Hindawi Publishing Corporation e Scientific World Journal Volume 2014, Article ID 521358, 6 pages http://dx.doi.org/10.1155/2014/521358
Research Article On the Higher Power Sums of Reciprocal Higher-Order Sequences Zhengang Wu1 and Jin Zhang2 1 2
Department of Mathematics, Northwest University, Xi’an, Shaanxi, China School of Mathematics and Computer Engineering, Xi’an University of Arts and Science, Xi’an, Shaanxi, China
Correspondence should be addressed to Zhengang Wu; [email protected] Received 21 November 2013; Accepted 29 January 2014; Published 10 March 2014 Academic Editors: A. Fiorenza and E. Kılıc¸ Copyright © 2014 Z. Wu and J. Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Let {𝑢𝑛 } be a higher-order linear recursive sequence. In this paper, we use the properties of error estimation and the analytic method to study the reciprocal sums of higher power of higher-order sequences. Then we establish several new and interesting identities relating to the infinite and finite sums.
1. Introduction The so-called Fibonacci zeta function and Lucas zeta function, defined by ∞
1 , 𝑠 𝐹 𝑛=1 𝑛
𝜁𝐹 (𝑠) = ∑
∞
1 , 𝑠 𝐿 𝑛=1 𝑛
𝜁𝐿 (𝑠) = ∑
(1)
where the 𝐹𝑛 and 𝐿 𝑛 denote the Fibonacci numbers and Lucas numbers, have been considered in several different ways; see [1, 2]. Ohtsuka and Nakamura [3] studied the partial infinite sums of reciprocal Fibonacci numbers and proved the following conclusions: ∞
⌊( ∑
𝑘=𝑛
∞
−1
1 𝐹 if 𝑛 is even, 𝑛 ≥ 2, ) ⌋ = { 𝑛−2 𝐹𝑘 𝐹𝑛−2 − 1 if 𝑛 is odd, 𝑛 ≥ 1, −1
(2)
1 𝐹 𝐹 − 1 if 𝑛 is even, 𝑛 ≥ 2, ⌊( ∑ 2 ) ⌋ = { 𝑛−1 𝑛 𝐹 𝐹𝑛−1 𝐹𝑛 if 𝑛 is odd, 𝑛 ≥ 1, 𝑘 𝑘=𝑛 where ⌊⋅⌋ denotes the floor function. Further, Wu and Zhang [4, 5] generalized these identities to the Fibonacci polynomials and Lucas polynomials. Various properties of the Fibonacci polynomials and Lucas polynomials have been studied by many authors; see [6–13]. Recently, some authors considered the nearest integer of the sums of reciprocal Fibonacci numbers and other wellknown sequences and obtained several meaningful results;
see [14–16]. In particular, in [16], Kılıc¸ and Arıkan studied a problem which is a little different from that of [3], namely, −1 that of determining the nearest integer to (∑∞ 𝑘=𝑛 (1/V𝑘 )) . Specifically, suppose that ‖𝑥‖ = ⌊𝑥 + (1/2)⌋ (the nearest integer function) and {V𝑛 }𝑛≥0 is an integer sequence satisfying the recurrence formula V𝑛 = 𝑝V𝑛−1 + 𝑞V𝑛−2 + V𝑛−3 + ⋅ ⋅ ⋅ + V𝑛−𝑘 ,
(3)
for any positive integer 𝑝 ≥ 𝑞 and 𝑛 > 𝑘. Then we can conclude that there exists a positive integer 𝑛0 such that ∞ −1 ( ∑ 1 ) = V − V , 𝑛 𝑛−1 𝑘=𝑛 V𝑘
(4)
for all 𝑛 > 𝑛0 . In [17], Wu and Zhang unified the above results by proving the following conclusion that includes all the results, [3–8, 15, 16], as special cases. Proposition 1. For any positive integer 𝑛 > 𝑚, the 𝑚th-order linear recursive sequence {𝑢𝑛 } is defined as follows: 𝑢𝑛 = 𝑎1 𝑢𝑛−1 + 𝑎2 𝑢𝑛−2 + ⋅ ⋅ ⋅ + 𝑎𝑚−1 𝑢𝑛−𝑚+1 + 𝑎𝑚 𝑢𝑛−𝑚 ,
(5)
with initial values 𝑢𝑖 ∈ N for 0 ≤ 𝑖 < 𝑚 and at least one of them not being zero. For any positive real number 𝛽 > 2 and any
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positive integer 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 1, there exists a positive integer 𝑛1 such that ⌊𝛽𝑛⌋ −1 1 ) = 𝑢𝑛 − 𝑢𝑛−1 , ( ∑ 𝑘=𝑛 𝑢𝑘
(𝑛 ≥ 𝑛1 ) .
(6)
In particular, taking 𝛽 → +∞, there exists a positive integer 𝑛2 such that ∞ −1 ( ∑ 1 ) = 𝑢 − 𝑢 , 𝑛 𝑛−1 𝑘=𝑛 𝑢𝑘
(𝑛 ≥ 𝑛2 ) .
(7)
Corollary 4. Let {𝑢𝑛 } be an 𝑚th-order sequence defined by (5) with the restrictions 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N and 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 2. For positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋, where 𝛼, 𝛼1 , . . . , 𝛼𝑚−1 are the roots of the characteristic equation of 𝑢𝑛 and 𝑑−1 = max{|𝛼1 |, |𝛼2 |, . . . , |𝛼𝑚−1 |}, then there exists a positive integer 𝑛4 such that ∞ 𝑠𝑘 −1 𝑠 𝑠 ( ∑ 𝑎1 ) − ( 𝑢𝑛 − 𝑢𝑛−1 ) = 0, 𝑠 𝑠𝑛−𝑠 𝑠𝑛 𝑎1 𝑎1 𝑘=𝑛 𝑢𝑘
Proposition 2. For any positive integer 𝑛 ≥ 1, we have the identity ∞
⌊( ∑
𝑘=𝑛
{ { { { { { ={ { { { { { {
−1
1 ) ⌋ 𝑃𝑘3
⌊𝛽𝑛⌋ −1 𝑎1𝑠𝑘 ) ( ∑ 𝑠 𝑢 𝑘=𝑛 𝑘
(11)
is an interesting open problem.
2. Several Lemmas To complete the proof of our theorem, we need two lemmas. Lemma 5. Let 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N with 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 1 and 𝑚 ∈ N with 𝑚 ≥ 2. Then for the polynomial 𝑓 (𝑥) = 𝑥𝑚 − 𝑎1 𝑥𝑚−1 − 𝑎2 𝑥𝑚−2 − ⋅ ⋅ ⋅ − 𝑎𝑚−1 𝑥 − 𝑎𝑚 ,
61 91 + + ⌊− 𝑃𝑛 − 𝑃𝑛−1 ⌋ 82 82 𝑖𝑓 𝑛 𝑖𝑠 𝑒V𝑒𝑛, 𝑛 ≥ 2, 61 91 2 + ⌊ 𝑃𝑛 + 𝑃𝑛−1 ⌋ 𝑃𝑛2 𝑃𝑛−1 + 3𝑃𝑛 𝑃𝑛−1 82 82 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑, 𝑛 ≥ 1.
𝑃𝑛2 𝑃𝑛−1
2 3𝑃𝑛 𝑃𝑛−1
(10)
For positive real number 1 < 𝛽 ≤ 2, whether there exits an identity for
−1
𝑠 It seems difficult to deal with (∑∞ for all 𝑘=𝑛 (1/𝑢𝑘 )) integers 𝑠 ≥ 2, because it is quite unclear a priori what the shape of the result might be. In [18], Xu and Wang applied the method of undetermined coefficients and constructed a number of delicate inequalities in order to study the infinite sum of the cubes of reciprocal Pell numbers and then obtained the following meaningful result.
(𝑛 ≥ 𝑛4 ) .
(12)
we have the following: (8)
(I) polynomial 𝑓(𝑥) has exactly one positive real zero 𝛼 with 𝑎1 < 𝛼 < 𝑎1 + 1; (II) other 𝑚 − 1 zeros of 𝑓(𝑥) lie within the unit circle in the complex plane. Proof. See Lemma 1 of [16].
To find and prove this result is a substantial achievement since such a complex formula would not be clear beforehand that a result would even be possible. However, there is no research considering the higher power (𝑠 > 2) of reciprocal sums of some recursive sequences. The main purpose of this paper is using the properties of error estimation and the analytic method to study the higher power of the reciprocal sums of {𝑢𝑛 } and obtain several new and interesting identities. The results are as follows. Theorem 3. Let {𝑢𝑛 } be an 𝑚th-order sequence defined by (5) with the restrictions 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N and 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 2. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋, where 𝛼, 𝛼1 , . . . , 𝛼𝑚−1 are the roots of the characteristic equation of 𝑢𝑛 and 𝑑−1 = max {|𝛼1 |, |𝛼2 |, . . . , |𝛼𝑚−1 |}, then there exists a positive integer 𝑛3 such that ⌊𝛽𝑛⌋ −1 𝑠 𝑢𝑛−1 𝑢𝑛𝑠 𝑎1𝑠𝑘 ( ∑ 𝑢𝑠 ) − ( 𝑎𝑠𝑛 − 𝑎𝑠𝑛−𝑠 ) = 0, 𝑘=𝑛 𝑘 1 1
(𝑛 ≥ 𝑛3 ) .
