On the Reciprocal Sums and Products of mth-order Linear Recursive Sequences

Tingting Du Research Center for Number Theory and Its Application Northwest University Xi’an, Shaanxi 710127 China

                     dutingtingaaa@163.com

Abstract The mth-order linear recursive sequence of (w_{n})_{n≥0} is defined by the recursion w_{n} = a_{1}w_{n−1} + a_{2}w_{n−2} + ⋯ + a_{m}w_{n−m} for n > m. In previous discussions of the reciprocal sums and products of (w_{n})_{n≥0}, the condition a_{1} ≥ a_{2} ≥ ⋯ ≥ a_{m} ≥ 1 was typically imposed. In this paper, we extend previous research and allow the coefficients to be arbitrary positive integers.

1 Introduction

For positive integers a_{1}, a_{2},…, a_{m} with a_{m} ≠ 0, the mth-order linear recursive sequence (w_{n})_{n≥0} is defined by

w_{n} = a_{1}w_{n−1} + a_{2}w_{n−2} + ⋯ + a_{m}w_{n−m}, n > m,  (1)

where the initial values w_{i} ∈ ℕ, and at least one is not zero. The characteristic polynomial of the sequence (w_{n})_{n≥0} is

φ(x) = x^{m} − a_{1}x^{m−1} − ⋯ − a_{m−1}x − a_{m} = (x − λ_{1})^{m_{1}} ⋯ (x − λ_{l})^{m_{l}},  (2)

where the λ_{i} are called the roots of the sequence. If the absolute value of a root of the sequence (w_{n})_{n≥0} is strictly largest, the root is called the dominant root.

If m = 2, a_{1} = a_{2} = 1, and w_{0} = 0, w_{1} = 1, the resulting sequence (w_{n})_{n≥0} is the famous Fibonacci sequence (f_{n})_{n≥0}. Ohtsuka and Nakamura [10] considered the reciprocal sums of the Fibonacci sequence and obtained the following result:

⌊(∑_{k=n}^{∞} \frac{1}{f_{k}} )^{−1}⌋ = \begin{cases}f_{n−2}, & if n is even and n ≥ 2; \\ f_{n−2} − 1, & if n is odd and n ≥ 1, \end{cases}

where ⌊z⌋ denotes the floor function.

Computing the floor function of ( ∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1} is a difficult problem. Some researchers have

studied the nearest integer to ( ∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1}. For example, Wu and Zhang [13] proved that there

exists a positive integer n_{1} such that

∥(∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1}∥ = w_{n} − w_{n−1}, n ≥ n_{1},

where a_{1} ≥ a_{2} ≥ ⋯ ≥ a_{m} ≥ 1 and ∥z∥ denotes the nearest integer function, defined by ∥z∥ = ⌊z + \frac{1}{2}⌋.

If two sequences (u_{n})_{n≥0} and (v_{n})_{n≥0} satisfy the condition that (u_{n}/v_{n})_{n≥0} tends to 1 as n → ∞, we call them asymptotically equivalent. Some researchers have continued the study

of reciprocal sums by finding a sequence that is asymptotically equivalent to (( ∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1}).

Specifically, Trojovský [12] proved that the sequences

((∑_{k=n}^{∞} \frac{1}{P (w_{k})})^{−1})_{n} and (P (w_{n}) − P (w_{n−1}))_{n}

are asymptotically equivalent, where P (z) is a non-constant polynomial with P (z) ∈ ℂ[z]. For more on reciprocal sums and products, see [3, 6, 15, 2, 1, 4, 9, 14, 16, 8].

In previous research concerning reciprocal sums and products of (w_{n})_{n≥0}, the condition a_{1} ≥ a_{2} ≥ ⋯ ≥ a_{m} ≥ 1 was typically imposed. In this paper, we allow the coefficients to be arbitrary positive integers, and obtain a series of sequences that are asymptotically

equivalent to (( ∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1}) and (1 − ∏_{k=n}^{∞}(1 − \frac{1}{w_{k}}))^{−1}). The main results are summarized in

the following theorem.

Theorem 1. Let (w_{n})_{n≥0} be an mth-order linear recursive sequence defined by (1). The sequences

((∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1}) and (w_{n} − w_{n−1})  (3)

are asymptotically equivalent, and the sequences

((1 − ∏_{k=n}^{∞}(1 − \frac{1}{w_{k}}))^{−1})_{n} and (w_{n} − w_{n−1})_{n}  (4)

are asymptotically equivalent.

