Next Article in Journal
A New Class of 2q-Point Nonstationary Subdivision Schemes and Their Applications
Previous Article in Journal
A Fast Derivative-Free Iteration Scheme for Nonlinear Systems and Integral Equations

Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article

# On the Sum of Reciprocal of Polynomial Applied to Higher Order Recurrences

by
Pavel Trojovský
Department of Mathematics, Faculty of Science, University of Hradec Králové, Hradec Králové 500 03, Czech Republic
Mathematics 2019, 7(7), 638; https://doi.org/10.3390/math7070638
Submission received: 24 June 2019 / Revised: 12 July 2019 / Accepted: 15 July 2019 / Published: 18 July 2019

## Abstract

:
Recently a lot of papers have been devoted to partial infinite reciprocal sums of a higher-order linear recursive sequence. In this paper, we continue this program by finding a sequence which is asymptotically equivalent to partial infinite sums, including a reciprocal of polynomial applied to linear higher order recurrences.
MSC:
11B39; 11C08

## 1. Introduction

A sequence $( u n ) n$ is a linear recurrence sequence with coefficients $c 0 , c 1 , … , c s − 1$, where $c 0 ≠ 0$ if:
$u n + s = c s − 1 u n + s − 1 + ⋯ + c 1 u n + 1 + c 0 u n ,$
for all non-negative integers n. A recurrence sequence is therefore completely determined by the initial values $u 0 , … , u s − 1 ∈ C$, and by the coefficients $c 0 , c 1 , … , c s − 1 ∈ C$. The integer s is called the ’order’ of the linear recurrence. The characteristic polynomial of the sequence $( u n ) n ≥ 0$ is given by:
$ψ ( x ) = x s − c s − 1 x s − 1 − ⋯ − c 1 x − c 0 = ( x − α 1 ) m 1 ⋯ ( x − α ℓ ) m ℓ ,$
where the $α j$ (which are distinct) are called the ’roots’ of the recurrence. Moreover, the recurrence $( u n ) n$ has a ’dominant root’ if one of its roots has strictly largest absolute value. A fundamental result in the theory of recurrence sequences asserts that there exist uniquely determined non-zero polynomials $g 1 , … , g ℓ ∈ Q ( { α j } j = 1 ℓ ) [ x ]$, with $deg g j ≤ m j − 1$ ($m j$ is the multiplicity of $α j$ as root of $ψ ( x )$), for $j = 1 , … , ℓ$, such that:
$u n = g 1 ( n ) α 1 n + ⋯ + g ℓ ( n ) α ℓ n , for all n .$
For more details, see [1] (Theorem C.1). The great importance of linear recurrence relations in practice can be seen, for example, from recently published papers [2,3,4,5].
The Lucas sequence $( U n ) n ≥ 0$ given by $U n + 2 = a U n + 1 + b U n$, for $n ≥ 0$, where $U 0 = 0$, $U 1 = 1$ and the values a and b are previously fixed, is an example of a linear recurrence of order 2 (also called ’binary’). For instance, if $a = b = 1$, then $( U n ) n ≥ 0 = ( F n ) n ≥ 0$ is the well-known Fibonacci sequence. The Fibonacci numbers are known for their amazing properties (see [6] for the history, properties and rich applications of the Fibonacci sequence and some of its generalizations).
In 2008, Ohtsuka and Nakamura [7] studied the partial infinite sums of reciprocal Fibonacci numbers and proved that:
$∑ k = n ∞ 1 F k − 1 = F n − 2 , if n ≡ 0 ( mod 2 ) , n ≥ 2 ; F n − 2 − 1 , if n ≡ 1 ( mod 2 ) , n ≥ 1 .$
When $a = 2$ and $b = 1$, we have another important sequence, namely, the sequence of the Pell numbers$( P n ) n$. In 2012, Wenpeng and Tingting [8] proved that:
$∑ k = n ∞ 1 P k − 1 = P n + P n − 2 , if n ≡ 0 ( mod 2 ) , n ≥ 2 ; P n − 1 + P n − 2 − 1 , if n ≡ 1 ( mod 2 ) , n ≥ 1 .$