(9)
Taking 𝛽 → +∞, from Theorem 3 we may immediately deduce the following.
Lemma 6. Let 𝑚 ≥ 2 and {𝑢𝑛 }𝑛≥0 be an integer sequence satisfying the recurrence formula (5). Then for any positive integer 𝑠, we have 𝑢𝑛𝑠 = 𝑐𝑠 𝛼𝑠𝑛 + 𝑂 (𝛼𝑠𝑛−𝑛 𝑑−𝑛 )
(𝑛 → ∞) ,
(13)
where 𝑐 > 0, 𝑑 > 1, and 𝑎1 < 𝛼 < 𝑎1 + 1 is the positive real zero of 𝑓(𝑥). Proof. From Lemma 2 of [16], the closed formula of 𝑢𝑛 is given by 𝑢𝑛 = 𝑐𝛼𝑛 + 𝑂 (𝑑−𝑛 )
(𝑛 → ∞) ,
(14)
where 𝑐 > 0, 𝑑 > 1, 𝑎1 < 𝛼 < 𝑎1 + 1, and 𝛼 is the positive real zero of 𝑓(𝑥). Now we prove Lemma 6 by mathematical induction. From formula (14), we have 𝑢𝑛2 = 𝑐2 𝛼2𝑛 + 𝑂 (𝑑−2𝑛 ) + 𝑂 (𝛼𝑛 𝑑−𝑛 ) = 𝑐2 𝛼2𝑛 + 𝑂 (𝛼𝑛 𝑑−𝑛 ) . (15) That is, the lemma holds for 𝑠 = 2. Suppose that for all integers 2 ≤ 𝑠 ≤ 𝑘 we have 𝑢𝑛𝑠 = 𝑐𝑠 𝛼𝑠𝑛 + 𝑂 (𝛼𝑠𝑛−𝑛 𝑑−𝑛 )
(𝑛 → ∞) .
(16)
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3
Then for 𝑠 = 𝑘 + 1 we have
Taking the reciprocal of this expression yields
𝑢𝑛𝑘+1 = (𝑐𝑘 𝛼𝑘𝑛 + 𝑂 (𝛼𝑘𝑛−𝑛 𝑑−𝑛 )) ⋅ (𝑐𝛼𝑛 + 𝑂 (𝑑−𝑛 ))
⌊𝛽𝑛⌋ 𝑠𝑘 𝑎 ( ∑ 1𝑠 ) 𝑢 𝑘=𝑛 𝑘
= 𝑐𝑘+1 𝛼𝑘𝑛+𝑛 + 𝑂 (𝛼𝑘𝑛 𝑑−𝑛 )
−1
(17)
+ 𝑂 (𝛼𝑘𝑛 𝑑−𝑛 ) + 𝑂 (𝛼𝑘𝑛−𝑛 𝑑−2𝑛 )
= (1) × (
= 𝑐𝑘+1 𝛼(𝑘+1)𝑛 + 𝑂 (𝛼𝑘𝑛 𝑑−𝑛 ) .
⋅ (1 + 𝑂 (
That is, Lemma 6 also holds for 𝑠 = 𝑘 + 1. This completes the proof of Lemma 6 by mathematical induction.
In this section, we shall complete the proof of Theorem 3. From the geometric series as 𝜖 → 0, we have
=
𝛼𝑠𝑛 𝑐𝑠 𝛼𝑠𝑛 𝑐𝑠 𝛼𝑠𝑛−𝑠 − + 𝑂 ( ⋅ ℎ) 𝑎1𝑠𝑛 𝑎1𝑠𝑛−𝑠 𝑎1𝑠𝑛
=
𝑠 𝑢𝑛−1 𝑢𝑛𝑠 𝛼𝑠𝑛 − 𝑠𝑛−𝑠 + 𝑂 ( 𝑠𝑛 ⋅ ℎ) . 𝑠𝑛 𝑎1 𝑎1 𝑎1
𝑠⌊𝛽𝑛⌋−𝑠𝑛
=
𝑎1𝑠𝑘 𝑎1𝑠𝑘 + 𝑂 ( 𝑠𝑘+𝑘 ). 𝑠 𝑠𝑘 𝑐𝛼 𝛼 𝑑𝑘
𝑛
𝛼𝑠𝑛 𝛼𝑠𝑛−𝑛 𝛼𝑠−1 ⋅ ℎ = = ( ) < 1. 𝑎1𝑠𝑛 𝑎1𝑠𝑛 𝑑𝑛 𝑎1𝑠 𝑑
⌊𝛽𝑛⌋
⌊𝛽𝑛⌋
𝑎1𝑠𝑘 𝑎 𝑠𝑘 1 ) ∑ ( 1 ) + 𝑂 ( ∑ 𝑠𝑘+𝑘 𝑠 𝑐 𝑘=𝑛 𝛼 𝑑𝑘 𝑘=𝑛 𝛼 𝑎 𝑠𝑛 𝑎 𝑠⌊𝛽𝑛⌋ 𝛼𝑠 1 ⋅ ( 1) − 𝑠 𝑠 ⋅ ( 1) 𝑠 𝑠 𝑠 𝛼 𝛼 (𝛼 − 𝑎1 ) 𝑐 (𝛼 − 𝑎1 )
+ 𝑂(
𝑎1𝑠𝑛 ) 𝛼𝑠𝑛+𝑛 𝑑𝑛 𝑠
𝑎𝑠𝑛 𝑂 ( 1𝑠𝑛 𝛼
⋅
𝑎1𝑠𝑛 𝑎1 𝑠𝑛 𝛼𝑠 + 𝑂 ( ⋅ ℎ) , ⋅ ( ) 𝛼 𝛼𝑠𝑛 𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 ) 𝑠⌊𝛽𝑛⌋−𝑠𝑛
In both cases, it follows that for any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋ there exists 𝑛 ≥ 𝑛3 sufficiently large so that the modulus of the last error term of identity (21) becomes less than 1/2. This completes the proof of Theorem 3.
𝑠⌊𝛽𝑛⌋−𝑠𝑛 𝑎1 ) 𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛
𝑎1s𝑘 𝑎1𝑠𝑘 𝑎1𝑠𝑘 = + 𝑂 ( ). 𝑢𝑘𝑠 𝑐𝑠 𝛼𝑠𝑘 𝛼𝑠𝑘+𝑘 𝑑𝑘
𝑎1𝑠𝑛 1 ⋅ ) 𝛼𝑠𝑛 𝛼𝑛 𝑑𝑛
where ℎ = max {(𝑎1
(24)
Proof of Corollary 4. From identity (19), we have 𝑠𝑛
𝑎 𝛼 ⋅ ( 1) + 𝑠 𝑠 𝛼 (𝛼 − 𝑎1 )
+ 𝑂( =
(23)
we have
Consequently,
𝑐𝑠
(22)
Case 2. If ℎ = (1/𝛼𝑛 𝑑𝑛 ), for any positive integer 𝑎1 ≥ 2, 1 < (𝛼/𝑎1 ) < 𝛼𝑑 holds. Then for any positive integer 𝑠 with 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋ ,
(19)
=
𝛼 ) ⋅ (1 + 𝑂 (ℎ)) 𝑎1
𝑎 𝑠⌊𝛽𝑛⌋−2𝑠𝑛 𝛼𝑠𝑛 𝛼𝑠𝑛 𝑎1 ⋅ ℎ = ⋅ 𝑠⌊𝛽𝑛⌋−𝑠𝑛 = ( 1 ) < 1. 𝑠𝑛 𝑠𝑛 𝑎1 𝑎1 𝛼 𝛼
𝑎1𝑠𝑘 𝑎1𝑠𝑘 (1 + 𝑂 (𝛼−𝑘 𝑑−𝑘 )) = 𝑐𝑠 𝛼𝑠𝑘 (1 + 𝑂 (𝛼−𝑘 𝑑−𝑘 )) 𝑐𝑠 𝛼𝑠𝑘
𝑐𝑠
⋅(
(21)
𝑠𝑛
𝑠⌊𝛽𝑛⌋−𝑠𝑛
=
=
𝑎1𝑠 )
−1
Case 1. If ℎ = (𝑎1 /𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛 ), then for any real number 𝛽 > 2 and positive integer 𝑠 we have
𝑎1𝑠𝑘 𝑎1𝑠𝑘 = 𝑠 𝑠𝑘 𝑠 𝑢𝑘 𝑐 𝛼 + 𝑂 (𝛼𝑠𝑘−𝑘 𝑑−𝑘 )
=
𝑠
𝑐 (𝛼 − 𝛼𝑠
(18)
Using Lemma 6, we have
⌊𝛽𝑛⌋ 𝑠𝑘 𝑎 ∑ 1𝑠 𝑢 𝑘=𝑛 𝑘
𝑠
𝑎1𝑠𝑛 ℎ 𝛼𝑠𝑛 ⋅ )) ) 𝛼𝑠𝑛 𝑎1𝑠𝑛
=
3. Proof of Theorem 3
1 = 1 ∓ 𝜖 + 𝑂 (𝜖2 ) = 1 + 𝑂 (𝜖) . 1±𝜖
𝑎1 𝑠𝑛 𝛼𝑠 ⋅ ( ) 𝛼 𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 )
/𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛 ), (1/𝛼𝑛 𝑑𝑛 )}.
