New Results for the Fibonacci Sequence Using Binet's ... - Hikari Ltd

International Mathematical Forum, Vol. 13, 2018, no. 6, 259 - 266 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/imf.2018.8321

New Results for the Fibonacci Sequence Using Binet’s Formula Reza Farhadian Department of Statistics, Lorestan University Khorramabad, Iran Rafael Jakimczuk Divisi´on Matem´atica, Universidad Nacional de Luj´an Buenos Aires, Argentina c 2018 Reza Farhadian and Rafael Jakimczuk. This article is distributed Copyright under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract Let k ≥ 1 and h = 0, . . . , k − 1. In this note we study the monotonicity of the sequences (Fkn+h )1/n , where Fn denotes the n-th Fibonacci number. In particular, we prove that the sequences (F2n )1/n and (F2n+1 )1/n are strictly increasing for n ≥ 1. We also obtain asympP Q totic formulae for nk=1 log Fk , nk=1 Fk and prove some limits.

Mathematics Subject Classification: 11B39, 11B99 Keywords: Sequences, monotonicity, Fibonacci numbers

1

Introduction and Preliminary Notes

In 1202, the Italian mathematician Leonardo Fibonacci (1175-1251), through a problem (rabbit problem) in his book Liber Abaci, he introduced the following sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...

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This sequence is called the Fibonacci sequence, and its terms are known as Fibonacci numbers. The Fibonacci sequence has a simple rule. In fact, starting with 0 and 1, every next number is found by adding up the two numbers before it. In mathematical terms, if Fn be the n-th Fibonacci number, then Fn = Fn−1 + Fn−2 . with F0 = 0 and F1 = 1. There are many methods and explicit formulas to finding the n-th Fibonacci number. For example, the well-known Binet’s formula (discovered by the French mathematician Jacques Philippe Marie Binet (1786-1856) in 1843) states that: √ !n # √ !n " 1− 5 1+ 5 1 − . Fn = √ 2 2 5 The Binet’s formula can also be written as ϕn − (− ϕ1 )n √ Fn = , 5

(1)

√

where ϕ = 1+2 5 (≈ 1.6180339887...), is the golden ratio. There are many papers and books on the Fibonacci numbers. See, for example, [1]. In this article, we are interested in studying the monotonicity of sequences consisting of Fibonacci numbers and also in asymptotic formulae and limits where the Fibonacci Numbers appear.

2

Main Results Lemma 2.1 The following limit holds log(1 + x) = 1. x−→0 x lim

Proof. Use L’ Hospital’s rule. The lemma is proved. Theorem 2.2 Let k ≥ 1 and h = 0, . . . , k − 1. If either h = 0 or h = 1 1 then the sequence (Fkn+h ) n is strictly increasing from a certain value of n and 1 its limit is ϕk . On the other hand, if h ≥ 2 then the sequence (Fkn+h ) n is strictly decreasing from a certain value of n and its limit is ϕk .

261

Results for the Fibonacci sequence using Binet’s formula

Proof. We have (Binet’s formula) 

ϕ

1

kn+h

(Fkn+h ) n = 



− ϕ1

− √ 5

kn+h  n1   !1

!1

n 1 (−1)kn+h+1 1 c = ϕ 1+ = ϕk a n 1 + n n 2h 2k ϕ (ϕ ) b    1 1 c = exp k log ϕ + log a + log 1 + n n n b ! 1 1 f (n)c = exp k log ϕ + log a + , n n bn

k

ϕh √ 5

n



1

n

(2)

kn+h+1

h

ϕ where a = √ , c = (−1)ϕ2h = O(1), b = ϕ2k > 1 and f (n) → 1 by Lemma 5 2.1. Note that an immediate consequence of the second line of equation (2) is the limit of the sequence is ϕk . Let us consider the sequence (see the last line of equation (2))

1 f (n)c log a 1 1 = k log ϕ + + O . An = k log ϕ + log a + n n bn n nbn 



We have − log a 1 An+1 − An = +O n(n + 1) nbn 



1 n+1 = − log a + O n(n + 1) bn 





.