In 2011, Holliday and Komatsu [9] obtained the infinite sum of the reciprocal of the generalized Fibonacci sequence ${ U n }$, when $a = p$ and $b = 1$
$∑ k = n ∞ 1 U k − 1 = U n − U n − 1 , if n ≡ 0 ( mod 2 ) , n ≥ 2 ; U n − 1 − U n − 1 − 1 , if n ≡ 1 ( mod 2 ) , n ≥ 1 .$
In 2016, Basbük and Yazlik [10] studied generalized bi-periodic Fibonacci numbers, defined by Edson and Yayenie [11] in the following manner:
$q n = a q n − 1 + q n − 2 , if n ≡ 0 ( mod 2 ) , n ≥ 2 ; b q n − 1 + q n − 2 , if n ≡ 1 ( mod 2 ) , n ≥ 2 ,$
with $q 0 = 0$ and $q 1 = 1$. They derived the following partial infinite sum of reciprocal for this sequence:
$∑ k = n ∞ a b Ψ ( k ) 1 q k − 1 = q n − q n − 1 , if n ≡ 0 ( mod 2 ) , n ≥ 2 ; q n − 1 − q n − 1 − 1 , if n ≡ 1 ( mod 2 ) , n ≥ 1 ,$
where $Ψ ( k ) = s ( k + 1 ) − s ( n + 1 ) − ( − 1 ) n ⌊ k − n 2 ⌋$, with:
$s ( n ) = 0 , if n ≡ 0 ( mod 2 ) ; 1 , if n ≡ 1 ( mod 2 ) .$
Choi and Choo [12] derived the formulas for the following sums of reciprocals of products of Fibonacci and Lucas numbers:
$∑ k = n ∞ 1 F k L k + m − 1 and ∑ k = n ∞ 1 L k F k + m − 1 .$
Choi and Choo [13] proceeded in study of sequence from [10] and found a formula for the following sum of reciprocals of the squares of generalized Fibonacci numbers:
$∑ k = n ∞ a b 1 − Ψ ( k ) 1 q k 2 − 1 .$
We point out to the reader the series of papers [14,15,16,17,18,19,20,21], where the authors deal, in particular, with partial sums related to the omnipresent Riemann zeta function $ζ ( s )$. For instance, Xin [17] proved that:
$∑ k = n ∞ 1 k 2 − 1 = n − 1 ,$
where n is any positive integer.
In 2013, Kiliç and Arikan [22] studied a problem which differs slightly from the one in [7], namely, they determined the nearest integer to $( ∑ k = n ∞ ( 1 / u k ) ) − 1$. Specifically, suppose that $∥ x ∥ = ⌊ x + 1 / 2 ⌋$ (the nearest integer formula) and let $( u n ) n$ be an integer sequence satisfying the recurrence formula:
$u n = p u n − 1 + q u n − 2 + u n − 3 + ⋯ + u n − k ,$
for any positive integer $p ≥ q$ and $n ≥ k$. Then, they proved the existence of a positive integer $n 0$ such that:
$∑ k = n ∞ 1 u k − 1 = u n − u n − 1 ,$
for all $n ≥ n 0$.
There are many generalizations of this result (see, for example, [15] and references therein). In all these cases, the authors considered sequences whose characteristic polynomial has only one root outside the closed unit disc and all the other roots lie inside this disc. To ensure that, they considered some particular recurrences and proved that this property holds by using a calculus approach to their characteristic polynomial.
Here we are interested in a huge class of recurrences. For this reason, we shall look at the problem from another viewpoint. Specifically, we recall that two sequences $( u n ) n$ and $( v n ) n$ are called ’asymptotically equivalent’ if $u n / v n$ tends to 1 as $n → ∞$ (we write $u n ∼ v n$).
In this paper, we shall prove the following result:
Theorem 1.
Let $( u n ) n$ be a linear recurrence having a simple dominant root. Let $P ( z ) ∈ C [ z ]$ be a non-constant polynomial. Then, the sequences:
are asymptotically equivalent.
Moreover, in the final section, we shall make some computations of ‘relative error of this asymptotic behavior’, which will be related to some special forms of recurrences $( u n ) n$. The calculations performed in this paper took several minutes using software Mathematica on a 2.5 GHz Intel Core i5 4GB Mac OSX.

