# Finding limit of the Fibonacci sequence

• member 731016
member 731016
Homework Statement
I am trying to understand whether part (c) of the problem below is wrong. I have also provided my working and the solutions.
Relevant Equations
Sequence notation ##\{a_n\}##
For this problem,

My working is

##\{\frac{a_{n + 2}}{a_{n+1}}\} = \{1 + \frac{a_n}{a_{n +1}}\}## using part (b)
Then

##\lim_{n \to \infty} b_n = \lim_{n \to \infty} (1 + \frac{a_n}{a_{n +1}})## which should equal ##L##.

However, does someone please know why it does this cause the sequence to converge to ##L##. The solution does not explain anything (just simply states the definition):

Any help greatly appreciated.

Many thanks!

The limit, n, in
##
\def\brack{\overwithdelims[]}

\lim_{n\rightarrow \infty} {\log {a_{n+2}} \brack \log {a_{n+1}}}
##
is a dummy index. We can replace ##n+1## with ##m## and so also ##n+2 = m+1##. Then it becomes
##
\def\brack{\overwithdelims[]}

\lim_{m\rightarrow \infty} {\log {a_{m+1}} \brack \log {a_{m}}}
##

member 731016
ChiralSuperfields said:
Homework Statement: I am trying to understand whether part (c) of the problem below is wrong. I have also provided my working and the solutions.
Relevant Equations: Sequence notation ##\{a_n\}##

For this problem,
View attachment 331096
My working is

##\{\frac{a_{n + 2}}{a_{n+1}}\} = \{1 + \frac{a_n}{a_{n +1}}\}## using part (b)
Then

##\lim_{n \to \infty} b_n = \lim_{n \to \infty} (1 + \frac{a_n}{a_{n +1}})## which should equal ##L##.

However, does someone please know why it does this cause the sequence to converge to ##L##. The solution does not explain anything (just simply states the definition):
View attachment 331100

Any help greatly appreciated.

Many thanks!
I am not sure what you are asking, partly because the question is bizarre.
Part c, as @FactChecker notes, is a somewhat trivial and general result unconnected with parts a and b.
I would have written the question as "using the trivial fact that ##\lim_{n \to \infty}(\frac{a_{n + 1}}{a_{n+2}}) = \lim_{n \to \infty} ( \frac{a_n}{a_{n +1}})##, show that ##L=1+1/L##". Or is that part d?

member 731016 and FactChecker
FactChecker said:
The limit, n, in
##
\def\brack{\overwithdelims[]}

\lim_{n\rightarrow \infty} {\log {a_{n+2}} \brack \log {a_{n+1}}}
##
is a dummy index. We can replace ##n+1## with ##m## and so also ##n+2 = m+1##. Then it becomes
##
\def\brack{\overwithdelims[]}

\lim_{m\rightarrow \infty} {\log {a_{m+1}} \brack \log {a_{m}}}
##
haruspex said:
I am not sure what you are asking, partly because the question is bizarre.
Part c, as @FactChecker notes, is a somewhat trivial and general result unconnected with parts a and b.
I would have written the question as "using the trivial fact that ##\lim_{n \to \infty}(\frac{a_{n + 1}}{a_{n+2}}) = \lim_{n \to \infty} ( \frac{a_n}{a_{n +1}})##, show that ##L=1+1/L##". Or is that part d?
Thank you for your replies @FactChecker and @haruspex !

Yeah I am still confused by this problem. I might have a chat to the professor who made since he might have idea what the tutorial question was trying to ask. Part d is:

Many thanks!

ChiralSuperfields said:
Thank you for your replies @FactChecker and @haruspex !

Yeah I am still confused by this problem. I might have a chat to the professor who made since he might have idea what the tutorial question was trying to ask. Part d is:
View attachment 331120

Many thanks!
I would say my proposed version of part c is the logical step towards part d.

member 731016
haruspex said:
I would say my proposed version of part c is the logical step towards part d.
Thank you for your help @haruspex! I think you are right.

You may also use the fact that if the sequence converges to L, all values for n large-enough will be as close to L as you wish.

member 731016
ChiralSuperfields said:
Homework Statement: I am trying to understand whether part (c) of the problem below is wrong. I have also provided my working and the solutions.
Relevant Equations: Sequence notation ##\{a_n\}##

For this problem,
View attachment 331096
My working is

##\{\frac{a_{n + 2}}{a_{n+1}}\} = \{1 + \frac{a_n}{a_{n +1}}\}## using part (b)
Then

##\lim_{n \to \infty} b_n = \lim_{n \to \infty} (1 + \frac{a_n}{a_{n +1}})## which should equal ##L##.

