\(\{a_{n}\}^{\infty}_{n=1}\)and\(\{b_{n}\}^{\infty}_{n=1}\)be sequences with\(\lim_{n\rightarrow\infty}a_{n}=0\). If\(a_{n}< b_{n}+a_{n+1}\)holds for any\(n\in\mathbb{N}\), then
holds for any\(n\in\mathbb{N}\).
We use the following relation when we calculate Fibonacci numbers.
Lemma 1.2
([2], Catalan’s identity)
For any\(n,k\in\mathbb{N}\), we have
The following lemma is useful when we get a lower bound of formulas containing Fibonacci numbers. It comes from the identity
Lemma 1.3
For any\(m,n\in\mathbb{N}\), we have\(f_{m+n}>f_{m+1}f_{n}\).
Remark 1.4
The Fibonacci numbers \(f_{n}\) can be written as the closed form by Binet’s formula [10]
where \(\alpha >\beta \) are two solutions of \(x^{2}-x-1=0\). The infinite sum \(\sum^{\infty}_{k=1}\frac{1}{f_{k}}\) is known to be irrational, but it is unknown whether \(\sum^{\infty}_{k=1}\frac{1}{f_{k}^{2}}\) is irrational or not.
2 Reciprocal sum of \(f_{k}^{2}\)
In [6], Ohtsuka and Nakamura proved the following inequalities:
- (i)
\(f_{n-1}f_{n}-1< (\sum^{\infty}_{k=n}\frac {1}{f_{k}^{2}} )^{-1}<f_{n-1}f_{n}\), when n is even,
- (ii)
\(f_{n-1}f_{n}< (\sum^{\infty}_{k=n}\frac {1}{f_{k}^{2}} )^{-1}<f_{n-1}f_{n}+1\), when n is odd.
Now we will prove the inequalities which are sharper than (i) and (ii). In this section, let
In fact, \(g_{n}\) can be written as
Theorem 2.1
For any\(n\in\mathbb{N}\), we have
where\(c_{n}=1/f_{n}\).
To prove Theorem 2.1, we need the following formula.
Proposition 2.2
For any\(n\in\mathbb{N}\), we have
Proof
Note that
By Catalan’s identity, we have
The desired result comes from the above two identities. □
Now we prove Theorem 2.1. By Proposition 2.2, we have
It follows that
for all \(n\in\mathbb{N}\). By Lemma 1.1, we obtain
For the proof of the converse inequality, we compute
where
By Proposition 2.2, we have
since \(c_{n}g_{n+1}=f_{n+1}-\frac{\frac{1}{3}(-1)^{n+1}}{f_{n}}\geq f_{n+1}-\frac{1}{3}\geq\frac{2}{3} \) for all \(n\in\mathbb{N}\). Thus we obtain
for all \(n\in\mathbb{N}\). By Lemma 1.1 again, we obtain
By (2.2) and (2.3), we complete the proof of Theorem 2.1.
Remark 2.3
The inequalities of Theorem 2.1 imply
See Table 1.
3 Reciprocal sum of \(f_{k}f_{k+m}\) when m is even
In Sects. 3 and 4, we deal with the value of
In fact, we can compute the explicit value when \(m=2\). Note that
Hence it holds that, for all \(n\in\mathbb{N}\),
However, it is difficult to find the explicit value of
except for \(m=2\).
Throughout this section, we assume that m is even, so that \(m=2\ell\) for some \(\ell\in\mathbb{N}\). In this case, we define
For simplicity, we write \(I_{1}:=f_{\ell}^{2}+(-1)^{\ell}\).
Proposition 3.1
For any\(n\in\mathbb{N}\), we have
Proof
Write \(g_{n}=f_{n+\ell-1}f_{n+\ell}-\frac{(-1)^{n}}{3}I_{1}\). Note that
Since \(f_{\ell}^{2}=I_{1}-(-1)^{\ell}\), we have
Note that
By Lemma 1.2, \(f_{n+\ell-1}f_{n+\ell+1}=f_{n+\ell }^{2}+(-1)^{n+\ell}\). Thus we have
By (3.1) and (3.2), we complete the proof. □
See Table 2. If \(m=4\), then \(g_{n,2}=f_{n+1}f_{n+2}-\frac {2}{3}(-1)^{n}\).
Theorem 3.2
If\(m=2\ell\)for some\(\ell\in\mathbb{N}\), then for any\(n\geq 2\ell-1\), we have
where\(c_{n}=1/f_{n}\).
For example, if \(\ell=4\), then \(g_{n}=f_{n+3}f_{n+4}-\frac {10(-1)^{n}}{3}\). Thus we have
Proof
(i) By Lemma 1.1, it is enough to show that
for all \(n\in\mathbb{N}\). By Proposition 3.1, we have
It follows that
which implies (3.3). In fact, the equality holds only when \(\ell=1\). This is the case when \(m=2\).
(ii) By Lemma 1.1, it is enough to show that
for all \(n\in\mathbb{N}\). Note that
Now we will show that, for \(n\geq2\ell-1\), we have
Note that
By Proposition 3.1, we have
By Lemma 1.3, we have
If \(n\geq2\ell-1\), then \(f_{n-2\ell+4}+f_{n-2\ell+2}\geq f_{3}+f_{1}=3\). Thus, we have
for \(n\geq2\ell-1\). From (3.4) and (3.5), we obtain
since \(f_{\ell}\geq1\). □
Remark 3.3
If \(\ell=0\), then \(I_{1}=f_{0}^{2}+1=1\). Thus in the proof of Theorem 3.2 we have
In fact, this is an identity in Proposition 2.2. Thus the inequality in Theorem 3.2 is opposite to Theorem 2.1.