## 2. Proof of the Theorem

In what follows, we shall use the usual notation O which denotes the ‘big-oh’ Landau symbol.
If $α 1 , … , α ℓ$ are the roots of $( u n ) n$ with $| α 1 | > | α 2 | ≥ | α 3 | ≥ ⋯ ≥ | α ℓ |$, we then write $α : = α 1$ and $β : = α 2$. By hypothesis, $α$ is a simple root of the function $ψ$ and so the polynomial $g 1 ( n )$ is a constant, say a. By the formula in Equation (2), we have that:
$u k = a α k + O ( k s β k ) .$
We may suppose that $| α | > 1$ (otherwise $u k$ tends to 0 as $k → ∞$ and the result clearly holds). Further, we assume that $| β | ≥ 1$ (if not, all the other roots of the recurrence would lie inside the open unit disc and the Kiliç-Arikan method could be used, since we can write $u k = a α k + O ( c − k )$, for some $c > 1$). Thus, we have:
$u k = a α k ( 1 + O ( ( β / α ) k / 2 ) ) ,$
where we used that $| α / β | k / 2 > k s$ for all sufficiently large k. Suppose, without loss of generality, that $P ( z ) = z m + ∑ j = 0 m − 1 b j z j$. Since:
$u k j = a j α k j ( 1 + O ( ( β / α ) k / 2 ) ) ,$
we obtain:
$P ( u k ) = P ( a α k ) ( 1 + O ( ( β / α ) k / 2 ) ) .$
Now, we shall use that $1 / ( 1 ± ϵ ) = 1 + O ( ϵ )$ (for all $ϵ$ sufficiently small) to write:
$1 P ( u k ) = 1 P ( a α k ) ( 1 + O ( ( β / α ) k / 2 ) ) .$
Thus:
$∑ k = n ∞ 1 P ( u k ) = ∑ k = n ∞ 1 P ( a α k ) + O α − n m ( β / α ) n / 2 ,$
where we used that $| P ( a α k ) | > | a m | | α k m | / 2$, for all sufficiently large k (when $| α | > 1$, for the case $| α | = 1$, the estimate in the previous formula is still valid). In addition, we use that $P ( a α k ) = a m α k m ( 1 + O ( α − k )$) to get:
$1 P ( a α k ) = 1 a m α k m ( 1 + O ( α − k ) ) = 1 a m α k m + O ( α − k ( m + 1 ) ) .$
So, by using the geometric series, we arrive at:
$∑ k = n ∞ 1 P ( a α k ) = α m a m α m n ( α m − 1 ) + O ( α − n ( m + 1 ) ) .$
Therefore:
$∑ k = n ∞ 1 P ( u k ) = α m a m α m n ( α m − 1 ) ( 1 + O ( ( β / α ) n / 2 ) )$
and so:
$∑ k = n ∞ 1 P ( u k ) − 1 = a m α m n ( α m − 1 ) α m ( 1 + O ( ( β / α ) n / 2 ) ) = ( a m ( α n ) m − a m α ( n − 1 ) m ) ( 1 + O ( ( β / α ) n / 2 ) ) .$
This implies that, if $A n : = ( ∑ k = n ∞ 1 / P ( u k ) ) − 1$, then:
$A n / ( a m ( α n ) m − a m ( α n − 1 ) m ) tends to 1 as n → ∞ .$
On the other hand:
$P ( u k ) = P ( a α k ) ( 1 + O ( ( β / α ) k / 2 ) ) = a m α k m ( 1 + O ( α − k ) ) ( 1 + O ( ( β / α ) k / 2 ) ) = a m α k m + α k m O ( ν k ) , = a m α k m + α k m O ( ν k − 1 ) ,$
where $ν : = max { | β / α | 1 / 2 , | α | − 1 }$. Hence:
$P ( u n ) − P ( u n − 1 ) = ( a m α m n − a m α m ( n − 1 ) ) ( 1 + O ( ν n − 1 ) )$
which yields that:
as $n → ∞$. So, by combining the previous facts, we conclude that $A n / ( P ( u n ) − P ( u n − 1 ) )$ tends to 1 as $n → ∞$ and then $( A n ) n ∼ ( P ( u n ) − P ( u n − 1 ) ) n$ which completes the proof. □

