On the Higher Power Sums of Reciprocal Higher-Order Sequences

Let {u n} 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.


Introduction
The so-called Fibonacci zeta function and Lucas zeta function, defined by where the and 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: 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][7][8][9][10][11][12][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][15][16]. In particular, in [16], Kılıç and Arıkan studied a problem which is a little different from that of [3], namely, that of determining the nearest integer to (∑ ∞ = (1/V )) −1 .
Specifically, suppose that ‖ ‖ = ⌊ + (1/2)⌋ (the nearest integer function) and {V } ≥0 is an integer sequence satisfying the recurrence formula for any positive integer ≥ and > . Then we can conclude that there exists a positive integer 0 such that for all > 0 .
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 > , the th-order linear recursive sequence { } is defined as follows: with initial values ∈ N for 0 ≤ < and at least one of them not being zero. For any positive real number > 2 and any 2 The Scientific World Journal positive integer 1 ≥ 2 ≥ ⋅ ⋅ ⋅ ≥ ≥ 1, there exists a positive integer 1 such that In particular, taking → +∞, there exists a positive integer 2 such that It seems difficult to deal with (∑ ∞ = (1/ )) −1 for all integers ≥ 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.

Proposition 2.
For any positive integer ≥ 1, we have the identity 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 ( > 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 { } and obtain several new and interesting identities. The results are as follows.

Corollary 4. Let { } be an th-order sequence defined by
For positive real number 1 < ≤ 2, whether there exits an identity for is an interesting open problem.

Several Lemmas
To complete the proof of our theorem, we need two lemmas.
Then for the polynomial we have the following: (I) polynomial ( ) has exactly one positive real zero with 1 < < 1 + 1; (II) other − 1 zeros of ( ) lie within the unit circle in the complex plane.
Proof. From Lemma 2 of [16], the closed formula of is given by where > 0, > 1, 1 < < 1 + 1, and is the positive real zero of ( ). Now we prove Lemma 6 by mathematical induction. From formula (14), we have That is, the lemma holds for = 2. Suppose that for all integers 2 ≤ ≤ we have The Scientific World Journal 3 Then for = + 1 we have That is, Lemma 6 also holds for = + 1. This completes the proof of Lemma 6 by mathematical induction.

Proof of Theorem 3
In this section, we shall complete the proof of Theorem 3. From the geometric series as → 0, we have Using Lemma 6, we have Consequently, where ℎ = max {( Taking the reciprocal of this expression yields Case 2. If ℎ = (1/ ), for any positive integer 1 ≥ 2, 1 < ( / 1 ) < holds. Then for any positive integer with we have In both cases, it follows that for any real number > 2 and positive integer 1 ≤ < ⌊log ( / 1 ) ⌋ there exists ≥ 3 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.
The Scientific World Journal Taking the reciprocal of this expression yields For any positive integer with we have So there exists ≥ 4 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.

Computation
We can determine the power of different sequence by MATHEMATICA as the following examples.

Example 7.
Let be the second-order linear recursive sequence (see Table 1).

Example 8. Let
be the third-order linear recursive sequence (see Table 2). Example 9. Let be the fifth-order linear recursive sequence (see Table 3). Therefore, we may immediately deduce the following corollaries.   Table 3 log / 1

Related Results
The following results are obtained similarly.     11 , 12 , and 13 depending on 1 , 2 , . . ., and such that the following hold.
Proof. We shall prove only (c) in Theorem 13 and other identities are proved similarly. From identity (19), we have Consequently, In both cases, it follows that for any real number > 2 and positive integer 1 ≤ < ⌊log ( / 1 ) ⌋, there exists ≥ 10 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).