2 Some lemmas

In this section, we shall give several lemmas that are useful for the proofs of the theorem.

Lemma 2. (Descartes rule of signs). Let ϕ(x) = a_{n_{1}}x^{n_{1}} +⋯+a_{n_{k}}x^{n_{k}} be a polynomial where the n_{i} are integers and n_{1} > n_{2} > ⋯ > n_{k} ≥ 0. The number of positive roots of ϕ(x) is at most the number of sign changes of adjacent nonzero coefficients.

Lemma 3. ([5, 7] Eneström-Kakeya theorem). Let ϕ′(x) = a_{0} + a_{1}x + ⋯ + a_{n}x^{n} be a polynomial of order n with real coefficients. If 0 ≤ a_{0} ≤ a_{1} ≤ ⋯ ≤ a_{n}, then all complex roots z of ϕ′(x) satisfy |z| ≤ 1.

Lemma 4. Let φ(x) = x^{m} − a_{1}x^{m−1} − ⋯ − a_{m−1}x − a_{m} be the characteristic polynomial of (w_{n})_{n≥0}, where the coefficients a_{1}, a_{2}, …, a_{m} are positive integers. Then the following hold:

• φ(x) has only one positive root λ_{1}, called λ, and λ > 1;

• The other m − 1 roots of φ(x) lie in the circle |z| < λ. Therefore λ is the dominant root of φ(x).

Proof. By Lemma 2, the characteristic polynomial φ(x) = x^{m} − a_{1}x^{m−1} − ⋯ − a_{m−1}x − a_{m} has at most one positive root λ_{1}, called λ. In addition, lim_{x→∞} φ(x) = +∞, and, by φ(1) < 0, we obtain that λ > 1 and λ is the only positive root of φ(x).

By using the relation

λ^{m} = a_{1}λ^{m−1} + a_{2}λ^{m−2} + ⋯ + a_{m−1}λ + a_{m},

we obtain that φ(x) = (x − λ)ψ(x), where

\begin{aligned}ψ(x) & = x^{m−1} + (λ − a_{1})x^{m−2} + (λ^{2} − a_{1}λ − a_{2})x^{m−3} + ⋯ \\ & + (λ^{m−2} − a_{1}λ^{m−3} − ⋯ − a_{m−2})x + λ^{m−1} − a_{1}λ^{m−2} − ⋯ − a_{m−1}. \end{aligned}

Claim 1. If z is a root of ψ(x), then |z| ≤ λ.

In fact, we prove that all the roots of Ψ(x) := ψ(λx) are in the closed unit ball. The polynomial

Ψ(x) = λ^{m−1}x^{m−1} + (λ − a_{1})λ^{m−2}x^{m−2} + (λ^{2} − a_{1}λ − a_{2})λ^{m−3}x^{m−3} + ⋯
+ (λ^{m−2} − a_{1}λ^{m−3} − ⋯ − a_{m−2})λx + λ^{m−1} − a_{1}λ^{m−2} − ⋯ − a_{m−1}

= λ^{m−1}x^{m−1} + (λ^{m−1} − a_{1}λ^{m−2})x^{m−2} + (λ^{m−1} − a_{1}λ^{m−2} − a_{2}λ^{m−3})x^{m−3} + ⋯

+ (λ^{m−1} − a_{1}λ^{m−2} − ⋯ − a_{m−2}λ)x + λ^{m−1} − a_{1}λ^{m−2} − ⋯ − a_{m−1}.

Since

\begin{matrix}λ^{m−1} > λ^{m−1} − a_{1}λ^{m−2} > λ^{m−1} − a_{1}λ^{m−2} − a_{2}λ^{m−3} > ⋯ \\ > λ^{m−1} − a_{1}λ^{m−2} − ⋯ − a_{m−2}λ > λ^{m−1} − a_{1}λ^{m−2} − ⋯ − a_{m−1} = a_{m}/λ > 0, \end{matrix}

which λ is positive root of φ(x). By Lemma 3, this completes the proof of Claim 2.

Claim 2. On the circle |z| = λ, the polynomial φ(x) has the unique root λ.

If φ(z) = 0, then

z^{m} = a_{1}z^{m−1} + a_{2}z^{m−2} + ⋯ + a_{m−1}z + a_{m}.