(25)
Consequently, ∞ 𝑎1𝑠𝑘 𝑎1𝑠𝑘 1 ∞ 𝑎1 𝑠𝑘 = ( + 𝑂 ( ) ) ∑ ∑ 𝑠𝑘+𝑘 𝑑𝑘 𝑢𝑠 𝑐𝑠 𝑘=𝑛 𝛼 𝑘=𝑛 𝑘 𝑘=𝑛 𝛼 ∞
∑
(20)
𝑎1𝑠𝑛 𝑎1 𝑠𝑛 𝛼𝑠 = 𝑠 𝑠 + 𝑂 ( ). ⋅ ( ) 𝛼 𝛼𝑠𝑛+𝑛 𝑑𝑛 𝑐 (𝛼 − 𝑎1𝑠 )
(26)
4
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Taking the reciprocal of this expression yields 𝑎1𝑠𝑘 ) 𝑢𝑠 𝑘=𝑛 𝑘 ∞
𝑢𝑛
−1
(∑
= =
(𝛼𝑠 /𝑐𝑠 𝑠
𝑠
(𝛼𝑠
𝑐 (𝛼 − 𝛼𝑠 𝑠 𝑠𝑛
−
𝑎1𝑠 )
𝑎1𝑠 )) ⋅(
𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛
1 𝑠𝑛 ⋅ (𝑎1 /𝛼) ⋅ (1 + 𝑂 (1/𝛼𝑛 𝑑𝑛 ))
𝛼 𝑠𝑛 1 ) ⋅ (1 + 𝑂 ( 𝑛 𝑛 )) 𝑎1 𝛼 𝑑
𝑠 𝑠𝑛−𝑠
𝑐𝛼 𝑐𝛼 𝛼 − 𝑠𝑛−𝑠 + 𝑂 ( 𝑠𝑛 𝑛 ) 𝑎1𝑠𝑛 𝑎1 𝑎1 𝑑
=
𝑠 𝑢𝑛−1 𝑢𝑛𝑠 𝛼𝑠𝑛−𝑛 − + 𝑂 ( ). 𝑎1𝑠𝑛 𝑎1𝑠𝑛−𝑠 𝑎1𝑠𝑛 𝑑𝑛
𝛼
𝑑
log𝛼/𝑎1 𝛼𝑑
2.4142 3.3028 4.2361 5.1926
2.4142 3.3028 4.2361 5.1926
9.3653 24.8500 50.3460 87.1630
𝛼
𝑑
log𝛼/𝑎1 𝛼𝑑
2.5468 3.6274 4.7246 5.8074
1.5959 1.9044 1.5370 1.2050
5.8020 10.1772 11.9086 12.9970
Table 2
(27) 𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛
𝑠𝑛−𝑛
=
= 2𝑢𝑛−1 + 𝑢𝑛−2 = 3𝑢𝑛−1 + 𝑢𝑛−2 = 4𝑢𝑛−1 + 𝑢𝑛−2 = 5𝑢𝑛−1 + 𝑢𝑛−2
= 2𝑢𝑛−1 + 𝑢𝑛−2 + 𝑢𝑛−3 = 3𝑢𝑛−1 + 2𝑢𝑛−2 + 𝑢𝑛−3 = 4𝑢𝑛−1 + 3𝑢𝑛−2 + 2𝑢𝑛−3 = 5𝑢𝑛−1 + 4𝑢𝑛−2 + 3𝑢𝑛−3
Table 3
For any positive integer 𝑠 with 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋ ,
(28)
we have 𝑛
𝛼𝑠𝑛−𝑛 𝛼𝑠−1 𝑠𝑛 𝑛 = ( 𝑠 ) < 1. 𝑎1 𝑑 𝑎1 𝑑
(29)
So there exists 𝑛 ≥ 𝑛4 sufficiently large so that the modulus of the last error term of identity (27) becomes less than 1/2. This completes the proof of Corollary 4.
4. Computation We can determine the power 𝑠 of different sequence 𝑢𝑛 by MATHEMATICA as the following examples. Example 7. Let 𝑢𝑛 be the second-order linear recursive sequence (see Table 1). Example 8. Let 𝑢𝑛 be the third-order linear recursive sequence (see Table 2). Example 9. Let 𝑢𝑛 be the fifth-order linear recursive sequence (see Table 3). Therefore, we may immediately deduce the following corollaries. Corollary 10. Let 𝑃𝑛 = 2𝑃𝑛−1 + 𝑃𝑛−2 be the Pell numbers. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < 9, there exists a positive integer 𝑛5 such that ⌊𝛽𝑛⌋ −1 𝑠 𝑃𝑛−1 𝑃𝑛𝑠 2𝑠𝑘 ( ∑ 𝑃𝑠 ) − ( 2𝑠𝑛 − 2𝑠𝑛−𝑠 ) = 0, (𝑛 ≥ 𝑛5 ) . (30) 𝑘=𝑛 𝑘 Corollary 11. Let T𝑛 = 4𝑇𝑛−1 +3𝑇𝑛−2 +2𝑇𝑛−3 be the generalized Tribonacci numbers. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < 11, there exists a positive integer 𝑛6 such that ⌊𝛽𝑛⌋ −1 𝑠 𝑇𝑛𝑠 𝑇𝑛−1 4𝑠𝑘 (31) ( ∑ 𝑠 ) − ( 𝑠𝑛 − 𝑠𝑛−𝑠 ) = 0, (𝑛 ≥ 𝑛6 ) . 𝑘=𝑛 𝑇𝑘 4 4
𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛
𝛼 = 2𝑢𝑛−1 + 𝑢𝑛−2 + 𝑢𝑛−3 + 𝑢𝑛−4 + 𝑢𝑛−5 2.6083 = 3𝑢𝑛−1 + 3𝑢𝑛−2 + 2𝑢𝑛−3 + 2𝑢𝑛−4 + 𝑢𝑛−5 3.9300 = 4𝑢𝑛−1 + 3𝑢𝑛−2 + 𝑢𝑛−3 + 𝑢𝑛−4 + 𝑢𝑛−5 4.6959 = 5𝑢𝑛−1 + 3𝑢𝑛−2 + 2𝑢𝑛−3 + 𝑢𝑛−4 + 𝑢𝑛−5 5.6055
𝑑
log𝛼/𝑎1 𝛼𝑑
1.1855 1.3189 1.4156 1.4828
4.2511 6.0936 11.8100 18.5257
Corollary 12. Let 𝑢𝑛 = 5𝑢𝑛−1 + 3𝑢𝑛−2 + 2𝑢𝑛−3 + 𝑢𝑛−4 + 𝑢𝑛−5 be a fifth-order sequence. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < 18, there exists a positive integer 𝑛7 such that ⌊𝛽𝑛⌋ −1 𝑠 𝑢𝑛−1 𝑢𝑛𝑠 5𝑠𝑘 ( ∑ 𝑠 ) − ( 𝑠𝑛 − 𝑠𝑛−𝑠 ) = 0, 𝑘=𝑛 𝑢𝑘 5 5
(𝑛 ≥ 𝑛7 ) .
(32)
5. Related Results The following results are obtained similarly. Theorem 13. Let {𝑢𝑛 } be an 𝑚th-order sequence defined by (5) with the restrictions 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N and 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 2. Let 𝑝 and 𝑞 be positive integers with 0 ≤ 𝑞 < 𝑝. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑝 𝑑𝑝 ⌋, where 𝛼, 𝛼1 , . . . , 𝛼𝑚−1 are the roots of the characteristic equation of 𝑢𝑛 and 𝑑−1 = max {|𝛼1 |, |𝛼2 |, . . . , |𝛼𝑚−1 |}, then there exist positive integers 𝑛8 , 𝑛9 , and 𝑛10 depending on 𝑎1 , 𝑎2 , . . ., and 𝑎𝑚 such that the following hold. −1
⌊𝛽𝑛⌋
𝑠 /𝑎1𝑠𝑛−𝑠 )‖ = (a) ‖(∑𝑘=𝑛 ((−𝑎1 )𝑠𝑘 /𝑢𝑘𝑠 )) −(−1)𝑠𝑛 (𝑢𝑛𝑠 /𝑎1𝑠𝑛 +𝑢𝑛−1 0, (𝑛 ≥ 𝑛8 ). ⌊𝛽𝑛⌋
𝑠𝑝𝑘+𝑠𝑞
−1
𝑠𝑝𝑛+𝑠𝑞
𝑠 𝑠 (b) ‖(∑𝑘=𝑛 (𝑎1 /𝑢𝑝𝑘+𝑞 )) − (𝑢𝑝𝑛+𝑞 /𝑎1 𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝 𝑎1 )‖ = 0, (𝑛 ≥ 𝑛9 ). ⌊𝛽𝑛⌋
𝑠 − 𝑢𝑝𝑛−𝑝+𝑞 /
−1
𝑠 𝑠 (c) ‖(∑𝑘=𝑛 ((−𝑎1 )𝑠𝑝𝑘+𝑠𝑞 /𝑢𝑝𝑘+𝑞 )) − (−1)𝑠𝑝𝑛+𝑠𝑞 (𝑢𝑝𝑛+𝑞 / 𝑠𝑝𝑛+𝑠𝑞 𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝 𝑠 𝑎1 + 𝑢𝑝𝑛−𝑝+𝑞 /𝑎1 )‖ = 0, (𝑛 ≥ 𝑛10 ).