## 3. Computational Aspects

We will now discuss ‘relative error of an asymptotic behavior of our result in Theorem 1’. Using Identities (3) and (4) we get
$∑ k = n ∞ ( P ( u k ) ) − 1 − 1 P ( u n ) − P ( u n − 1 ) − 1 = 1 + O ( ( β / α ) n / 2 ) 1 + O ( ν n − 1 ) − 1 = O ( ν n − 1 ) .$
Example 1.
Determine the dominant root α and the second dominant root β as well as the magnitude of the relative error of asymptotic equivalence by Identity (5) for $n = 100$, for the following two sequences $( u n ) n$ defined by linear recurrences of the fourth-order:
$u n = 2 u n − 1 + 7 u n − 2 + u n − 3 + u n − 4 , u n = 2 u n − 1 + 7 u n − 2 + 5 u n − 3 + u n − 4 .$
Requested computations are in Table 1. We used software Mathematica.
Example 2.
Let $( u n ) n$ be the sequence defined by the fourth-order linear recurrence:
$u n = u n − 1 + 2 u n − 2 + u n − 3 + u n − 4 ,$
with initial values $u 1 = u 2 = u 3 = 1$ and $u 4 = 2$. Let $P ( z ) = z 2 + z + 1$, $A n : = ∑ k = n ∞ 1 / P ( u k )$ and $B n : = P ( u n ) − P ( u n − 1 )$. Examine the behavior of values $A n / B n$, the magnitude of the relative error of asymptotic equivalence given by Identity (5), for $n = 10 , 100 , 500$.
Using software Wolfram Mathematica, we can find that the two biggest roots (in the absolute value) of the characteristic polynomial of $( u n ) n$ are $α = 2.2026$ and $β = − 1$. Table 2 shows the results.

## 4. A Generalization for Higher Dimensional Recurrences

In this section, we shall point out what can be done for partial infinite reciprocal sums of higher dimensional recurrences $( u → n ) n$, with $u → n ∈ C d$, where d is a positive integer. As the proof is very similar to the proof of Theorem 1 in Section 2, we shall only remark on the main points of the proof in order to avoid many technical details. Suppose that we have the recurrence relation:
$u → n + s = c s − 1 u → n + s − 1 + ⋯ + c 0 u → n ,$
with the initial values $u → 0 , … , u → s − 1 ∈ C d$. We can write the sequence $( u → n ) n$ in the following form:
$( u → n ) n = ( ( u n ( 1 ) ) n , ( u n ( 2 ) ) n , … , ( u n ( d ) ) n ) ,$
thus as a d-tuple of complex linear recurrences (or a $2 d$-tuple by considering their real and imaginary parts). Note that each sequence $( u n ( i ) ) n$, where $i ∈ { 1 , 2 , … , d }$, satisfies the same linear recurrence relation as the sequence $( u → n ) n$ (the initial values may differ). Hence, the characteristic polynomial of each of them is $ψ ( x ) = x s − c s − 1 x s − 1 − ⋯ − c 1 − c 0$, with roots $α 1 , … , α ℓ$. If we suppose that this polynomial has a dominant root, then we can write (for all $1 ≤ i ≤ d$):
$u n ( i ) = a i α 1 n + O ( n s β k ) ,$
where $a i ∈ C$ and $| α | > | β | = max 2 ≤ j ≤ ℓ { | α j | }$. Therefore, if $u n ( i ) = ( x n ( i ) , y n ( i ) ) ∈ R 2$, we have:
$u → n = ∑ i = 1 d x n ( i ) e 2 i − 1 + ∑ i = 1 d y n ( i ) e 2 i = α n v → + O ( n s β n ) w → ,$
where $v → = ( a 1 , … , a d ) , w → = ( 1 , … , 1 )$ and ${ e 1 , … , e 2 d }$ is the standard basis of $R 2 d ≅ C d$. The validity of Identity (6) follows from the fact that $x n ( i )$ and $y n ( i )$ have the same asymptotic order (thus, $x n ( i ) ≍ y n ( i )$). Since $∥ w ∥ = 2 d = O ( 1 )$, we can simply mimic the proof of Theorem 1 to obtain the following result (with notation as above):
Theorem 2.
Let $( u → n ) n$ be a linear recurrence in $C d$ with a simple dominant root (i.e., $ψ ( x )$ has a simple dominant root). Let $P ( z 1 , … , z d ) ∈ C [ z 1 , … , z d ]$ be a polynomial with a degree of at least one in each variable $z i$ ($1 ≤ i ≤ d$). Then, the sequences:
$∑ k = n ∞ 1 P ( u → k ) − 1 n a n d ( P ( u → n ) − P ( u → n − 1 ) ) n$
are asymptotically equivalent (here, as usual, $P ( u → k )$ means $P ( u n ( 1 ) , … , u n ( d ) )$).