The triangle inequality is satisfied:

|z|^{m} ≤ a_{1} |z|^{m−1} + a_{2} |z|^{m−2} + ⋯ + a_{m−1} |z| + a_{m}.  (5)

If z = λ, then φ(z) = 0. So (5) must be an equality. Therefore, a_{1}z^{m−1}, a_{2}z^{m−2}, …, a_{m−1}z, a_{m} all on the same ray leaving the origin. Since a_{1}, a_{2}, …, a_{m} are all the elements of ℝ^{+}, z^{m−1}, z^{m−2}, …, z must be elements of ℝ^{+}. Therefore we obtain φ(z) ∈ ℝ^{+}. On the circle |z| = λ, we obtain z = λ. This completes the proof of Claim 2.

We define f(x) = O(g(x)) to mean that the quotient f(x)/g(x) is bounded for x ≥ a.

Lemma 5. Let (w_{n})_{n≥0} be mth-order linear recursive sequence defined by (1). Then the asymptotic formula of (w_{n})_{n≥0} is as follows:

w_{n} = cλ^{n} + O((λμ)^{\frac{n}{2}} )  (6)

where c is a constant. The λ defined by Lemma 4 is the dominant root of φ(x), and μ is the largest absolute value of the remaining m − 1 roots of φ(x).

Proof. By [11], there exist unique nonzero polynomials ℓ_{1}, …, ℓ_{l} ∈ ℚ({λ_{i}}_{i=1}^{l}) [x], with deg ℓ_{i} ≤ m_{i} − 1 (where m_{i} is the multiplicity of λ_{i} as a root of the characteristic polynomial φ(x)) for 1 ≤ i ≤ l, such that

w_{n} = ℓ_{1}(n)λ_{1}^{n} + ⋯ + ℓ_{l}(n)λ_{l}^{n}, for all n.

By Lemma 4, we get

w_{n} = cλ^{n} + ∑_{i=2}^{l} ℓ_{i}(n)λ_{i}^{n}, ℓ_{i}(n) ∈ ℝ[n],

where λ is the dominant root of φ(x), c is a nonzero constant, and

deg ℓ_{i}(n) ≤ m_{i} − 1, for i = 2, …, l, m_{2} + ⋯ + m_{l} = m − 1,

Therefore, we obtain that

w_{n} = cλ^{n} + O(n^{m}μ^{n}) = cλ^{n} + O((λμ)^{\frac{n}{2}} ),

where n^{m} < |\frac{λ}{μ} |^{\frac{n}{2}} for all sufficiently large n. Remark 6. For discussing the reciprocal product of mth-order linear recursive sequences, we assume that μ > 1. When μ < 1, as in [3], we write w_{n} = cλ^{n} + O(c^{−n}), for some c > 1). Lemma 7. Let λ > |μ| > 1. We get

∏_{k=n}^{∞}(1 − \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{k}{2}} )) = 1 − ∑_{k=n}^{∞} \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} ),  (7)

where c is a constant. Proof. First we prove the following equality:

∏_{k=n}^{n+m}(1 − \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{k}{2}} )) = 1 − ∑_{k=n}^{n+m} \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} ),  (8)

We prove (8) by mathematical induction. When m = 1, we have

∏_{k=n}^{n+1}(1 − \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{k}{2}} ))

= (1 − \frac{1}{cλ^{n}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} )) × (1 − \frac{1}{cλ^{n+1}} + O((\frac{μ}{λ^{3}} )^{\frac{n+1}{2}} ))

= 1 − \frac{1}{cλ^{n}} − \frac{1}{cλ^{n+1}} + \frac{1}{c^{2}λ^{2n+1}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} )

+ O((\frac{μ}{λ^{3}} )^{\frac{n+1}{2}} ) + O((\frac{μ}{λ^{5}} )^{\frac{n}{2}} ) + O((\frac{μ}{λ^{3}} )^{\frac{2n+1}{2}} )

= 1 − \frac{1}{cλ^{n}} − \frac{1}{cλ^{n+1}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} )

= 1 − ∑_{k=n}^{n+1} \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} ).

That is, (8) holds when m = 1. Now suppose that for every integer m, we have

∏_{k=n}^{n+m}(1 − \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{k}{2}} )) = 1 − ∑_{k=n}^{n+m} \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} ).