For 𝛽 → +∞, we deduce the following identity of infinite sum as special case of Theorem 13.
The Scientific World Journal
5
Corollary 14. Let {𝑢𝑛 } be an 𝑚th-order sequence defined by (5) with the restrictions 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N and 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 2. Let 𝑝 and 𝑞 be positive integers with 0 ≤ 𝑞 < 𝑝. For any positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋, where 𝛼, 𝛼1 , . . . , 𝛼𝑚−1 are the roots of the characteristic equation of 𝑢𝑛 and 𝑑−1 = max {|𝛼1 |, |𝛼2 |, . . . , |𝛼𝑚−1 |}, then there exist positive integers 𝑛11 , 𝑛12 , and 𝑛13 depending on 𝑎1 , 𝑎2 , . . ., and 𝑎𝑚 such that the following hold. −1
𝑠𝑘 𝑠 𝑠𝑛 𝑠 𝑠𝑛 𝑠 𝑠𝑛−𝑠 )‖ = (d) ‖(∑∞ 𝑘=𝑛 ((−𝑎1 ) /𝑢𝑘 )) −(−1) (𝑢𝑛 /𝑎1 +𝑢𝑛−1 /𝑎1 0, (𝑛 ≥ 𝑛11 ). −1
𝑠𝑝𝑘+𝑠𝑞
𝑠𝑝𝑛+𝑠𝑞
𝑠 𝑠 /𝑢𝑝𝑘+𝑞 )) − (𝑢𝑝𝑛+𝑞 /𝑎1 (e) ‖(∑∞ 𝑘=𝑛 (𝑎1 𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝 𝑎1 )‖ = 0, (𝑛 ≥ 𝑛12 ).
𝑠 − 𝑢𝑝𝑛−𝑝+𝑞 /
Proof. We shall prove only (c) in Theorem 13 and other identities are proved similarly. From identity (19), we have 𝑠𝑝𝑘+𝑠𝑞
(−𝑎1 ) (∑ 𝑠 𝑢𝑝𝑘+𝑞 𝑘=𝑛
𝑠 𝑢𝑝𝑛+𝑞
𝑠𝑝𝑛+𝑠𝑞
𝑎1
+
𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 ) 𝛼 𝑠𝑝𝑛+𝑠𝑞 ⋅ ( ) ⋅ (1 + 𝑂 (ℎ𝑝 )) 𝛼𝑠 𝑎1 𝑠 𝑢𝑝𝑛−𝑝+𝑞
𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝
𝑎1
+ 𝑂(
𝛼𝑠𝑝𝑛 𝑝 𝑠𝑝𝑛 ⋅ ℎ ) . 𝑎1
(35)
𝑠⌊𝛽𝑛⌋−𝑠𝑛
)/(𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛 ), then for any real number Case 1. If ℎ = (𝑎1 𝛽 > 2 and positive integer 𝑠 we have 𝑎1 𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛−𝑠𝑛 𝛼𝑠𝑛 𝑝 𝛼𝑠𝑛 𝑎1 ⋅ ℎ = ⋅ = ( < 1. ) 𝑎1𝑠𝑛 𝑎1𝑠𝑛 𝛼𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛 𝛼
(−𝑎1 ) (1 + 𝑂 (𝛼−𝑝𝑘−𝑞 𝑑−𝑝𝑘−𝑞 )) . 𝑐𝑠 𝛼𝑠𝑝𝑘+𝑠𝑞
(36)
Case 2. If ℎ = (1/𝛼𝑛 𝑑𝑛 ), for any positive integer 𝑎1 ≥ 2, 1 < (𝛼/𝑎1 ) < 𝛼𝑑 holds. Then for any positive integer 𝑠 with
𝑠𝑝𝑘+𝑠𝑞
=
)
= (−1)𝑠𝑝𝑛+𝑠𝑞 ⋅ =
−1
𝑠𝑝𝑘+𝑠𝑞
⌊𝛽𝑛⌋
𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛
−1
𝑠𝑝𝑘+𝑠𝑞 𝑠 𝑠 (f) ‖(∑∞ /𝑢𝑝𝑘+𝑞 )) − (−1)𝑠𝑝𝑛+𝑠𝑞 (𝑢𝑝𝑛+𝑞 / 𝑘=𝑛 ((−𝑎1 ) 𝑠𝑝𝑛+𝑠𝑞 𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝 𝑠 𝑎1 + 𝑢𝑝𝑛−𝑝+𝑞 /𝑎1 )‖ = 0, (𝑛 ≥ 𝑛13 ).
(−𝑎1 ) 𝑠 𝑢𝑝𝑘+𝑞
Taking the reciprocal of this expression yields
1 ≤ s < ⌊log(𝛼/𝑎1 ) 𝛼𝑝 𝑑𝑝 ⌋ ,
(33)
(37)
we have 𝑛
𝛼𝑠𝑛 𝑝 𝛼𝑠𝑛−𝑝𝑛 𝛼𝑠−𝑝 ( 𝑠 𝑝 ) < 1. 𝑠𝑛 ⋅ ℎ = 𝑠𝑛 𝑝𝑛 = 𝑎1 𝑎1 𝑑 𝑎1 𝑑
Consequently, ⌊𝛽𝑛⌋
In both cases, it follows that for any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑝 𝑑𝑝 ⌋, there exists 𝑛 ≥ 𝑛10 sufficiently large so that the modulus of the last error term of identity (35) becomes less than 1/2. This completes the proof of Theorem 13(c).
𝑠𝑝𝑘+𝑠𝑞
(−𝑎1 ) ∑ 𝑠 𝑢𝑠𝑝𝑘+𝑠𝑞 𝑘=𝑛
𝑠𝑝𝑘+𝑠𝑞
=
⌊𝛽𝑛⌋ ⌊𝛽𝑛⌋ 𝑎1 −𝑎1 𝑠𝑝𝑘+𝑠𝑞 1 ( + 𝑂 ( ) ) ∑ ∑ 𝑠𝑝𝑘+𝑠𝑞+𝑝𝑘+𝑞 𝑠 𝑐 𝑘=𝑛 𝛼 𝑑𝑝𝑘+𝑞 𝑘=𝑛 𝛼
Conflict of Interests
𝑎1 𝑠𝑝𝑛+𝑠𝑞 (−1)𝑠𝑝𝑛+𝑠𝑞 (−1)𝑠𝑝𝑛+𝑠𝑞 ⋅ 𝛼𝑠 − 𝑠 𝑠 = 𝑠 𝑠 ⋅ ( ) 𝛼 𝑐 (𝛼 − 𝑎1𝑠 ) 𝑐 (𝛼 − 𝑎1𝑠 )
The authors declare that there is no conflict interests regarding the publication of this paper.
𝑠𝑝𝑛
𝑎1 𝑎 𝑠𝑝⌊𝛽𝑛⌋+𝑠𝑞 ⋅ ( 1) + 𝑂 ( 𝑠𝑝𝑛+𝑝𝑛 ) 𝛼 𝛼 𝑑𝑝𝑛
Acknowledgments 𝑠𝑝𝑛
=
𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛
𝑎 𝑎 𝑎1 𝑠𝑝𝑛+𝑠𝑞 (−1)𝑠𝑝𝑛+𝑠𝑞 ⋅ 𝛼𝑠 + 𝑂 ( 1𝑠𝑝𝑛 ⋅ 1𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛 ) ⋅ ( ) 𝛼 𝛼 𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 ) 𝛼 𝑠𝑝𝑛
+ 𝑂(
𝑎1 1 ⋅ 𝑝𝑛 𝑝𝑛 ) 𝑠𝑝𝑛 𝛼 𝛼 𝑑 𝑠𝑝𝑛
=
𝑎 𝑎1 𝑠𝑝𝑛+𝑠𝑞 (−1)𝑠𝑝𝑛+𝑠𝑞 ⋅ 𝛼𝑠 + 𝑂 ( 1𝑠𝑝𝑛 ⋅ ℎ𝑝 ) , ⋅ ( ) 𝛼 𝛼 𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 )
𝑠⌊𝛽𝑛⌋−𝑠𝑛
where ℎ = max {(𝑎1
(38)
/𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛 ), (1/𝛼𝑛 𝑑𝑛 )}.
The authors express their gratitude to the referee for very helpful and detailed comments. This work is supported by the N.S.F. (11371291), the S.R.F.D.P. (20136101110014), the N.S.F. (2013JZ001) of Shaanxi Province, and the G.I.C.F. (YZZ12062) of NWU, China.