4 Reciprocal sum of \(f_{k}f_{k+m}\) when m is odd
Throughout this section, we assume that m is odd, so that \(m=2\ell -1\) for some \(\ell\in\mathbb{N}\). In this case, we define
For simplicity, we write \(I_{2}:=f_{\ell-1}f_{\ell}+(-1)^{\ell}\).
Proposition 4.1
For any\(n\in\mathbb{N}\), we have
Proof
By Catalan’s identity, we have
Note that
Note that
It follows that
Since \(f_{n+\ell-1}f_{n+\ell}=f_{n+\ell+1}f_{n+\ell-2}+(-1)^{n+\ell }\), we have
Since \(I_{2}-f_{\ell+1}f_{\ell-2}=f_{\ell-1}f_{\ell}-f_{\ell+1}f_{\ell -2}+(-1)^{\ell}=2(-1)^{\ell}\),
□
See Table 3. If \(m=5\), then \(h_{n,3}=f_{n+2}^{2}-\frac{1}{3}(-1)^{n}\).
Theorem 4.2
If\(m=2\ell-1\), then for any\(n\geq2\ell-2\), we have
where\(c_{n}=1/f_{n}\).
For example, if \(\ell=4\), then \(h_{n}=f_{n+3}^{2}-\frac{7(-1)^{n}}{3}\). Thus we have
Proof
(i) We will show that
Note that
By Proposition 4.1, we have
(ii) We will show that
Note that
It is enough to show that
Note that
By Proposition 4.1, we have
Note that
If \(n\geq2\ell-2\), then \(f_{n-2\ell+5}+f_{n-2\ell+4}\geq f_{3}+f_{2}=3\). It follows that
Thus we have
□
5 Reciprocal sum of \(f_{3k}^{2}\)
In this final section, we discuss the reciprocal sum of \(f_{mk}^{2}\) for any \(m\geq2\). Similar to (2.1), when \(m=1\), we expected to find a suitable constant \(C_{m}>0\) such that
In Sect. 2, we proved that \(C_{1}=\frac{2}{3}\). The integer part when \(m=3\) has been obtained as follows.
Theorem 5.1
([5])
Now we prove that \(C_{3}=\frac{4}{9}\). See Table 4. Here
Our theorem contains more optimal inequality than (5.2).
Theorem 5.2
For all\(n\in\mathbb{N}\), we have
where\(\widetilde{g}_{n}:=f_{3n}^{2}-f_{3n-3}^{2}+ \frac{4}{9}(-1)^{n}\). Thus we have
Lemma 5.3
For any\(n\geq1\), we have
Proof
Since \(f_{3n+2}+f_{3n-2}=3f_{3n}\), we have
It follows that
which completes the proof. □
Proof
(i) We will show that
Note that
where \(A:=(\widetilde{g}_{n+1}-\widetilde{g}_{n})f_{3n}^{2}-\widetilde {g}_{n}\widetilde{g}_{n+1}\). It is enough to show that \(A>0\).
Note that
Note also that
It follows that
By Lemma 5.3, we have
(ii) We will prove that
Note that
where \(B=(\widetilde{g}_{n+1}-\widetilde{g}_{n})f_{3n}^{2}-(\widetilde {g}_{n}+c_{n})(\widetilde{g}_{n+1}+c_{n+1})\). Now it is enough to show that \(B<0\).
By using (5.3), we have
Note that
It follows that \(B<0\). □
Remark 5.4
It looks not easy to find the explicit values of \(C_{m}\) satisfying (5.1) except for \(m=1,3\). By using a computer software program (Maple17 and wolframalpha.com), we found the following. If m is even, then
If m is odd, then
We might expect that \(C_{m}\) tends to \(\frac{2}{5}\) as \(n\rightarrow \infty\).
6 Conclusion
We summarize all the results that have been proved in this paper.
- (i)
\(( \sum^{\infty}_{k=n}\frac {1}{f_{k}^{2}} )^{-1}\sim f_{n}^{2}-f_{n-1}^{2}+\frac{2}{3}(-1)^{n}\) as \(n\rightarrow\infty\),
- (ii)
\(( \sum^{\infty}_{k=n}\frac {1}{f_{k}f_{k+2\ell}} )^{-1}\sim f_{n+\ell-1}f_{n+\ell}-(f_{\ell}^{2}+(-1)^{\ell})\frac{(-1)^{n}}{3}\) as \(n\rightarrow\infty\),
- (iii)
\(( \sum^{\infty}_{k=n}\frac {1}{f_{k}f_{k+2\ell-1}} )^{-1}\sim f_{n+\ell-1}^{2}-(f_{\ell-1}f_{\ell}+(-1)^{\ell})\frac{(-1)^{n}}{3}\) as \(n\rightarrow\infty\),
- (iv)
\(( \sum^{\infty}_{k=n}\frac {1}{f_{3k}^{2}} )^{-1}\sim f_{3n}^{2}-f_{3n-3}^{2}+\frac{4}{9}(-1)^{n} \) as \(n\rightarrow\infty\).
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This work was supported by NRF-2018R1D1A1B07050044 from the National Research Foundation of Korea.
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Lee, HH., Park, JD. Asymptotic behavior of reciprocal sum of two products of Fibonacci numbers. J Inequal Appl 2020, 91 (2020). https://doi.org/10.1186/s13660-020-02359-z
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DOI: https://doi.org/10.1186/s13660-020-02359-z