Then for m + 1, we have

\begin{matrix}∏_{k=n}^{n+m+1}(1 − \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{k}{2}} )) \\ = (1 − ∑_{k=n}^{n+m} \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} )) × (1 − \frac{1}{cλ^{n+m+1}} + O((\frac{μ}{λ^{3}} )^{\frac{n+m+1}{2}} )) \\ = 1 − ∑_{k=n}^{n+m} \frac{1}{cλ^{k}} − \frac{1}{cλ^{n+m+1}} + \frac{1}{cλ^{n+m+1}} (∑_{k=n}^{n+m} \frac{1}{cλ^{k}}) + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} ) \\ + O((\frac{μ}{λ^{3}} )^{\frac{n+m+1}{2}} ) + O((\frac{μ}{λ^{5}} )^{\frac{n}{2}} ) + O((\frac{μ}{λ^{3}} )^{\frac{2n+m+1}{2}} ) \\ = 1 − ∑_{k=n}^{n+m+1} \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} ). \end{matrix}  (9)

Now (9) follows from (9) and mathematical induction. Taking m → ∞, we have

∏_{k=n}^{∞}(1 − \frac{1}{cλ^{k}} + O(\frac{μ}{λ^{3}} )^{\frac{k}{2}} ) = 1 − ∑_{k=n}^{∞} \frac{1}{cλ^{k}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} ),

which completes the proof.

3 Proof of the theorem

Proof of Theorem 1. Using the geometric series, we get

\frac{1}{1 ± ϵ} = 1 + O(ϵ),  (10)

where ϵ → 0. Using Lemma 5, we obtain

\begin{aligned}\frac{1}{w_{k}} & = \frac{1}{cλ^{k} + O((λμ)^{\frac{k}{2}} )} = \frac{1}{cλ^{k}(1 + O((\frac{μ}{λ})^{\frac{k}{2}} ))} \\ & = \frac{1}{cλ^{k}} (1 + O(\frac{μ}{λ})^{\frac{k}{2}} ) = \frac{1}{cλ^{k}} + O(\frac{μ}{λ^{3}} )^{\frac{k}{2}} . \end{aligned}  (11)

Thus

∑_{k=n}^{∞} \frac{1}{w_{k}} = \frac{1}{c} ∑_{k=n}^{∞} \frac{1}{λ^{k}} + O(∑_{k=n}^{∞}(\frac{μ}{λ^{3}} )^{\frac{k}{2}} ) = \frac{λ}{c(λ − 1)λ^{n}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} ).

Taking the reciprocal, we get

\begin{aligned}(∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1} & = \frac{1}{\frac{λ}{c(λ−1)λ^{n}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} )} = \frac{1}{\frac{λ}{c(λ−1)λ^{n}} (1 + O((\frac{μ}{λ})^{\frac{n}{2}} ))} \\ & = \frac{c(λ − 1)λ^{n}}{λ} (1 + O((\frac{μ}{λ})^{\frac{n}{2}} )) = (cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} )), \end{aligned}

which yields that

\begin{aligned}\frac{( ∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1}}{w_{n} − w_{n−1}} & = \frac{(cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} ))}{(cλ^{n} + O((λμ)^{\frac{n}{2}} )) − (cλ^{n−1} + O((λμ)^{\frac{n−1}{2}} ))} \\ & = \frac{(cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} ))}{(cλ^{n} − cλ^{n−1})(1 + O(\frac{(λμ)^{\frac{n}{2}}}{λ^{n}−λ^{n−1}} ) + O(^{\frac{(λμ)^{\frac{n−1}{2}}}{λ^{n}−λ^{n−1}}} ))} \\ & = \frac{(cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} ))}{(cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} ) + O((\frac{μ}{λ})^{\frac{n−1}{2}} ))}. \end{aligned}  (12)

We obtain that

\frac{( ∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1}}{w_{n} − w_{n−1}} tends to 1, as n → ∞.

In addition, by Lemma 7 and identity (11), we obtain

\begin{matrix}∏_{k=n}^{∞}(1 − \frac{1}{w_{k}}) = ∏_{k=n}^{∞}(1 − \frac{1}{cλ^{k}} + O(\frac{μ}{λ^{3}} )^{\frac{k}{2}} ) \\ = 1 − ∑_{k=n}^{∞} \frac{1}{cλ^{k}} + O(\frac{μ}{λ^{3}} )^{n} = 1 − \frac{λ}{c(λ − 1)λ^{n}} + O(\frac{μ}{λ^{3}} )^{n}. \end{matrix}