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The Scientific World Journal
Research Article On the Higher Power Sums of Reciprocal Higher-Order Sequences Zhengang Wu1 and Jin Zhang2 1 2
Department of Mathematics, Northwest University, Xi’an, Shaanxi, China School of Mathematics and Computer Engineering, Xi’an University of Arts and Science, Xi’an, Shaanxi, China
Correspondence should be addressed to Zhengang Wu; [email protected] Received 21 November 2013; Accepted 29 January 2014; Published 10 March 2014 Academic Editors: A. Fiorenza and E. Kılıc¸ Copyright © 2014 Z. Wu and J. Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Let {𝑢𝑛 } be a higher-order linear recursive sequence. In this paper, we use the properties of error estimation and the analytic method to study the reciprocal sums of higher power of higher-order sequences. Then we establish several new and interesting identities relating to the infinite and finite sums.
1. Introduction The so-called Fibonacci zeta function and Lucas zeta function, defined by ∞
1 , 𝑠 𝐹 𝑛=1 𝑛
𝜁𝐹 (𝑠) = ∑
∞
1 , 𝑠 𝐿 𝑛=1 𝑛
𝜁𝐿 (𝑠) = ∑
(1)
where the 𝐹𝑛 and 𝐿 𝑛 denote the Fibonacci numbers and Lucas numbers, have been considered in several different ways; see [1, 2]. Ohtsuka and Nakamura [3] studied the partial infinite sums of reciprocal Fibonacci numbers and proved the following conclusions: ∞
⌊( ∑
𝑘=𝑛
∞
−1
1 𝐹 if 𝑛 is even, 𝑛 ≥ 2, ) ⌋ = { 𝑛−2 𝐹𝑘 𝐹𝑛−2 − 1 if 𝑛 is odd, 𝑛 ≥ 1, −1
(2)
1 𝐹 𝐹 − 1 if 𝑛 is even, 𝑛 ≥ 2, ⌊( ∑ 2 ) ⌋ = { 𝑛−1 𝑛 𝐹 𝐹𝑛−1 𝐹𝑛 if 𝑛 is odd, 𝑛 ≥ 1, 𝑘 𝑘=𝑛 where ⌊⋅⌋ denotes the floor function. Further, Wu and Zhang [4, 5] generalized these identities to the Fibonacci polynomials and Lucas polynomials. Various properties of the Fibonacci polynomials and Lucas polynomials have been studied by many authors; see [6–13]. Recently, some authors considered the nearest integer of the sums of reciprocal Fibonacci numbers and other wellknown sequences and obtained several meaningful results;
see [14–16]. In particular, in [16], Kılıc¸ and Arıkan studied a problem which is a little different from that of [3], namely, −1 that of determining the nearest integer to (∑∞ 𝑘=𝑛 (1/V𝑘 )) . Specifically, suppose that ‖𝑥‖ = ⌊𝑥 + (1/2)⌋ (the nearest integer function) and {V𝑛 }𝑛≥0 is an integer sequence satisfying the recurrence formula V𝑛 = 𝑝V𝑛−1 + 𝑞V𝑛−2 + V𝑛−3 + ⋅ ⋅ ⋅ + V𝑛−𝑘 ,
(3)
for any positive integer 𝑝 ≥ 𝑞 and 𝑛 > 𝑘. Then we can conclude that there exists a positive integer 𝑛0 such that ∞ −1 ( ∑ 1 ) = V − V , 𝑛 𝑛−1 𝑘=𝑛 V𝑘
(4)
for all 𝑛 > 𝑛0 . In [17], Wu and Zhang unified the above results by proving the following conclusion that includes all the results, [3–8, 15, 16], as special cases. Proposition 1. For any positive integer 𝑛 > 𝑚, the 𝑚th-order linear recursive sequence {𝑢𝑛 } is defined as follows: 𝑢𝑛 = 𝑎1 𝑢𝑛−1 + 𝑎2 𝑢𝑛−2 + ⋅ ⋅ ⋅ + 𝑎𝑚−1 𝑢𝑛−𝑚+1 + 𝑎𝑚 𝑢𝑛−𝑚 ,
(5)
with initial values 𝑢𝑖 ∈ N for 0 ≤ 𝑖 < 𝑚 and at least one of them not being zero. For any positive real number 𝛽 > 2 and any
2
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positive integer 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 1, there exists a positive integer 𝑛1 such that ⌊𝛽𝑛⌋ −1 1 ) = 𝑢𝑛 − 𝑢𝑛−1 , ( ∑ 𝑘=𝑛 𝑢𝑘
(𝑛 ≥ 𝑛1 ) .
(6)
In particular, taking 𝛽 → +∞, there exists a positive integer 𝑛2 such that ∞ −1 ( ∑ 1 ) = 𝑢 − 𝑢 , 𝑛 𝑛−1 𝑘=𝑛 𝑢𝑘
(𝑛 ≥ 𝑛2 ) .
(7)
Corollary 4. Let {𝑢𝑛 } be an 𝑚th-order sequence defined by (5) with the restrictions 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N and 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 2. For positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋, where 𝛼, 𝛼1 , . . . , 𝛼𝑚−1 are the roots of the characteristic equation of 𝑢𝑛 and 𝑑−1 = max{|𝛼1 |, |𝛼2 |, . . . , |𝛼𝑚−1 |}, then there exists a positive integer 𝑛4 such that ∞ 𝑠𝑘 −1 𝑠 𝑠 ( ∑ 𝑎1 ) − ( 𝑢𝑛 − 𝑢𝑛−1 ) = 0, 𝑠 𝑠𝑛−𝑠 𝑠𝑛 𝑎1 𝑎1 𝑘=𝑛 𝑢𝑘
Proposition 2. For any positive integer 𝑛 ≥ 1, we have the identity ∞
⌊( ∑
𝑘=𝑛
{ { { { { { ={ { { { { { {
−1
1 ) ⌋ 𝑃𝑘3
⌊𝛽𝑛⌋ −1 𝑎1𝑠𝑘 ) ( ∑ 𝑠 𝑢 𝑘=𝑛 𝑘
(11)
is an interesting open problem.
2. Several Lemmas To complete the proof of our theorem, we need two lemmas. Lemma 5. Let 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N with 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 1 and 𝑚 ∈ N with 𝑚 ≥ 2. Then for the polynomial 𝑓 (𝑥) = 𝑥𝑚 − 𝑎1 𝑥𝑚−1 − 𝑎2 𝑥𝑚−2 − ⋅ ⋅ ⋅ − 𝑎𝑚−1 𝑥 − 𝑎𝑚 ,
61 91 + + ⌊− 𝑃𝑛 − 𝑃𝑛−1 ⌋ 82 82 𝑖𝑓 𝑛 𝑖𝑠 𝑒V𝑒𝑛, 𝑛 ≥ 2, 61 91 2 + ⌊ 𝑃𝑛 + 𝑃𝑛−1 ⌋ 𝑃𝑛2 𝑃𝑛−1 + 3𝑃𝑛 𝑃𝑛−1 82 82 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑, 𝑛 ≥ 1.
𝑃𝑛2 𝑃𝑛−1
2 3𝑃𝑛 𝑃𝑛−1
(10)
For positive real number 1 < 𝛽 ≤ 2, whether there exits an identity for
−1
𝑠 It seems difficult to deal with (∑∞ for all 𝑘=𝑛 (1/𝑢𝑘 )) integers 𝑠 ≥ 2, because it is quite unclear a priori what the shape of the result might be. In [18], Xu and Wang applied the method of undetermined coefficients and constructed a number of delicate inequalities in order to study the infinite sum of the cubes of reciprocal Pell numbers and then obtained the following meaningful result.
(𝑛 ≥ 𝑛4 ) .
(12)
we have the following: (8)
(I) polynomial 𝑓(𝑥) has exactly one positive real zero 𝛼 with 𝑎1 < 𝛼 < 𝑎1 + 1; (II) other 𝑚 − 1 zeros of 𝑓(𝑥) lie within the unit circle in the complex plane. Proof. See Lemma 1 of [16].
To find and prove this result is a substantial achievement since such a complex formula would not be clear beforehand that a result would even be possible. However, there is no research considering the higher power (𝑠 > 2) of reciprocal sums of some recursive sequences. The main purpose of this paper is using the properties of error estimation and the analytic method to study the higher power of the reciprocal sums of {𝑢𝑛 } and obtain several new and interesting identities. The results are as follows. Theorem 3. Let {𝑢𝑛 } be an 𝑚th-order sequence defined by (5) with the restrictions 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N and 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 2. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋, where 𝛼, 𝛼1 , . . . , 𝛼𝑚−1 are the roots of the characteristic equation of 𝑢𝑛 and 𝑑−1 = max {|𝛼1 |, |𝛼2 |, . . . , |𝛼𝑚−1 |}, then there exists a positive integer 𝑛3 such that ⌊𝛽𝑛⌋ −1 𝑠 𝑢𝑛−1 𝑢𝑛𝑠 𝑎1𝑠𝑘 ( ∑ 𝑢𝑠 ) − ( 𝑎𝑠𝑛 − 𝑎𝑠𝑛−𝑠 ) = 0, 𝑘=𝑛 𝑘 1 1
(𝑛 ≥ 𝑛3 ) .
(9)
Taking 𝛽 → +∞, from Theorem 3 we may immediately deduce the following.