## 5. Conclusions

In this paper, we found a sequence which is asymptotically equivalent to partial infinite sums, including a reciprocal of polynomial applied to linear higher order recurrences with a simple dominant root. Further, we included in this article a possible generalization of our main result to a higher dimensional recurrence, which the reviewer pointed out to us.

## Funding

The author was supported by Project of Excelence PrF UHK, University of Hradec Králové, Czech Republic 01/2019.

## Acknowledgments

The author thanks the anonymous referees for their careful corrections and very helpful and detailed comments, which have significantly improved the presentation of this paper.

## Conflicts of Interest

The author declares no conflict of interest.

## References

1. Shorey, T.N.; Tijdeman, R. Exponential Diophantine Equations; Cambridge Tracts in Mathematics 87; Cambridge University Press: Cambridge, UK, 1986. [Google Scholar]
2. Bednařík, D.; Freitas, G.; Marques, D.; Trojovský, P. On the sum of squares of consecutive k-bonacci numbers which are l-bonacci numbers. Colloq. Math. 2019, 156, 153–164. [Google Scholar] [CrossRef]
3. Brison, O.; Nogueira, J.E. Linear recurring sequence subgroups in the complex field—II. Fibonacci Quart. 2019, 57, 148–154. [Google Scholar]
4. Buse, C.; O’Regan, D.; Saierli, O. Hyers-Ulam Stability for Linear Differences with Time Dependent and Periodic Coefficients. Symmetry 2019, 11, 512. [Google Scholar] [CrossRef]
5. Costabile, F.; Gualtieri, M.; Napoli, A. Recurrence relations and determinant forms for general polynomial sequences. Application to Genocchi polynomials. Integral Transforms Spec. Funct. 2019, 30, 112–127. [Google Scholar] [CrossRef]
6. Koshy, T. Fibonacci and Lucas Numbers with Applications; Wiley: New York, NY, USA, 2001. [Google Scholar]
7. Ohtsuka, H.; Nakamura, S. On the sum of reciprocal Fibonacci numbers. Fibonacci Quart. 2008, 46–47, 153–159. [Google Scholar]
8. Zhang, W.; Wang, T. The infinite sum of reciprocal Pell numbers. Appl. Math. Comput. 2012, 218, 6164–6167. [Google Scholar]
9. Holliday, S.; Komatsu, T. On the sum of reciprocal generalized Fibonacci numbers. Integers 2011, 11, 441–455. [Google Scholar] [CrossRef]
10. Basbük, M.; Yazlik, Y. On the sum of reciprocal of generalized bi-periodic Fibonacci numbers. Miskolc Math. Notes 2016, 17, 35–41. [Google Scholar] [CrossRef]
11. Edson, M.; Yayenie, O. A new generalization of Fibonacci sequences and extended Binet‘s formula. Integers 2009, 9, 639–654. [Google Scholar] [CrossRef]
12. Choi, G.; Choo, Y. On the Reciprocal Sums of Products of Fibonacci and Lucas Numbers. Filomat 2018, 32, 2911–2920. [Google Scholar] [CrossRef]