However, does someone please know why it does this cause the sequence to converge to ##L##. The solution does not explain anything (just simply states the definition):
View attachment 331100

Any help greatly appreciated.

Many thanks!
c) Is just saying that if ##\lim_{n \to \infty} a_n = L## then ##\lim_{n \to \infty} a_{n+1} = L##.
This is a true statement, and in fact true for every fixed-length shift (positive & negative).
The proof idea / sketch would be notice that if for ##n > N## we have ##|a_n - L| < \delta##, then for ##n > N## we must trivially have ##|a_{n+1} - L| < \delta##, as ##n+1>N##.

There's no need for fancy arguments.

member 731016 and e_jane
ChiralSuperfields said:
Thank you for your replies @FactChecker and @haruspex !

Yeah I am still confused by this problem. I might have a chat to the professor who made since he might have idea what the tutorial question was trying to ask. Part d is:
View attachment 331120

Many thanks!
Suppose we have proved convergence of the limit ##\lim_{n\to \infty} {a_{n+1} \over a_n}## (I'm not sure if that was proven beforehand? and I'm also too lazy to do a proof of convergence if I don't know if it has).
Even if you don't know what this limit is, but call it ##L## for now.
You can rewrite the recurrence relation as ##{a_{n+2} \over a_{n+1}} = 1 + {a_n \over a_{n+1}}##.
We can apply the limit on both sides, notice by c) the limit of LHS is just ##L##
The probably known laws of limits (though mention me if you don't know) tell you:
##\lim_{n \to \infty} 1 + {a_n \over a_{n+1}} = 1 + 1 /\left(\lim_{n \to \infty} {a_{n+1} \over a_n}\right) = 1+ {1 \over L}##
Finally you arrive at ##L = 1 + {1 \over L}##, and so ##L^2 - L - 1 = 0##.
Solving this leads to two solutions, those being:
##L \in \left\{ {1-\sqrt{5} \over 2}, {1 + \sqrt{5} \over 2}\right\}##
The one with the ##-## is negative, but the terms of the fibonacci sequence are always positive (can you prove / convince yourself why?) so the limit must be ##{1 + \sqrt{5} \over 2}##, ending the exercise.
I think this highlights the ideas of a proof, again if you have any question feel free to mention me.

member 731016 and e_jane

## What is the limit of the ratio of consecutive Fibonacci numbers?

The limit of the ratio of consecutive Fibonacci numbers is the golden ratio, approximately 1.61803398875. Mathematically, if F(n) represents the nth Fibonacci number, then the limit of F(n+1)/F(n) as n approaches infinity is (1 + √5) / 2.

## How do you prove that the limit of the Fibonacci sequence ratio is the golden ratio?

To prove this, consider the Fibonacci sequence defined by F(n+2) = F(n+1) + F(n). Dividing both sides by F(n+1), we get F(n+2)/F(n+1) = 1 + F(n)/F(n+1). Letting the limit of F(n+1)/F(n) be L, we get L = 1 + 1/L. Solving the quadratic equation L^2 - L - 1 = 0, we find L = (1 + √5) / 2, the golden ratio.

## Does the Fibonacci sequence itself have a limit?

No, the Fibonacci sequence itself does not have a limit as it tends to infinity. However, the ratio of consecutive Fibonacci numbers converges to the golden ratio.

## Can the limit of the ratio of Fibonacci numbers be derived using linear algebra?

Yes, the limit can be derived using linear algebra by considering the eigenvalues of the matrix transformation that defines the Fibonacci sequence. The dominant eigenvalue of the matrix [[1, 1], [1, 0]] is the golden ratio, which explains why the ratio of consecutive Fibonacci numbers converges to this value.

## Is there a connection between the Fibonacci sequence and continued fractions?

Yes, there is a connection. The continued fraction representation of the golden ratio is [1; 1, 1, 1, ...], which means it can be expressed as an infinite continued fraction with all 1s. This continued fraction converges to the golden ratio, the same limit as the ratio of consecutive Fibonacci numbers.

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