Lemma 6. Let 𝑚 ≥ 2 and {𝑢𝑛 }𝑛≥0 be an integer sequence satisfying the recurrence formula (5). Then for any positive integer 𝑠, we have 𝑢𝑛𝑠 = 𝑐𝑠 𝛼𝑠𝑛 + 𝑂 (𝛼𝑠𝑛−𝑛 𝑑−𝑛 )
(𝑛 → ∞) ,
(13)
where 𝑐 > 0, 𝑑 > 1, and 𝑎1 < 𝛼 < 𝑎1 + 1 is the positive real zero of 𝑓(𝑥). Proof. From Lemma 2 of [16], the closed formula of 𝑢𝑛 is given by 𝑢𝑛 = 𝑐𝛼𝑛 + 𝑂 (𝑑−𝑛 )
(𝑛 → ∞) ,
(14)
where 𝑐 > 0, 𝑑 > 1, 𝑎1 < 𝛼 < 𝑎1 + 1, and 𝛼 is the positive real zero of 𝑓(𝑥). Now we prove Lemma 6 by mathematical induction. From formula (14), we have 𝑢𝑛2 = 𝑐2 𝛼2𝑛 + 𝑂 (𝑑−2𝑛 ) + 𝑂 (𝛼𝑛 𝑑−𝑛 ) = 𝑐2 𝛼2𝑛 + 𝑂 (𝛼𝑛 𝑑−𝑛 ) . (15) That is, the lemma holds for 𝑠 = 2. Suppose that for all integers 2 ≤ 𝑠 ≤ 𝑘 we have 𝑢𝑛𝑠 = 𝑐𝑠 𝛼𝑠𝑛 + 𝑂 (𝛼𝑠𝑛−𝑛 𝑑−𝑛 )
(𝑛 → ∞) .
(16)
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3
Then for 𝑠 = 𝑘 + 1 we have
Taking the reciprocal of this expression yields
𝑢𝑛𝑘+1 = (𝑐𝑘 𝛼𝑘𝑛 + 𝑂 (𝛼𝑘𝑛−𝑛 𝑑−𝑛 )) ⋅ (𝑐𝛼𝑛 + 𝑂 (𝑑−𝑛 ))
⌊𝛽𝑛⌋ 𝑠𝑘 𝑎 ( ∑ 1𝑠 ) 𝑢 𝑘=𝑛 𝑘
= 𝑐𝑘+1 𝛼𝑘𝑛+𝑛 + 𝑂 (𝛼𝑘𝑛 𝑑−𝑛 )
−1
(17)
+ 𝑂 (𝛼𝑘𝑛 𝑑−𝑛 ) + 𝑂 (𝛼𝑘𝑛−𝑛 𝑑−2𝑛 )
= (1) × (
= 𝑐𝑘+1 𝛼(𝑘+1)𝑛 + 𝑂 (𝛼𝑘𝑛 𝑑−𝑛 ) .
⋅ (1 + 𝑂 (
That is, Lemma 6 also holds for 𝑠 = 𝑘 + 1. This completes the proof of Lemma 6 by mathematical induction.
In this section, we shall complete the proof of Theorem 3. From the geometric series as 𝜖 → 0, we have
=
𝛼𝑠𝑛 𝑐𝑠 𝛼𝑠𝑛 𝑐𝑠 𝛼𝑠𝑛−𝑠 − + 𝑂 ( ⋅ ℎ) 𝑎1𝑠𝑛 𝑎1𝑠𝑛−𝑠 𝑎1𝑠𝑛
=
𝑠 𝑢𝑛−1 𝑢𝑛𝑠 𝛼𝑠𝑛 − 𝑠𝑛−𝑠 + 𝑂 ( 𝑠𝑛 ⋅ ℎ) . 𝑠𝑛 𝑎1 𝑎1 𝑎1
𝑠⌊𝛽𝑛⌋−𝑠𝑛
=
𝑎1𝑠𝑘 𝑎1𝑠𝑘 + 𝑂 ( 𝑠𝑘+𝑘 ). 𝑠 𝑠𝑘 𝑐𝛼 𝛼 𝑑𝑘
𝑛
𝛼𝑠𝑛 𝛼𝑠𝑛−𝑛 𝛼𝑠−1 ⋅ ℎ = = ( ) < 1. 𝑎1𝑠𝑛 𝑎1𝑠𝑛 𝑑𝑛 𝑎1𝑠 𝑑
⌊𝛽𝑛⌋
⌊𝛽𝑛⌋
𝑎1𝑠𝑘 𝑎 𝑠𝑘 1 ) ∑ ( 1 ) + 𝑂 ( ∑ 𝑠𝑘+𝑘 𝑠 𝑐 𝑘=𝑛 𝛼 𝑑𝑘 𝑘=𝑛 𝛼 𝑎 𝑠𝑛 𝑎 𝑠⌊𝛽𝑛⌋ 𝛼𝑠 1 ⋅ ( 1) − 𝑠 𝑠 ⋅ ( 1) 𝑠 𝑠 𝑠 𝛼 𝛼 (𝛼 − 𝑎1 ) 𝑐 (𝛼 − 𝑎1 )
+ 𝑂(
𝑎1𝑠𝑛 ) 𝛼𝑠𝑛+𝑛 𝑑𝑛 𝑠
𝑎𝑠𝑛 𝑂 ( 1𝑠𝑛 𝛼
⋅
𝑎1𝑠𝑛 𝑎1 𝑠𝑛 𝛼𝑠 + 𝑂 ( ⋅ ℎ) , ⋅ ( ) 𝛼 𝛼𝑠𝑛 𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 ) 𝑠⌊𝛽𝑛⌋−𝑠𝑛
In both cases, it follows that for any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋ there exists 𝑛 ≥ 𝑛3 sufficiently large so that the modulus of the last error term of identity (21) becomes less than 1/2. This completes the proof of Theorem 3.
𝑠⌊𝛽𝑛⌋−𝑠𝑛 𝑎1 ) 𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛
𝑎1s𝑘 𝑎1𝑠𝑘 𝑎1𝑠𝑘 = + 𝑂 ( ). 𝑢𝑘𝑠 𝑐𝑠 𝛼𝑠𝑘 𝛼𝑠𝑘+𝑘 𝑑𝑘
𝑎1𝑠𝑛 1 ⋅ ) 𝛼𝑠𝑛 𝛼𝑛 𝑑𝑛
where ℎ = max {(𝑎1
(24)
Proof of Corollary 4. From identity (19), we have 𝑠𝑛
𝑎 𝛼 ⋅ ( 1) + 𝑠 𝑠 𝛼 (𝛼 − 𝑎1 )
+ 𝑂( =
(23)
we have
Consequently,
𝑐𝑠
(22)
Case 2. If ℎ = (1/𝛼𝑛 𝑑𝑛 ), for any positive integer 𝑎1 ≥ 2, 1 < (𝛼/𝑎1 ) < 𝛼𝑑 holds. Then for any positive integer 𝑠 with 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋ ,
(19)
=
𝛼 ) ⋅ (1 + 𝑂 (ℎ)) 𝑎1
𝑎 𝑠⌊𝛽𝑛⌋−2𝑠𝑛 𝛼𝑠𝑛 𝛼𝑠𝑛 𝑎1 ⋅ ℎ = ⋅ 𝑠⌊𝛽𝑛⌋−𝑠𝑛 = ( 1 ) < 1. 𝑠𝑛 𝑠𝑛 𝑎1 𝑎1 𝛼 𝛼
𝑎1𝑠𝑘 𝑎1𝑠𝑘 (1 + 𝑂 (𝛼−𝑘 𝑑−𝑘 )) = 𝑐𝑠 𝛼𝑠𝑘 (1 + 𝑂 (𝛼−𝑘 𝑑−𝑘 )) 𝑐𝑠 𝛼𝑠𝑘
𝑐𝑠
⋅(
(21)
𝑠𝑛
𝑠⌊𝛽𝑛⌋−𝑠𝑛
=
=
𝑎1𝑠 )
−1
Case 1. If ℎ = (𝑎1 /𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛 ), then for any real number 𝛽 > 2 and positive integer 𝑠 we have
𝑎1𝑠𝑘 𝑎1𝑠𝑘 = 𝑠 𝑠𝑘 𝑠 𝑢𝑘 𝑐 𝛼 + 𝑂 (𝛼𝑠𝑘−𝑘 𝑑−𝑘 )
=
𝑠
𝑐 (𝛼 − 𝛼𝑠
(18)
Using Lemma 6, we have
⌊𝛽𝑛⌋ 𝑠𝑘 𝑎 ∑ 1𝑠 𝑢 𝑘=𝑛 𝑘
𝑠
𝑎1𝑠𝑛 ℎ 𝛼𝑠𝑛 ⋅ )) ) 𝛼𝑠𝑛 𝑎1𝑠𝑛
=
3. Proof of Theorem 3
1 = 1 ∓ 𝜖 + 𝑂 (𝜖2 ) = 1 + 𝑂 (𝜖) . 1±𝜖
𝑎1 𝑠𝑛 𝛼𝑠 ⋅ ( ) 𝛼 𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 )
/𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛 ), (1/𝛼𝑛 𝑑𝑛 )}.