13. Choi, G.; Choo, Y. On the Reciprocal Sums of Square of Generalized Bi-Periodic Fibonacci Numbers. Miskolc Math. Notes 2018, 19, 201–209. [Google Scholar] [CrossRef]
14. Wu, Z.; Zhang, H. On the reciprocal sums of higher-order sequences. Adv. Differ. Equ. 2013, 2013, 189. [Google Scholar] [CrossRef] [Green Version]
15. Wu, Z.; Zhang, J. On the Higher Power Sums of Reciprocal Higher-Order Sequences. Sci. World J. 2014, 2014, 521358. [Google Scholar] [CrossRef] [PubMed]
16. Wu, Z. Several identities relating to Riemann zeta-Function. Bull. Math. Soc. Sci. Math. Roum. 2016, 59, 285–294. [Google Scholar]
17. Xin, L. Some identities related to Riemann zeta-function. J. Inequal. Appl. 2016, 2016, 32. [Google Scholar] [CrossRef]
18. Xin, L. Partial reciprocal sums of the Mathieu series. J. Inequal. Appl. 2017, 2017, 60. [Google Scholar]
19. Lin, X.; Li, X. A reciprocal sum related to the Riemann zeta-function. J. Math. Inequal. 2017, 11, 209–215. [Google Scholar]
20. Xu, H. Some computational formulas related the Riemann zeta-function tails. J. Inequal. Appl. 2016, 2016, 132. [Google Scholar] [CrossRef]
21. Zhang, H.; Wu, Z. On the reciprocal sums of the generalized Fibonacci sequences. Adv. Diff. Equ. 2013, 2013, 377. [Google Scholar] [CrossRef] [Green Version]
22. Kiliç, E.; Arikan, T. More on the infinite sum of reciprocal Fibonacci, Pell and higher order recurrences. Appl. Math. Comput. 2016, 219, 7783–7788. [Google Scholar]
Table 1. Fourth order linear recurrences.
Table 1. Fourth order linear recurrences.
$u n$$α$$β$$| ν | n − 1$
$u n = 2 u n − 1 + 7 u n − 2 + u n − 3 + u n − 4$$3.8851$$1.7845$$1.8847 × 10 − 17$
$u n = 2 u n − 1 + 7 u n − 2 + 5 u n − 3 + u n − 4$$4.0489$$− 1$$8.6442 × 10 − 31$
Table 2. Values $A n / B n$ the magnitude of the relative error of asymptotic equivalence.
Table 2. Values $A n / B n$ the magnitude of the relative error of asymptotic equivalence.
n$A n$$B n$$A n / B n$$| ν | n − 1$
10≈66,125.65$65 , 940$$1.0028$$0.01916$
100$≈ 4.5 × 10 66$$≈ 4.5 × 10 66$$1 + 3 × 10 − 34$$6.6 × 10 − 18$
500$≈ 2.9 × 10 341$$≈ 2.9 × 10 341$$1 + 10 − 100$$1.3 × 10 − 86$

## Share and Cite

MDPI and ACS Style

Trojovský, P. On the Sum of Reciprocal of Polynomial Applied to Higher Order Recurrences. Mathematics 2019, 7, 638. https://doi.org/10.3390/math7070638

AMA Style

Trojovský P. On the Sum of Reciprocal of Polynomial Applied to Higher Order Recurrences. Mathematics. 2019; 7(7):638. https://doi.org/10.3390/math7070638

Chicago/Turabian Style

Trojovský, Pavel. 2019. "On the Sum of Reciprocal of Polynomial Applied to Higher Order Recurrences" Mathematics 7, no. 7: 638. https://doi.org/10.3390/math7070638

Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.