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Let $\left\{u_n\right\}$ sequence satisfy:


Find $$\lim_{n\to \infty} u_n$$

My tried:

I prove that $\left\{u_n\right\}$ sequence decreasing and $u_n>\frac{1+\sqrt{5}}{2}$. Therefore $\lim_{n\to \infty}u_n=\frac{1+\sqrt{5}}{2}$. But, I can't prove to $$u_n>\frac{1+\sqrt{5}}{2}$$

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Showing that a sequence is decreasing and bounded below by $m$, doesn't mean that the limit is $m$. For example, $1+1/n$ is bounded below by $0$ and decreasing. – Gaffney Dec 29 '13 at 14:15
You say "I prove that ... $u_n > \alpha$. But I can't prove to $u_n > \alpha$". I don't understand what you know and what you would like to know. – Emanuele Paolini Dec 29 '13 at 14:16
It's $$u_{n+1}=1+\dfrac{1}{u_{n}}?$$ – math110 Dec 29 '13 at 14:17
@gaffney : it would be enough here as $\frac{(1+\sqrt{5})}{2}$ is the only positive fixpoint of the reccurence formula. However this sequence is in fact not decreasing, and not minored as suggested : $u_1=1<\frac{1+\sqrt{5}}{2}$ – imj Dec 29 '13 at 14:18

The sequence $\langle u_n\rangle$ is not decreasing sequence : $u_0=1$ and $u_1=2>1$. However, you can prove following facts:

  1. $\langle u_{2n}\rangle$ is decreasing.
  2. $\langle u_{2n-1}\rangle$ is increasing.
  3. $1\le u_n \le 2$ for all $n$.
  4. $u_{2n+1}-u_{2n}\to 0$ as $n$ goes to infinity.

By the 1, 2 and 3, you can prove the convergence of $\langle u_{2n}\rangle$ and $\langle u_{2n-1}\rangle$. By 4, you can check the limit of $\langle u_{2n}\rangle$ and $\langle u_{2n-1}\rangle$ have same value (and $\langle u_n\rangle$ converges.) By 3, the limit of $\langle u_n\rangle$ is greater or equal than $1$. Take $u=\lim_{n\to\infty} u_n$, then the relation between $u_n$ and $u_{n+1}$, you get $u=1+\frac{1}{u}$. Therefore $u=\frac{1}{2}(1\pm\sqrt{5})$. However $u\ge 1$ so $u=\frac{1}{2}(1+\sqrt{5})$.

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You can use this trick that if the limit exists, then "at infinity" $u_{n+1}=u_n=u$, where $u$ is the limit. So $$ u=1+\frac{1}{u} $$

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let $$A=1+\dfrac{1}{A},A=\dfrac{1+\sqrt{5}}{2}>1$$ so $$|u_{n+1}-A|=\left|1+\dfrac{1}{u_{n}}-\left(1+\dfrac{1}{A}\right)\right|=\dfrac{|u_{n}-A|}{u_{n}A}<\dfrac{1}{A}|u_{n}-A|<\cdots\dfrac{|u_{1}-A|}{A^{n}}\to 0,n\to\infty$$ and

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