(25)
Consequently, ∞ 𝑎1𝑠𝑘 𝑎1𝑠𝑘 1 ∞ 𝑎1 𝑠𝑘 = ( + 𝑂 ( ) ) ∑ ∑ 𝑠𝑘+𝑘 𝑑𝑘 𝑢𝑠 𝑐𝑠 𝑘=𝑛 𝛼 𝑘=𝑛 𝑘 𝑘=𝑛 𝛼 ∞
∑
(20)
𝑎1𝑠𝑛 𝑎1 𝑠𝑛 𝛼𝑠 = 𝑠 𝑠 + 𝑂 ( ). ⋅ ( ) 𝛼 𝛼𝑠𝑛+𝑛 𝑑𝑛 𝑐 (𝛼 − 𝑎1𝑠 )
(26)
4
The Scientific World Journal Table 1
Taking the reciprocal of this expression yields 𝑎1𝑠𝑘 ) 𝑢𝑠 𝑘=𝑛 𝑘 ∞
𝑢𝑛
−1
(∑
= =
(𝛼𝑠 /𝑐𝑠 𝑠
𝑠
(𝛼𝑠
𝑐 (𝛼 − 𝛼𝑠 𝑠 𝑠𝑛
−
𝑎1𝑠 )
𝑎1𝑠 )) ⋅(
𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛
1 𝑠𝑛 ⋅ (𝑎1 /𝛼) ⋅ (1 + 𝑂 (1/𝛼𝑛 𝑑𝑛 ))
𝛼 𝑠𝑛 1 ) ⋅ (1 + 𝑂 ( 𝑛 𝑛 )) 𝑎1 𝛼 𝑑
𝑠 𝑠𝑛−𝑠
𝑐𝛼 𝑐𝛼 𝛼 − 𝑠𝑛−𝑠 + 𝑂 ( 𝑠𝑛 𝑛 ) 𝑎1𝑠𝑛 𝑎1 𝑎1 𝑑
=
𝑠 𝑢𝑛−1 𝑢𝑛𝑠 𝛼𝑠𝑛−𝑛 − + 𝑂 ( ). 𝑎1𝑠𝑛 𝑎1𝑠𝑛−𝑠 𝑎1𝑠𝑛 𝑑𝑛
𝛼
𝑑
log𝛼/𝑎1 𝛼𝑑
2.4142 3.3028 4.2361 5.1926
2.4142 3.3028 4.2361 5.1926
9.3653 24.8500 50.3460 87.1630
𝛼
𝑑
log𝛼/𝑎1 𝛼𝑑
2.5468 3.6274 4.7246 5.8074
1.5959 1.9044 1.5370 1.2050
5.8020 10.1772 11.9086 12.9970
Table 2
(27) 𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛
𝑠𝑛−𝑛
=
= 2𝑢𝑛−1 + 𝑢𝑛−2 = 3𝑢𝑛−1 + 𝑢𝑛−2 = 4𝑢𝑛−1 + 𝑢𝑛−2 = 5𝑢𝑛−1 + 𝑢𝑛−2
= 2𝑢𝑛−1 + 𝑢𝑛−2 + 𝑢𝑛−3 = 3𝑢𝑛−1 + 2𝑢𝑛−2 + 𝑢𝑛−3 = 4𝑢𝑛−1 + 3𝑢𝑛−2 + 2𝑢𝑛−3 = 5𝑢𝑛−1 + 4𝑢𝑛−2 + 3𝑢𝑛−3
Table 3
For any positive integer 𝑠 with 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋ ,
(28)
we have 𝑛
𝛼𝑠𝑛−𝑛 𝛼𝑠−1 𝑠𝑛 𝑛 = ( 𝑠 ) < 1. 𝑎1 𝑑 𝑎1 𝑑
(29)
So there exists 𝑛 ≥ 𝑛4 sufficiently large so that the modulus of the last error term of identity (27) becomes less than 1/2. This completes the proof of Corollary 4.
4. Computation We can determine the power 𝑠 of different sequence 𝑢𝑛 by MATHEMATICA as the following examples. Example 7. Let 𝑢𝑛 be the second-order linear recursive sequence (see Table 1). Example 8. Let 𝑢𝑛 be the third-order linear recursive sequence (see Table 2). Example 9. Let 𝑢𝑛 be the fifth-order linear recursive sequence (see Table 3). Therefore, we may immediately deduce the following corollaries. Corollary 10. Let 𝑃𝑛 = 2𝑃𝑛−1 + 𝑃𝑛−2 be the Pell numbers. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < 9, there exists a positive integer 𝑛5 such that ⌊𝛽𝑛⌋ −1 𝑠 𝑃𝑛−1 𝑃𝑛𝑠 2𝑠𝑘 ( ∑ 𝑃𝑠 ) − ( 2𝑠𝑛 − 2𝑠𝑛−𝑠 ) = 0, (𝑛 ≥ 𝑛5 ) . (30) 𝑘=𝑛 𝑘 Corollary 11. Let T𝑛 = 4𝑇𝑛−1 +3𝑇𝑛−2 +2𝑇𝑛−3 be the generalized Tribonacci numbers. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < 11, there exists a positive integer 𝑛6 such that ⌊𝛽𝑛⌋ −1 𝑠 𝑇𝑛𝑠 𝑇𝑛−1 4𝑠𝑘 (31) ( ∑ 𝑠 ) − ( 𝑠𝑛 − 𝑠𝑛−𝑠 ) = 0, (𝑛 ≥ 𝑛6 ) . 𝑘=𝑛 𝑇𝑘 4 4
𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛 𝑢𝑛
𝛼 = 2𝑢𝑛−1 + 𝑢𝑛−2 + 𝑢𝑛−3 + 𝑢𝑛−4 + 𝑢𝑛−5 2.6083 = 3𝑢𝑛−1 + 3𝑢𝑛−2 + 2𝑢𝑛−3 + 2𝑢𝑛−4 + 𝑢𝑛−5 3.9300 = 4𝑢𝑛−1 + 3𝑢𝑛−2 + 𝑢𝑛−3 + 𝑢𝑛−4 + 𝑢𝑛−5 4.6959 = 5𝑢𝑛−1 + 3𝑢𝑛−2 + 2𝑢𝑛−3 + 𝑢𝑛−4 + 𝑢𝑛−5 5.6055
𝑑
log𝛼/𝑎1 𝛼𝑑
1.1855 1.3189 1.4156 1.4828
4.2511 6.0936 11.8100 18.5257
Corollary 12. Let 𝑢𝑛 = 5𝑢𝑛−1 + 3𝑢𝑛−2 + 2𝑢𝑛−3 + 𝑢𝑛−4 + 𝑢𝑛−5 be a fifth-order sequence. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < 18, there exists a positive integer 𝑛7 such that ⌊𝛽𝑛⌋ −1 𝑠 𝑢𝑛−1 𝑢𝑛𝑠 5𝑠𝑘 ( ∑ 𝑠 ) − ( 𝑠𝑛 − 𝑠𝑛−𝑠 ) = 0, 𝑘=𝑛 𝑢𝑘 5 5
(𝑛 ≥ 𝑛7 ) .
(32)
5. Related Results The following results are obtained similarly. Theorem 13. Let {𝑢𝑛 } be an 𝑚th-order sequence defined by (5) with the restrictions 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N and 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 2. Let 𝑝 and 𝑞 be positive integers with 0 ≤ 𝑞 < 𝑝. For any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑝 𝑑𝑝 ⌋, where 𝛼, 𝛼1 , . . . , 𝛼𝑚−1 are the roots of the characteristic equation of 𝑢𝑛 and 𝑑−1 = max {|𝛼1 |, |𝛼2 |, . . . , |𝛼𝑚−1 |}, then there exist positive integers 𝑛8 , 𝑛9 , and 𝑛10 depending on 𝑎1 , 𝑎2 , . . ., and 𝑎𝑚 such that the following hold. −1
⌊𝛽𝑛⌋
𝑠 /𝑎1𝑠𝑛−𝑠 )‖ = (a) ‖(∑𝑘=𝑛 ((−𝑎1 )𝑠𝑘 /𝑢𝑘𝑠 )) −(−1)𝑠𝑛 (𝑢𝑛𝑠 /𝑎1𝑠𝑛 +𝑢𝑛−1 0, (𝑛 ≥ 𝑛8 ). ⌊𝛽𝑛⌋
𝑠𝑝𝑘+𝑠𝑞
−1
𝑠𝑝𝑛+𝑠𝑞
𝑠 𝑠 (b) ‖(∑𝑘=𝑛 (𝑎1 /𝑢𝑝𝑘+𝑞 )) − (𝑢𝑝𝑛+𝑞 /𝑎1 𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝 𝑎1 )‖ = 0, (𝑛 ≥ 𝑛9 ). ⌊𝛽𝑛⌋
𝑠 − 𝑢𝑝𝑛−𝑝+𝑞 /
−1
𝑠 𝑠 (c) ‖(∑𝑘=𝑛 ((−𝑎1 )𝑠𝑝𝑘+𝑠𝑞 /𝑢𝑝𝑘+𝑞 )) − (−1)𝑠𝑝𝑛+𝑠𝑞 (𝑢𝑝𝑛+𝑞 / 𝑠𝑝𝑛+𝑠𝑞 𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝 𝑠 𝑎1 + 𝑢𝑝𝑛−𝑝+𝑞 /𝑎1 )‖ = 0, (𝑛 ≥ 𝑛10 ).
