Finding limit of the Fibonacci sequence

[ChiralSuperfields]

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!

 

[FactChecker]

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]

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?

 

[ChiralSuperfields]

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
$\def\brack{\overwithdelims[]} \lim_{m\rightarrow \infty} {\log {a_{m+1}} \brack \log {a_{m}}}$

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!

 

[haruspex]

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.

 

[ChiralSuperfields]

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.

 

[WWGD]

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.

 

[TurtleKrampus]

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.

 

[TurtleKrampus]

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.