Several Identities Involving the Fibonacci Polynomials and Lucas Polynomials

In this paper, the authors consider infinite sums derived from the reciprocals of the Fibonacci polynomials and Lucas polynomials. Then applying the floor function to the reciprocals of these sums, the authors obtain several new identities involving the Fibonacci polynomials and Lucas polynomials.


Introduction
For any variable quantity x, the Fibonacci polynomials F n (x) and Lucas polynomials L n (x) are defined by F n+ (x) = xF n+ (x) + F n (x), n ≥ , with the initial values F  (x) =  and Ohtsuka and Nakamura [] studied the partial infinite sums of reciprocal Fibonacci numbers and proved the following conclusions: if n is even and n ≥ ; F n- - if n is odd and n ≥ . u n = pu n- + qu n- + u n- + · · · + u n-k , and they proved that there exists a positive integer n  such that where · denotes the nearest integer. (Clearly, x = x +   .) In this paper, we consider the subseries of infinite sums derived from the reciprocals of the Fibonacci polynomials and Lucas polynomials and prove the following.
Theorem  For any positive integer x, n and even a ≥ , we have Theorem  For any positive integer x and odd b ≥ , we have if n is even and n ≥ ; if n is odd and n ≥ .
if n is even and n ≥ ; Particularly, for x ≥ , we have

Proof of theorems
To complete the proof of our theorems, we need the following lemma.
Lemma For any positive integer x, m, and n, This proves identities () and ().
Now we shall complete the proof of our theorems. We shall prove only Theorems (), (), (), and (), and other identities are proved similarly and omitted. First, we prove Theorem (). http://www.journalofinequalitiesandapplications.com/content/2013/1/205 For any positive integer x, k and even a ≥ , using identity (), we have Since F n (x) and L n (x) are monotone increasing for n and a fixed positive integer x, we have L a (x)- > , F ak (x)-F ak-a (x) > , and F ak+a (x)-F ak (x) >  for any positive integer x, k and even a ≥ . Hence the numerator of the right-hand side of the above identity is positive for any positive integer x, k and even a ≥ , so we get On the other hand, we prove that for any positive integer x, k and even a ≥ , . http://www.journalofinequalitiesandapplications.com/content/2013/1/205 Using identity (), the above inequality is equivalent to Since F n (x) and L n (x) are monotone increasing for n and a fixed positive integer x, we have any positive integer x, k and even a ≥ . Using identity (), we have Hence the numerator of each part in parentheses of the left-hand side of inequality () is positive, so inequality () holds for any positive integer x, k and even a ≥ . Hence inequality () is true. Using () repeatedly, we have Now inequality () follows from () and (). This proves Theorem ().
Proof of Theorem () Now we prove Theorem (). Theorem () is equivalent to For any positive integer x, k and even a ≥ , using identities () and (), we have .
() http://www.journalofinequalitiesandapplications.com/content/2013/1/205 Since F n (x) and L n (x) are monotone increasing for n and a fixed positive integer x, we have >  for any positive integer x, k and even a ≥ . Hence the numerator of the right-hand side of the above identity is positive for any positive integer x, k and even a ≥ , so we get On the other hand, we prove that for any positive integer x, k and even a ≥ , Using identities () and (), the above inequality is equivalent to Since L n (x) are monotone increasing for n and a fixed positive integer x, we have L a (x) -L  (x) >  for any positive integer x and even a ≥ . On the other hand, we have Hence the numerator of the left-hand side of inequality () is positive, so inequality () holds for any positive integer x, k and even a ≥ . Hence inequality () is true. Using () repeatedly, we have Proof of Theorem () First we consider the case that n = m ≥  is even. At this time, for any odd b ≥ , Theorem () is equivalent to Now we prove that for any positive integer x, k and odd b ≥ , .
Using identities () and (), the above inequality is equivalent to Since F n (x) is monotone increasing for n and a fixed positive integer x, we have F bk+b (x) -F bk+b (x) > , F bk+b (x) -F bk-b (x) > , and F bk (x) -F bk-b (x) >  for any positive integer x, k and odd b ≥ . Hence the numerator of each part in parentheses of the left-hand side of inequality () is positive, so inequality () holds for any positive integer x, k and odd b ≥ . Hence inequality () is true. Using () repeatedly, we have On the other hand, we prove that for any positive integer x, k and odd b ≥ , Using identities () and (), the above inequality is equivalent to Since F n (x) and L n (x) are monotone increasing for n and a fixed positive integer x, using identity (), we have Hence the numerator of each part in parentheses of the left-hand side of inequality () is positive, so inequality () holds for any positive integer x, k and odd b ≥ . Hence inequality () is true. Using () repeatedly, we have

Now inequality () follows from () and ().
Similarly, we can consider the case that n = m +  ≥  is odd. At this time, for any odd b ≥ , Theorem () is equivalent to the inequality First we can prove that for any positive integer x, k and odd b ≥ , () http://www.journalofinequalitiesandapplications.com/content/2013/1/205

Inequality () is equivalent to
Using identities () and (), the above inequality is equivalent to Since F n (x) and L n (x) are monotone increasing for n and a fixed positive integer x, using identity (), we have and Hence the numerator of each part in parentheses of the left-hand side of inequality () is positive, so inequality () holds for any positive integer x, k and odd b ≥ . Hence inequality () is true. Using () repeatedly, we have On the other hand, we prove that for any positive integer x, k and odd b ≥ , .
Using identities () and (), the above inequality is equivalent to () http://www.journalofinequalitiesandapplications.com/content/2013/1/205 Since F n (x) is monotone increasing for n and a fixed positive integer x, using identity (), we have Hence the numerator of each part in parentheses of the left-hand side of inequality () is positive, so inequality () holds for any positive integer x, k and odd b ≥ . Hence inequality () is true. Using () repeatedly, we have

Combining () and (), we deduce inequality (). This proves Theorem ().
Proof of Theorem () First we consider the case that n = m ≥  is even. At this time, for any odd b ≥ , Theorem () is equivalent to Now we prove that for any positive integer k, x ≥  and odd b ≥ , .
Using identity (), the above inequality is equivalent to Hence the numerator of the left-hand side of inequality () is positive, so inequality () holds for any positive integer k, x ≥  and odd b ≥ . Hence inequality () is true. Using () repeatedly, we have On the other hand, we prove that for any positive integer k, x ≥  and odd b ≥ , .
Using identity (), the above inequality is equivalent to Since L n (x) is monotone increasing for n and a fixed positive integer x, for any positive integer k, x ≥  and odd b ≥ , we have > L bk+b- (x) -L bk+b (x) > , and L bk (x) -L bk-b (x) > , L bk (x) -L bk-b (x) > . Hence the numerator of the left-hand side of inequality () is positive, so inequality () holds for any positive integer k, x ≥  and even b ≥ . Hence inequality () is true. http://www.journalofinequalitiesandapplications.com/content/2013/1/205 Now inequality () follows from () and (). Similarly, we can consider the case that n = m +  ≥  is odd. At this time, for any odd b ≥ , Theorem () is equivalent to the inequality Now we prove that for any positive integer k, x ≥  and odd b ≥ , .
Using identity (), the above inequality is equivalent to On the other hand, we prove that for any positive integer k, x ≥  and odd b ≥ , Inequality () is equivalent to .
Using identity (), the above inequality is equivalent to