For 𝛽 → +∞, we deduce the following identity of infinite sum as special case of Theorem 13.
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Corollary 14. Let {𝑢𝑛 } be an 𝑚th-order sequence defined by (5) with the restrictions 𝑎1 , 𝑎2 , . . . , 𝑎𝑚 ∈ N and 𝑎1 ≥ 𝑎2 ≥ ⋅ ⋅ ⋅ ≥ 𝑎𝑚 ≥ 2. Let 𝑝 and 𝑞 be positive integers with 0 ≤ 𝑞 < 𝑝. For any positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑑⌋, where 𝛼, 𝛼1 , . . . , 𝛼𝑚−1 are the roots of the characteristic equation of 𝑢𝑛 and 𝑑−1 = max {|𝛼1 |, |𝛼2 |, . . . , |𝛼𝑚−1 |}, then there exist positive integers 𝑛11 , 𝑛12 , and 𝑛13 depending on 𝑎1 , 𝑎2 , . . ., and 𝑎𝑚 such that the following hold. −1
𝑠𝑘 𝑠 𝑠𝑛 𝑠 𝑠𝑛 𝑠 𝑠𝑛−𝑠 )‖ = (d) ‖(∑∞ 𝑘=𝑛 ((−𝑎1 ) /𝑢𝑘 )) −(−1) (𝑢𝑛 /𝑎1 +𝑢𝑛−1 /𝑎1 0, (𝑛 ≥ 𝑛11 ). −1
𝑠𝑝𝑘+𝑠𝑞
𝑠𝑝𝑛+𝑠𝑞
𝑠 𝑠 /𝑢𝑝𝑘+𝑞 )) − (𝑢𝑝𝑛+𝑞 /𝑎1 (e) ‖(∑∞ 𝑘=𝑛 (𝑎1 𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝 𝑎1 )‖ = 0, (𝑛 ≥ 𝑛12 ).
𝑠 − 𝑢𝑝𝑛−𝑝+𝑞 /
Proof. We shall prove only (c) in Theorem 13 and other identities are proved similarly. From identity (19), we have 𝑠𝑝𝑘+𝑠𝑞
(−𝑎1 ) (∑ 𝑠 𝑢𝑝𝑘+𝑞 𝑘=𝑛
𝑠 𝑢𝑝𝑛+𝑞
𝑠𝑝𝑛+𝑠𝑞
𝑎1
+
𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 ) 𝛼 𝑠𝑝𝑛+𝑠𝑞 ⋅ ( ) ⋅ (1 + 𝑂 (ℎ𝑝 )) 𝛼𝑠 𝑎1 𝑠 𝑢𝑝𝑛−𝑝+𝑞
𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝
𝑎1
+ 𝑂(
𝛼𝑠𝑝𝑛 𝑝 𝑠𝑝𝑛 ⋅ ℎ ) . 𝑎1
(35)
𝑠⌊𝛽𝑛⌋−𝑠𝑛
)/(𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛 ), then for any real number Case 1. If ℎ = (𝑎1 𝛽 > 2 and positive integer 𝑠 we have 𝑎1 𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛−𝑠𝑛 𝛼𝑠𝑛 𝑝 𝛼𝑠𝑛 𝑎1 ⋅ ℎ = ⋅ = ( < 1. ) 𝑎1𝑠𝑛 𝑎1𝑠𝑛 𝛼𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛 𝛼
(−𝑎1 ) (1 + 𝑂 (𝛼−𝑝𝑘−𝑞 𝑑−𝑝𝑘−𝑞 )) . 𝑐𝑠 𝛼𝑠𝑝𝑘+𝑠𝑞
(36)
Case 2. If ℎ = (1/𝛼𝑛 𝑑𝑛 ), for any positive integer 𝑎1 ≥ 2, 1 < (𝛼/𝑎1 ) < 𝛼𝑑 holds. Then for any positive integer 𝑠 with
𝑠𝑝𝑘+𝑠𝑞
=
)
= (−1)𝑠𝑝𝑛+𝑠𝑞 ⋅ =
−1
𝑠𝑝𝑘+𝑠𝑞
⌊𝛽𝑛⌋
𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛
−1
𝑠𝑝𝑘+𝑠𝑞 𝑠 𝑠 (f) ‖(∑∞ /𝑢𝑝𝑘+𝑞 )) − (−1)𝑠𝑝𝑛+𝑠𝑞 (𝑢𝑝𝑛+𝑞 / 𝑘=𝑛 ((−𝑎1 ) 𝑠𝑝𝑛+𝑠𝑞 𝑠𝑝𝑛+𝑠𝑞−𝑠𝑝 𝑠 𝑎1 + 𝑢𝑝𝑛−𝑝+𝑞 /𝑎1 )‖ = 0, (𝑛 ≥ 𝑛13 ).
(−𝑎1 ) 𝑠 𝑢𝑝𝑘+𝑞
Taking the reciprocal of this expression yields
1 ≤ s < ⌊log(𝛼/𝑎1 ) 𝛼𝑝 𝑑𝑝 ⌋ ,
(33)
(37)
we have 𝑛
𝛼𝑠𝑛 𝑝 𝛼𝑠𝑛−𝑝𝑛 𝛼𝑠−𝑝 ( 𝑠 𝑝 ) < 1. 𝑠𝑛 ⋅ ℎ = 𝑠𝑛 𝑝𝑛 = 𝑎1 𝑎1 𝑑 𝑎1 𝑑
Consequently, ⌊𝛽𝑛⌋
In both cases, it follows that for any real number 𝛽 > 2 and positive integer 1 ≤ 𝑠 < ⌊log(𝛼/𝑎1 ) 𝛼𝑝 𝑑𝑝 ⌋, there exists 𝑛 ≥ 𝑛10 sufficiently large so that the modulus of the last error term of identity (35) becomes less than 1/2. This completes the proof of Theorem 13(c).
𝑠𝑝𝑘+𝑠𝑞
(−𝑎1 ) ∑ 𝑠 𝑢𝑠𝑝𝑘+𝑠𝑞 𝑘=𝑛
𝑠𝑝𝑘+𝑠𝑞
=
⌊𝛽𝑛⌋ ⌊𝛽𝑛⌋ 𝑎1 −𝑎1 𝑠𝑝𝑘+𝑠𝑞 1 ( + 𝑂 ( ) ) ∑ ∑ 𝑠𝑝𝑘+𝑠𝑞+𝑝𝑘+𝑞 𝑠 𝑐 𝑘=𝑛 𝛼 𝑑𝑝𝑘+𝑞 𝑘=𝑛 𝛼
Conflict of Interests
𝑎1 𝑠𝑝𝑛+𝑠𝑞 (−1)𝑠𝑝𝑛+𝑠𝑞 (−1)𝑠𝑝𝑛+𝑠𝑞 ⋅ 𝛼𝑠 − 𝑠 𝑠 = 𝑠 𝑠 ⋅ ( ) 𝛼 𝑐 (𝛼 − 𝑎1𝑠 ) 𝑐 (𝛼 − 𝑎1𝑠 )
The authors declare that there is no conflict interests regarding the publication of this paper.
𝑠𝑝𝑛
𝑎1 𝑎 𝑠𝑝⌊𝛽𝑛⌋+𝑠𝑞 ⋅ ( 1) + 𝑂 ( 𝑠𝑝𝑛+𝑝𝑛 ) 𝛼 𝛼 𝑑𝑝𝑛
Acknowledgments 𝑠𝑝𝑛
=
𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛
𝑎 𝑎 𝑎1 𝑠𝑝𝑛+𝑠𝑞 (−1)𝑠𝑝𝑛+𝑠𝑞 ⋅ 𝛼𝑠 + 𝑂 ( 1𝑠𝑝𝑛 ⋅ 1𝑠𝑝⌊𝛽𝑛⌋−𝑠𝑝𝑛 ) ⋅ ( ) 𝛼 𝛼 𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 ) 𝛼 𝑠𝑝𝑛
+ 𝑂(
𝑎1 1 ⋅ 𝑝𝑛 𝑝𝑛 ) 𝑠𝑝𝑛 𝛼 𝛼 𝑑 𝑠𝑝𝑛
=
𝑎 𝑎1 𝑠𝑝𝑛+𝑠𝑞 (−1)𝑠𝑝𝑛+𝑠𝑞 ⋅ 𝛼𝑠 + 𝑂 ( 1𝑠𝑝𝑛 ⋅ ℎ𝑝 ) , ⋅ ( ) 𝛼 𝛼 𝑐𝑠 (𝛼𝑠 − 𝑎1𝑠 )
𝑠⌊𝛽𝑛⌋−𝑠𝑛
where ℎ = max {(𝑎1
(38)
/𝛼𝑠⌊𝛽𝑛⌋−𝑠𝑛 ), (1/𝛼𝑛 𝑑𝑛 )}.
The authors express their gratitude to the referee for very helpful and detailed comments. This work is supported by the N.S.F. (11371291), the S.R.F.D.P. (20136101110014), the N.S.F. (2013JZ001) of Shaanxi Province, and the G.I.C.F. (YZZ12062) of NWU, China.
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