Therefore,  h  from a certain value of n the sign of An+1 −An is the sign of − log a = ϕ . If either h = 0 or h = 1 then − log a > 0 and consequently the − log √ 5 sequence An is strictly increasing (from a certain value of n). On the other hand, if h ≥ 2 then − log a < 0 and consequently the sequence An is strictly decreasing (from a certain value of n). The theorem is proved. In the case k = 2 we can to prove that both sequences are strictly increasing from n = 1. Lemma 2.3 Let a > 1.2841 is a constant. Then for every x > 0 we have log(a) 2x a log(a) + 2x a 2x

!

> (a2x −

1 1 ) log(a2x − 2x ). 2x a a

Proof. Let us consider the real function h(x) = a2x − a12x . The derivative of 4x +1) h(x) is h0 (x) = 2 log(a)(a that is positive for a > 1 and every real number a2x x, which means that the function h(x) (with a > 1) is strictly increasing. In

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Reza Farhadian and Rafael Jakimczuk

addition by calculation we have h(x) > 1 for a ≥ 1.2841 and x > 0. Now let’s continue the proof. We have 1 2xa2x log(a) = a2x log(a2x ) > a2x log(a2x − 2x ) = a2x log(h(x)). a We have log(h(x)) > 0 for a ≥ 1.2841 and x > 0 (since h(x) > 1). Hence, we have 2xa2x log(a) − a2x log(a2x −

1 ) > 0, a2x

a ≥ 1.2841, ∀x > 0.

(3)

and a12x log(a2x − a12x ) both are positive for a > 1 and x > 0, Since 2x alog(a) 2x by adding these to the left side of inequality (3), we have 2xa2x log(a)−(a2x −

1 1 2x log(a) ) log(a2x − 2x )+ > 0, 2x a a a2x

∀x > 0,

a ≥ 1.2841.

∀x > 0,

a ≥ 1.2841.

Consequently log(a) 2x a2x log(a) + 2x a

!

> (a2x −

1 1 ) log(a2x − 2x ), 2x a a

The lemma is proved. 1

Theorem 2.4 The sequence (F2n ) n is strictly increasing for n ≥ 1. Proof. Using Binet’s formula we have 1



1



ϕ2n − (− ϕ1 )2n n ϕ2n − ( ϕ1 )2n n 1     . n √ √ (F2n ) = = 5 5

(4)

Let us consider the following continuous function corresponding to (4): 1



ϕ2x − ( ϕ1 )2x x   , √ f (x) = 5

x > 0.

We obtain the derivative of f (x) as follows: 

1/x 

ϕ2x − ϕ12x 0 f (x) = √ 1 x2 ( 5) x



2x(ϕ2x log(ϕ) + (ϕ2x − 

log(ϕ) ) ϕ2x

1 ) ϕ2x

1/x



1 − log (ϕ2x − 2x ) ϕ

log(5) ϕ2x − ϕ12x √ + . 2x2 ( 5)1/x Using Lemma 2.3 we know that f 0 (x) > 0 for x > 0. Since the function f (x) has a positive derivative for x > 0, so f (x) is strictly increasing for every 1 x > 0, consequently the sequence (F2n ) n is strictly increasing. The theorem is proved.

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Results for the Fibonacci sequence using Binet’s formula

Lemma 2.5 If x > 0 then the following inequality holds 0
0 and f (0) = 0. The lemma is proved. 1

Theorem 2.6 The sequence (F2n+1 ) n is strictly increasing for n ≥ 1. Proof. If k = 2 and h = 1 equation (2) gives !

1 ϕ 1 1 1 (F2n+1 ) = exp 2 log ϕ + log √ + f (n) 2 4 n n n ϕ (ϕ ) 5 1 n

!

(n ≥ 1)

(5)

where, by Lemma 2.5, we have 0 < f (n) < 1. Let us consider the sequence (see (5)) !

1 ϕ 1 1 1 An = 2 log ϕ + log √ + f (n) 2 4 n . n n ϕ (ϕ ) 5 We have !

!

1 ϕ An+1 − An = − log √ + F (n) , n(n + 1) 5

(6)

where F (n) = f (n + 1)

1 1 n+1 n − f (n) , n+1 2 ϕ (ϕ4 ) ϕ2 (ϕ4 )n

and consequently 1 |F (n)| ≤ 2 ϕ

n+1 n + n+1 (ϕ4 )n (ϕ4 )

Since the function

x (ϕ4 )x

!

