The Fibonacci series defined recursively by $x(1) = 1, x(2) = 2$ and $x(n+1) = x(n) + x(n-1)$
Find$$\lim_{n\rightarrow\infty}\frac{x(n+1)}{x(n)}$$
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There's a nice explicit formula for the $n$th term of the Fibonacci sequence. http://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression – Ray Yang Jul 04 '13 at 14:00
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Let $L= \lim_{x\rightarrow \infty} \dfrac{x(n+1)}{x(n)}= L$. By the recursive definition we also have that $\lim_{x\rightarrow \infty} \dfrac{x(n+1)}{x(n)}=\lim_{x\rightarrow \infty}\dfrac{x(n)+x(n-1)}{x(n)}= 1+\dfrac{1}{L}$. Therefore,
$$\begin{align*}1+\dfrac{1}{L}=L\\ L^2 - L - 1=0 \end{align*}$$
So $$\dfrac{L=1\pm\sqrt{5}}{2}$$.
A.E
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Thanks Orangutango. I was stuck at 1 + (1/L) but your solution did the rest. – Shravan40 Jul 04 '13 at 14:11
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Hint:
If we denote $\lambda = \lim_{n \to \infty}\frac{x(n+1)}{x(n)} \neq 0$ then
$$\lambda = \lim_{n \to \infty}\frac{x(n)+x(n-1)}{x(n)} = 1+\frac{1}{\lambda}.$$
Also, observe that $x(n) < x(n+1) < 2x(n)$, which gives you convergence (see the Bolzano-Weierstrass theorem) and $1 < \lambda < 2$.
I hope this helps ;-)
dtldarek
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I came up to this. But can't proceed further. Anyway i got the solution. Thanks for your help. – Shravan40 Jul 04 '13 at 14:13
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$\cfrac{x_{n+1}}{x_n}=\cfrac{x_n+x_{n-1}}{x_n}=1+\cfrac{x_{n-1}}{x_n}$
Now take $l=\lim\limits_{n\to +\infty}\cfrac{x_{n+1}}{x_n}$
$l=1+\cfrac{1}{l}$
xavierm02
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I came up to this. But can't proceed further. Anyway i got the solution. Thanks for your help. – Shravan40 Jul 04 '13 at 14:14