Taking the reciprocal, we get

\begin{matrix}(1 − ∏_{k=n}^{∞}(1 − \frac{1}{w_{k}}))^{−1} = \frac{1}{\frac{λ}{c(λ−1)λ^{n}} + O((\frac{μ}{λ^{3}} )^{\frac{n}{2}} )} = \frac{1}{\frac{λ}{c(λ−1)λ^{n}} (1 + O((\frac{μ}{λ})^{\frac{n}{2}} ))} \\ = \frac{c(λ − 1)λ^{n}}{λ} (1 + O((\frac{μ}{λ})^{\frac{n}{2}} )) = (cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} )), \end{matrix}

which yields that

\begin{aligned}\frac{(1 − ∏_{k=n}^{∞}(1 − \frac{1}{w_{k}}))^{−1}}{w_{n} − w_{n−1}} & = \frac{(cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} ))}{(cλ^{n} + O((λμ)^{\frac{n}{2}} )) − (cλ^{n−1} + O((λμ)^{\frac{n−1}{2}} ))} \\ & = \frac{(cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} ))}{(cλ^{n} − cλ^{n−1})(1 + O(\frac{(λμ)^{\frac{n}{2}}}{λ^{n}−λ^{n−1}} ) + O(^{\frac{(λμ)^{\frac{n−1}{2}}}{λ^{n}−λ^{n−1}}} ))} \\ & = \frac{(cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} ))}{(cλ^{n} − cλ^{n−1})(1 + O((\frac{μ}{λ})^{\frac{n}{2}} ) + O((\frac{μ}{λ})^{\frac{n−1}{2}} ))}. \end{aligned}  (13)

We obtain that

\frac{(1 − ∏_{k=n}^{∞}(1 − \frac{1}{w_{k}}))^{−1}}{w_{n} − w_{n−1}} tends to 1, as n → ∞,

which completes the proof.

Remark 8. We now discuss the relative error of the asymptotic behavior of the result in Theorem 1. By identities (12) and (13), we obtain

\begin{aligned}|\frac{( ∑_{k=n}^{∞} \frac{1}{w_{k}})^{−1}}{w_{n} − w_{n−1}} − 1| & = |\frac{1 + O((\frac{μ}{λ})^{\frac{n}{2}} )}{1 + O((\frac{μ}{λ})^{\frac{n}{2}} ) + O((\frac{μ}{λ})^{\frac{n−1}{2}} )} − 1| \\ & = O((\frac{μ}{λ})^{\frac{n−1}{2}} ) + O((\frac{μ}{λ})^{\frac{2n−1}{2}} ) = O((\frac{μ}{λ})^{\frac{n−1}{2}} ), \end{aligned}  (14)

and

\begin{aligned}|\frac{(1 − ∏_{k=n}^{∞}(1 − \frac{1}{w_{k}}))^{−1}}{w_{n} − w_{n−1}} − 1| & = |\frac{1 + O((\frac{μ}{λ})^{\frac{n}{2}} )}{1 + O((\frac{μ}{λ})^{\frac{n}{2}} ) + O((\frac{μ}{λ})^{\frac{n−1}{2}} )} − 1| \\ & = O((\frac{μ}{λ})^{\frac{n−1}{2}} ) + O((\frac{μ}{λ})^{\frac{2n−1}{2}} ) = O((\frac{μ}{λ})^{\frac{n−1}{2}} ). \end{aligned}  (15)

When n = 101, by the identities (14) and (15), we can determine the magnitude of the relative error of the asymptotic behavior for the following two sequences (w_{n})_{n_{1}≥0} and (w_{n})_{n_{2}≥0}:

w_{n_{1}} = 2w_{n−1} + 3w_{n−2} + 7w_{n−3} + w_{n−4} + w_{n−5},

w_{n_{2}} = 3w_{n−1} + 5w_{n−2} + w_{n−3} + 6w_{n−4} + w_{n−5}.

\begin{aligned}w_{n} & λ & μ & |\frac{μ}{λ} |^{\frac{n−1}{2}} \\ w_{n_{1}} & ≈ 3.2421 & ≈ −1.1509 & ≈ 3.2401 × 10^{−23} \\ w_{n_{2}} & ≈ 4.2965 & ≈ −1.4946 & ≈ 1.1762 × 10^{−23} \end{aligned}

Table 1: The higher-order linear recurrences.

The computations are given in Table 1. We used the software Mathematica.

4 Acknowledgments

The authors would like to thank the referees for their very helpful and detailed comments. This work is supported by the N. S. F. (12126357) of China.

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2020 Mathematics Subject Classification: Primary 11B39; Secondary 11B37. Keywords: mth-order linear recursive sequence, reciprocal sum, reciprocal product, characteristic polynomial.

Received October 25 2023; revised versions received October 27 2023; January 11 2024; January 17 2024. Published in Journal of Integer Sequences, January 17 2024.

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