A NOTE ON TWO FUNDAMENTAL RECURSIVE SEQUENCES

. In this note, we establish some general results for two fundamental recursive sequences that are the basis of many well-known recursive sequences, as the Fibonacci sequence, Lucas sequence, Pell sequence, Pell-Lucas sequence, etc. We establish some general limit formulas, where the product of the ﬁrst n terms of these sequences appears. Furthermore, we prove some general limits that connect these sequences to the number e ( ≈ 2 . 71828 ... ) .


Introduction
In mathematics, a recursive sequence is a sequence in which terms are defined using one or more previous terms which are given. There are many recursive sequences and the most well-known of them are the following two fundamental and primordial sequences:  if n = 1, F n−1 + F n−2 otherwise, The Binet forms for F n and L n are Further examples of sequences in recursive forms (1.1) and (1.2) are the Pell and Pell-Lucas sequences that the first is of the recursive form (1.1) and the second is of the recursive form (1.2). These sequences are defined as follows (for n ≥ 0), respectively: Thus, the following closed-form solutions (according to (1.3) and (1.4)) exist for the Pell and Pell-Lucas numbers, respectively: For more information about the above sequences see [1]- [6]. The aim of this note is to establish some general results for fundamental recursive sequences (1.1) and (1.2). We prove some general limit formulas, where the product of the first n terms of sequences (1.1) and (1.2) appears. We also prove some limit formulas that connect sequences (1.1) and (1.2) to the number e(≈ 2.71828...).

Preliminaries
In this section we present some preliminary results. Note that throughout this paper, the symbol ∼ means asymptotic equivalence.
Lemma 2.1. If {U n } n≥0 is a sequence in recursive form (1.1) corresponding to m > 0, then and if {V n } n≥0 is a sequence in recursive form (1.2), then is absolutely convergent.
Proof. By simple calculations we have the following inequalities: We have also the following well-known geometric power series:

Inequality (2.2) and identity (2.4) give
i odd Therefore the series of positive terms i odd log 1 + 1 α 2i s , where the sum runs over all odd numbers i, converges. Inequality (2.3) gives where the series Therefore the series of positive terms i even − log 1 − 1 α 2i s , where the sum runs over all even numbers i, also converges. Now, we have where the two series in the right hand side of (2.5) we have proved are convergent. Therefore the series is absolutely convergent. The lemma is proved.

Main results
We start this section with the following theorem.
Theorem 3.1. Let {U n } n≥0 be a sequence in recursive form (1.1) corresponding to m ≥ 1. Then for any s ∈ N the following limit holds: Proof. For a sequence in recursive form (1.1), we have (see (1.3)) Therefore, Hence, Consequently, for any s ∈ N, we have We know that if m > 0, then α = m+ √ m 2 +4 2 > 1. Since x s log α (with s ∈ N and α > 1) is strictly increasing and positive in the interval [1, ∞), we find that This completes the proof.  is called the golden ratio and is usually denoted by the Greek letter ϕ (phi). It is well-known that the ratio of two consecutive Fibonacci numbers tends to the golden ratio, i.e., The golden ratio ϕ, can also be expressed exactly by the following infinite series of continued fractions and that of nested square roots (see, for example, [8]): Moreover, if ℘(x) denotes the counting function of the Fibonacci numbers, i.e., the number of F n not exceeding x, then (see [3]) lim n→∞ ℘(n) ℘(1) + ℘(2) + · · · + ℘(n) = ϕ.
Here, the following example gives a new expansion of the golden ratio ϕ.
Example 3.3. The Fibonacci sequence {F n } n≥0 is of the recursive form (1.1) corresponding to m = 1. Hence, by Theorem 3.1 the following limit holds: In particular, if s = 1, 2, and 3, we have respectively, Theorem 3.4. If {U n } n≥0 is a sequence in recursive form (1.1) corresponding to m ≥ 1, then and if {V n } n≥0 is a sequence in recursive form (1.2) corresponding to m ≥ 1, then Proof. We first prove (3.5). For a sequence in recursive form (1.1) we have (see (3.2)) Hence, By subtracting n 2 2 log α from both sides of (3.7), then multiplying by 1 n , we obtain Now, we take the exponential of both sides of (3.8) to obtain Taking the limit of (3.9) as n → ∞ we get We know that for any recursive sequence of the form (1.
converges absolutely, hence, 1 tends to zero as n → ∞, consequently (3.11) Hence, (3.10) and (3.11) give Combining the property U n ∼ α n ∆ (see (2.1)) with (3.12), this completes the proof of (3.5). For a sequence {V n } n≥0 in recursive form (1.2) the proof of relation (3.6) is the same as above and easier. The theorem is proved. Next, we prove some limit formulas that connect sequences in recursive forms (1.1) and (1.2) to the number e.
Therefore if we put, for sake of simplicity, c n = log ∆ −1 +log 1+  Here, we shall recall the well-known prime number theorem (PNT), which states that the n-th prime number p n is asymptotically equivalent to n ln n (i.e., p n ∼ n ln n). We use the PNT to prove the next theorem.
, that is, the U n+1 -th prime number.
Proof. The function ln x is continuous on the interval [U n , U n+1 ] for all n ∈ N. By the integral mean value theorem, we have U n+1 U n ln x dx = (U n+1 − U n ) ln c = U n−1 ln c for some c with U n < c < U n+1 . Hence Since ln U n+1 ∼ ln U n (by Lemma 2.1), we have that is, Now, let us recall the well-known proposition (see [7, page 332]) that states for two series of positive terms Using this fact and by use of (3.15), we have (use also PNT),

This gives
This completes the proof.
Remark 3.9. By a proof similar to the proof of Theorem 3.8, it can be shown that relation (3.14) holds for a sequence {V n } n≥0 in recursive form (1.2) corresponding to m = 1. where φ n is the L n+1 -th prime number.