≤

2 n+1 4 n 4 n ≤ n ≤ 2 4 2 4 ϕ (ϕ ) ϕ (ϕ ) ϕ6

(n ≥ 1) (7)

has negative derivative for x ≥ 1. 



On the other hand, we have the inequality − log √ϕ5 > ϕ46 . Therefore equations (6) and (7) give An+1 − An > 0 if n ≥ 1. The theorem is proved. Theorem 2.7 The following asymptotic formulae and limits hold. n X

n2 n ϕ log Fi = log ϕ + log + C + o(1), 2 2 5 i=1 



(8)

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Reza Farhadian and Rafael Jakimczuk

where ∞ X

(−1)k+1 . C= log 1 + ϕ2k k=1 n Y

!

0

Fi ∼ C ϕ

n2 2



i=1

ϕ 5

(9)

n 2

,

(10)

where ∞ Y

0

C =

(−1)k+1 1+ . ϕ2k !

k=1

1

lim (F1 F2 · · · Fn ) n2 =

n→∞

√ n lim

n→∞

√ n lim n→∞ √ n lim

n→∞

(11)

√ ϕ.

(12)

F1 F2 · · · Fn = 0. Fn

F1 F2 · · · Fn √ = Fn

s

(13)

ϕ √ . 5

(14)

log F3 log F4 · · · log Fn 1 = log Fn e

(15)

Let k be a positive integer, we have 

lim

k

k

k)

(log F3 )(3 ) (log F4 )(4 ) · · · (log Fn )(n



k+1 nk+1

log Fn

n→∞

log Fn+1 log Fn

lim

n→∞

=

1 ek+1

(16)

!n

=e

Proof. We have (Binet’s formula) 1 1 Fn = √ ϕn + (−1)n+1 n ϕ 5

!

1 (−1)n+1 = √ ϕn 1 + ϕ2n 5

!

(n ≥ 1)

Therefore 1 Fn ∼ √ ϕn , 5

(17)

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Results for the Fibonacci sequence using Binet’s formula

and 1 (−1)n+1 log Fn = n log ϕ − log 5 + log 1 + . 2 ϕ2n !

(18)

Hence ∞ X n ϕ (−1)k+1 n2 log ϕ + log + log 1 + log Fi = 2 2 5 ϕ2k i=1 k=1

n X





!

+ o(1).

Note that the series ∞ X

(−1)k+1 log 1 + ϕ2k k=1

!

=

∞ X

f (k)

k=1

(−1)k+1 ϕ2k

where f (k) → 1 (by Lemma 2.1) converges absolutely. This proves equations (8) and (9). Equations (10) and (11) are an immediate consequence of equations (8) and (9). Limit (12) is an immediate consequence of equation (10). Limits (13) and (14) are an immediate consequence of equations (10) and (16). Equation (17) gives log log Fn = log n + log log ϕ + o(1) √

From the Stirling’s formula n! ∼ n X

√ 2πnn n n e

(19)

we obtain

log i = n log n − n + o(n)

(20)

i=1

Therefore (see (18) and (19)) ! √ n log F3 log F4 · · · log Fn 1 log = (log log F3 + · · · + log log Fn ) log Fn n ! n 1 X − log log Fn = log i + n log log ϕ + o(n) n i=1 − (log n + log log ϕ + o(1)) = −1 + o(1) That is, equation (15). The proof of equation (16) is the same as the proof of equation (15). Note that we have the well-known asymptotic formula 1k + 2k + · · · + nk ∼

nk+1 k+1

and since the function xk log x is strictly increasing and integration by parts we have n X i=1

k

i log i =

Z n 1

  nk+1 1 k+1 k+1 x log x dx+O n log n = log n− n +o n k+1 (k + 1)2 k



k



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Reza Farhadian and Rafael Jakimczuk

The theorem is proved. Acknowledgements. Rafael Jakimczuk is very grateful to Universidad Nacional de Luj´an.

References [1] N. N. Vorob`ev, Fibonacci Numbers, Springer, Birkh¨auser Verlag, Basel, 2002. https://doi.org/10.1007/978-3-0348-8107-4 Received: April 9, 2018; Published: April 24, 2018