What is the Convergence of the Fibonacci Sequence?

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In summary: Sorry if I miss something.In summary, the conversation revolved around proving the convergence of the sequence a_n/a_(n-1) as n approaches infinity. Different methods were proposed, such as using induction and setting up a quadratic equation, but it was pointed out that these methods do not prove convergence, only that there is a limit. It was also discussed that the limit must be independent of n and that the sequence must be increasing for all n > 1. The conversation concluded with the suggestion to solve for the limit by setting L = 1 + 1/L.
  • #1
al-mahed
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Hi,

I would like some help in this problem:

given the fibonacci sequence 1,2,3,5,8,13...

we know that a_n/a_(n-1) --> phi when n --> [tex]\infty[/tex]

I have tried a new proof in these terms:

My goal is to prove that when n --> [tex]\infty[/tex], a_n/a_(n-1) = a_(n+1)/a_n in the limit

I have notice that a_n^2= a_(n-1)*a_(n+1) [tex]\pm[/tex] 1 for some values

Dividing both sides of the equation by a_n*a_(n-1)

[tex]\frac{a_n}{a_n_-_1}[/tex] = [tex]\frac{a_n_+_1}{a_n}[/tex][tex]\pm[/tex] [tex]\frac{1}{a_n*a_n_-_1}[/tex]

I want to find the limits, because the second term in the right side of the equation --> 0 when n --> [tex]\infty[/tex]

But I don't know if the equation above is valid for all n, so I'll prove this by induction as follows:

supose that the expression above is valid for some n

by definition a_(n+1)=a_n + a_(n-1)

a_n^2= a_(n-1)*[a_n + a_(n-1)] [tex]\pm[/tex] 1 ==> a_n^2 - a_(n-1)^2= a_(n-1)*a_n [tex]\pm[/tex] 1 ==>

==> [a_n + a_(n-1)]*[a_n - a_(n-1)] = a_(n-1)*a_n [tex]\pm[/tex] 1

by definition a_n - a_(n-1)= a_(n-2) ==>

==> [a_n - a_(n-1)]* a_(n-2) = a_(n-1)*a_n [tex]\pm[/tex] 1

by definition a_n = a_(n-1) + a_(n-2) ==>

==> [a_n - a_(n-1)]* a_(n-2) = a_(n-1)*[a_(n-1) + a_(n-2)] [tex]\pm[/tex] 1 ==>

==> a_(n-2)*a_n + a_(n-1)*a_(n-2) = a_(n-1)^2 + a_(n-1)*a_(n-2) [tex]\pm[/tex] 1 ==>

==> a_(n-1)^2 = a_(n-2)*a_n [tex]\pm[/tex] 1

compare the two expressions

a_n^2= a_(n-1)*a_(n+1) [tex]\pm[/tex] 1

a_(n-1)^2 = a_(n-2)*a_n [tex]\pm[/tex] 1

most generally we have

[tex]\{a}{_i}{^2}[/tex]} = [tex]\{a_i_-_1*a_i_+_1}[/tex]} [tex]\pm[/tex] 1, with i=2,3,4,5,6...,n

proving by induction (sorry about the english and notation)

Hence,


Lim[tex]\{_n_-_>_i_n_f_i_n_i_t_y}[/tex][tex]\frac{a_n}{a_n_-_1}[/tex]} = Lim[tex]\{_n_-_>_i_n_f_i_n_i_t_y}[/tex][tex]\frac{a_n_+_1}{a_n}[/tex][tex]\pm[/tex] [tex]\frac{1}{a_n*a_n_-_1}[/tex]}


Lim[tex]\{_n_-_>_i_n_f_i_n_i_t_y}[/tex][tex]\pm[/tex] [tex]\frac{1}{a_n*a_n_-_1}[/tex]} = 0

in the limit [tex]\frac{a_n}{a_n_-_1}[/tex] = [tex]\frac{a_n_+_1}{a_n}[/tex]

Why this do not prove the convergence?
 
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  • #2
I'm not sure what you're trying to prove. When you say convergence, are you trying to prove that in the [tex] \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} [/tex] = 1?

That would imply that the sequence of numbers goes to some final value, although the sum of those numbers would still diverge since that number isn't 0.

The sequence grows uniformly, a_n+1 > a_n, so a_n+1/a_n > 1 always. What you've done is prove that

[tex] \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} [/tex] = constant

With further math you'll be able to solve for that constant as phi, the golden ratio. You can set up a quadratic equation, solve for it, and get phi.

Starter:
[tex] a_{n+1} = a_n + a_{n-1} [/tex] divide both sides by a_n

[tex] \frac{ a_{n+1} }{a_n} = 1 + \frac{ a_{n-1} }{a_n} } [/tex]

Take the limit as n goes to infinity, define
[tex] \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = x [/tex], and go.
 
  • #3
BoTemp, I am not so good in english. What I'm trying to prove is that the sequence of the terms a_n/a_(n-1), ... is convergent to a value, not the sum of these ratios (not series, like 1 + 1/2 + 1/3 + ...).

I know the formulas to find the golden ratio (1 + 5^1/2)/2, I would like to know if that proof is correct, if I could conclude that there is a limit or not.

thanks
 
  • #4
"My goal is to prove that when n --> , a_n/a_(n-1) = a_(n+1)/a_n in the limit"

That's simply adjusting the indices. Let j= n-1. Then an-1= aj and an= aj+1. The sequence an/an-1 is exactly the same as aj+1/aj which is, in turn, exactly the same as an+1/an just labled differently. In the limit as n goes to infinity, j also goes to infinity and the two sequences have the same limit.
 
