Consider the recursive relation $ a_0=1, \ a_1=1 \ \ and \ \ a_{n+1}=a_n+a_{n-1} $.

Question

Assuming that $ \ \ a=\lim |\frac{a_n}{a_{n+1}}| \ $ exists , find the limit $ \ a \ $.

Answer:

$ \ \ a=\lim |\frac{a_n}{a_{n+1}}|=\lim |\frac{a_n}{a_{n}+a_{n-1}}| $ , (since $a_{n+1}=a_n+a_{n-1}) , $

or, $ a=\lim |\frac{\frac{a_n}{a_{n-1}}}{\frac{a_n} {a_{n-1}}+1}| $

or, $ a=\frac{\lim|\frac{a_n}{a_{n-1}}|}{\lim|\frac{a_n}{a_{n-1}}+1|}=\frac{a}{\lim|\frac{a_n}{a_{n-1}}+1|}$ , (Since $ \lim |a_n /a_{n-1}|=\lim |a_{n+1}/a_n| $ ) .

Or, $ \lim|\frac{a_n}{a_{n-1}}+1|=1 $

But I can't proceed furher . Any help is really ppreciating .

On the right hand side above you have $\lim \lvert a_n/a_{n-1} \rvert$ which is $1/a$, not $a$. — User8128, Jul 10 '17 at 02:02
A tangential observation: Since all $a_n$ are positive, you can omit all those absolute value signs from your calculation. — Andreas Blass, Jul 10 '17 at 04:29

score 1 · Accepted Answer · answered Jul 10 '17 at 02:04

The ratio of the Fibonacci sequence converges to the golden ratio.

Let $\phi = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$.

Now, $\frac{a_{n+1}}{a_n}=\frac{a_n + a_{n-1}}{a_n} = 1 + \frac{1}{\frac{a_n}{a_{n-1}}}$.

So if we let $b_n = \frac{a_n}{a_{n-1}}$ then we have that $b_{n+1}=1 + \frac{1}{b_n}$ and $b_n \to \phi$ so $\phi$ must satisfy the equation

$\phi = 1 + \frac{1}{\phi}$.

Solving this quadratic yields the golden ratio.

Consider the recursive relation $ a_0=1, \ a_1=1 \ \ and \ \ a_{n+1}=a_n+a_{n-1} $.

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