The sums of the reciprocals of Fibonacci polynomials and Lucas polynomials

In this article, we consider infinite sums derived from the reciprocals of the Fibonacci polynomials and Lucas polynomials, and infinite sums derived from the reciprocals of the square of the Fibonacci polynomials and Lucas polynomials. Then applying the floor function to these sums, we obtain several new equalities involving the Fibonacci polynomials and Lucas polynomials.

Mathematics Subject Classification (2010): Primary, 11B39.

For any variable quantity x, the Fibonacci polynomials F_n(x) and Lucas polynomials L_n(x) are defined by F_n+2(x) = xF_n+1(x) + F_n(x), n ≥ 0 with the initial values F₀(x) = 0 and F₁(x) = 1; L_n+2(x) = xL_n+1(x) + L_n(x), n ≥ 0 with the initial values L₀(x) = 2 and L₁(x) = x. For x = 1 we obtain the usual Fibonacci numbers and Lucas numbers. Let and , then from the properties of the second-order linear recurrence sequences we have

Various authors studied the properties of Fibonacci polynomials and Lucas polynomials, and obtained many interesting results, see [1-3]. Recently, several authors studied the infinite sums derived from the reciprocals of the Fibonacci numbers and Pell numbers, and obtained some important results. For example, Ohtsuka and Nakamura [4] studied the properties of the Fibonacci numbers, and proved the following conclusions:

Wenpeng and Tingting [5] studied the infinite sums derived from the Pell numbers, and obtained two similar results.

In this article, we considered infinite sums derived from the reciprocal of the Fibonacci polynomials and Lucas polynomials, and proved the following:

Theorem 1. For any positive integer x, we have

Theorem 2. For any positive integer x, we have

Theorem 3. For any positive integer x, we have

Theorem 4. For any positive integer x ≥ 2, we have

If x = 1 and x = 2, then from our theorems we can deduce the conclusions of [4,5]. Especially, we also have the following:

Corollary 1. For any positive integer n, we have the identities

Corollary 2. For any positive integer n, we have the identities

Corollary 3. For any positive integer n, we have the identities

In this section, we shall complete the proof of our theorems. We shall prove only Theorems 1 and 2, and the other two theorems are proved similarly and omitted. First we prove Theorem 1. We consider the case that n = 2m is even. At this time, Theorem 1 equivalent to

Now we prove that for any positive integers x and k ≥ 1,

This inequality equivalent to

or

applying the identities

the inequality (2.2) equivalent to

or

It is easy to check that the inequality (2.3) holds for any positive integers x and k ≥ 1. So the inequality (2.2) is true. Using (2.2) repeatedly, we have

On the other hand, we prove that for any positive integers x and k ≥ 1,

The inequality (2.5) equivalent to

or

It is easy to check that the inequality (2.6) holds for all positive integers x and k ≥ 1. So the inequality (2.5) is true. Using (2.5) repeatedly, we have

Now the inequality (2.1) follows from (2.4) and (2.7).

Similarly, we can consider the case that n = 2m + 1 is odd. Note that F₁(x) - F₀(x) - 1 = 0 and

So Theorem 1 is true if m = 0. If m ≥ 1, then our Theorem 1 equivalent to the inequality

First we can prove that for any positive integers x and k ≥ 1,

The inequality (2.9) equivalent to

or

It is easy to check that the inequality (2.10) holds for all positive integers x and k ≥ 1. So the inequality (2.9) is true. Using (2.9) repeatedly, we have

On the other hand, we prove that for any positive integers x and k ≥ 1,

The inequality (2.12) equivalent to

or

It is easy to check that the inequality (2.13) holds for all positive integers x and k ≥ 1. So the inequality (2.12) is true. Using (2.12) repeatedly, we have

Combining (2.11) and (2.14) we deduce the inequality (2.8). This proves Theorem 1.

Proof of Theorem 2. First we consider the case that n = 2m is even. At this time, Theorem 2 equivalent to

Now we prove that for any positive integers x and k ≥ 1,

So the inequality (2.16) equivalent to

or

It is clear that the inequality (2.17) holds for all positive integers x and k ≥ 1. So the inequality (2.16) is true. Using (2.16) repeatedly, we have

On the other hand, we prove that for any positive integers x and k ≥ 1,

So the inequality (2.19) equivalent to

or

It is clear that the inequality (2.20) holds for all positive integers x and k ≥ 1.

So the inequality (2.19) is true. Using (2.19) repeatedly, we have

Now the inequality (2.15) follows from (2.18) and (2.21).

Similarly, we can consider the case that n = 2m + 1 is odd. Note that xF₀(x)F₁(x) = 0 and

So Theorem 2 is true if m = 0. If m ≥ 1, then Theorem 2 equivalent to

First we prove that for any positive integers x and k ≥ 1,

So the inequality (2.23) equivalent to

or

It is clear that the inequality (2.24) is correct. So the inequality (2.23) is true.

Using (2.23) repeatedly, we have

On the other hand, we prove that for any positive integers x and k ≥ 1,

So the inequality (2.26) equivalent to

or

It is clear that inequality (2.27) is correct. So the inequality (2.26) is true.

Using (2.26) repeatedly, we have

Now the inequality (2.22) follows from (2.25) and (2.28). This proves Theorem 2.

The authors declare that they have no competing interests.

ZW proposed if we could obtain some similarly results in [4,5], and proved a part of the theorems. Wu Zhengang obtained the theorems and completed the proof. All authors read and approved the final manuscript.

The authors express their gratitude to the referee for very helpful and detailed comments. This work is supported by the N.S.F. of P.R.China (11071194).

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