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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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3 Answers 3

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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 $$\dfrac{L=1\pm\sqrt{5}}{2}$$.

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  • $\begingroup$ Thanks Orangutango. I was stuck at 1 + (1/L) but your solution did the rest. $\endgroup$
    – Shravan40
    Jul 4, 2013 at 14:11
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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 $l=\lim\limits_{n\to +\infty}\cfrac{x_{n+1}}{x_n}$

$l=1+\cfrac{1}{l}$

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  • $\begingroup$ I came up to this. But can't proceed further. Anyway i got the solution. Thanks for your help. $\endgroup$
    – Shravan40
    Jul 4, 2013 at 14:14
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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 ;-)

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  • $\begingroup$ I came up to this. But can't proceed further. Anyway i got the solution. Thanks for your help. $\endgroup$
    – Shravan40
    Jul 4, 2013 at 14:13
  • $\begingroup$ @Shravan40 Don't forget about the convergence (as I did, fixed now). $\endgroup$
    – dtldarek
    Jul 4, 2013 at 14:22

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