  • #5
a_n/a_(n-1) [tex]\neq[/tex] a_(n+1)/a_n for all n that you can test by yourself, is quite different to say
[tex] \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} [/tex] = [tex] \frac{a_n}{a_{n-1}}[/tex]
 
  • #6
I may be misunderstanding, but the limit value cannot include subscripts "n" or "n-1."
 
  • #7
EnumaElish said:
I may be misunderstanding, but the limit value cannot include subscripts "n" or "n-1."

what you mean EnumaElish?
 
  • #8
In your statement

[tex] \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{a_n}{a_{n-1}}[/tex]

the right-hand side cannot have a_n or a_{n-1}. The limit must be independent of n.

Did you mean [itex] \lim_{n \rightarrow \infty} \left({a_{n+1}}\left/{a_n}\right.\right) = \lim_{n \rightarrow \infty} \left({a_n}\left/{a_{n-1}}\right.\right)[/itex]? [1]

If so, [1] does not prove convergence because it presumes that a limit exists in the first place. If there is no limit then [1] has no meaning.
 
Last edited:
  • #9
I meant [1] as you said, sorry my mistake.

I see what you mean, my dificult is:

if [1] does not prove the convergence, why the procedure below do?

[tex] \frac{ a_{n+1} }{a_n} = 1 + \frac{ a_{n-1} }{a_n} } [/tex] ==>

==> phi = 1 + 1/phi ==> phi = (1 + 5^1/2)/2

Here I'm suposing that there is a limit like the other solution, right?

Thanks for your help!
 
  • #10
Since a(n) > 0 for all n > 0 and a(n+1) = a(n) + a(n-1), {a(n)} is an increasing sequence for all n > 1.

Let r(n+1) = a(n+1)/a(n), then r(n+1) > 1 for all n > 1.

Suppose lim r(n) = infinity. Then lim 1/r(n) = 0. But since lim r(n+1) = 1 + lim 1/r(n), this implies infinity = 1, a contradiction. Therefore I can conclude that lim r(n) = L < infinity.

Then you can solve L = 1 + 1/L.
 
  • #11
"Therefore I can conclude that lim r(n) = L < infinity."

as 0 < a(n-1)/a_n < 1 ==> 1 < a(n-1)/a_n + 1 < 2 ==> 1 < a(n+1)/a_n < 2

and if the values oscillate between 1 and 2? you prove that the limit do not diverges to infinity, but this not implies that there is a limit, am I right?

f(x) = sen(x), lim sen(x) is not infinite when x --> infinity, but there is no limit, the function oscillate between -1 and 1
 
  • #12
Yes, you are right; which makes it clear that one cannot presume a limit exists.
 
  • #13
I see, you're right...

We could prove convergence starting with this fact: 1 < a(n+1)/a_n < 2 , showing (I presume) that the sequence is bounded

to prove that it is monotonic, we should work with [tex]\frac{a_n}{a_n_-_1}[/tex] = [tex]\frac{a_n_+_1}{a_n}[/tex][tex]\pm[/tex] [tex]\frac{1}{a_n*a_n_-_1}[/tex]

for n even [tex]\frac{a_n}{a_n_-_1}[/tex] = [tex]\frac{a_n_+_1}{a_n}[/tex] - [tex]\frac{1}{a_n*a_n_-_1}[/tex]

for n odd [tex]\frac{a_n}{a_n_-_1}[/tex] = [tex]\frac{a_n_+_1}{a_n}[/tex] + [tex]\frac{1}{a_n*a_n_-_1}[/tex]

a_n < a_(n+1) < a_(n+2) < a_(n+3) < ... ==> [tex]\frac{1}{a_n*a_n_-_1}[/tex] --> 0 ==> the sequence [tex]\frac{a_n}{a_n_-_1}[/tex] increase in monotonic fashion, bounded above for n odd and decrease in monotonic fashion, bounded below for n even
 

1. Why is there a limit in mathematics?

The concept of a limit in mathematics is used to define the behavior of a function as it approaches a particular point or value. It helps us understand the continuity of a function and how it behaves in the vicinity of a given value. Without limits, many mathematical concepts and calculations would not be possible.

2. What does it mean when a limit doesn't exist?

A limit not existing means that the function does not approach a specific value or approaches different values from different directions. This can happen when there is a discontinuity in the function or when there is an asymptote. It indicates that the function does not have a well-defined behavior at that particular point.

3. Can a function have a limit at one point but not at another?

Yes, a function can have a limit at one point but not at another. This is because the behavior of a function at a particular point is dependent on the behavior of the function in its surrounding points. So, even if a function has a limit at one point, it may not have a limit at a different point due to a different behavior in its vicinity.

4. How can we determine if a limit exists?

In order for a limit to exist, the function must approach the same value from both the left and right sides of the point in question. This means that the limit can be evaluated by taking the limit from the left side and the limit from the right side and seeing if they are equal. If they are, then the limit exists. If they are not, then the limit does not exist.

5. What are the different types of limits?

The different types of limits include finite limits, infinite limits, and limit at infinity. A finite limit is one where the function approaches a particular value. An infinite limit is one where the function approaches either positive or negative infinity. A limit at infinity is one where the function approaches a particular value as the independent variable goes to infinity.

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