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Let $a_{n+2}=\frac{2}{a_{n+1}+a_{n}}$, where $a_0, a_1>0$.

Prove that:

  1. $ \lim_{n \to \infty}a_n=1$

  2. $ \exists \lambda \in (0,1 ), \exists c, \forall n, \| a_n -1\|<c\cdot\lambda^n$

  3. Find the smallest $\lambda$ in 2.

It's obvious that if the limit exists, it must be 1 (fix point). So the problem changed to whether the limit exists or not?

I don't think the formula of general term with a close form exists, and it's hard for me to determine by some regular tools, like Cauchy sequence, the bounded monotone sequence(some example show this sequence is not a monotone one) and so on.

In a word, I have no idea on this problem. :-(

Useful Link How show that $|a_{n}-1|\le c\lambda ^n,\lambda\in (0,1)$

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  • $\begingroup$ Welcome to MathSE. When you pose a question here, it is expected that you include your own attempt to solve the problem and indicate where you are stuck so that you receive responses that address the specific difficulties you are encountering. $\endgroup$ – N. F. Taussig Oct 13 '15 at 9:20
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For the first question :

Step 1. $(a_n)_{n \geqslant 0}$ is bounded.

Proof. We are looking for $m,M$ such that $m \leqslant a_n \leqslant M$ for every $n \geqslant 0$. If it's true for $n,n+1$ then : $$\frac{1}{2M}\leqslant \frac{2}{a_n+a_{n+1}} \leqslant \frac{1}{2m}$$ thus we need $mM=1$. Then we just chose $m$ wisely so that $m \leqslant a_0,a_1$.

Since the sequence is bounded, we set $\ell$ the inferior limit and $L$ the superior one.

Step 2. $\ell L=1$

Proof. Let $\phi$ be an extraction such that $a_{\phi (n)} \to \ell$ and by successive extractions we can assume that $a_{\phi (n)-1)} \to \ell _1 \in [\ell, L]$ and $a_{\phi (n)-2} \to \ell _2 \in [\ell, L]$. We obtain : $$\ell = \frac{2}{\ell _1 + \ell _2} \geqslant \frac{1}{L}$$ and then $\ell L \geqslant 1$. By replacing $\ell$ with $L$ we have $\ell L=1$

Step 3. $\ell = L=1$

Proof. It is the same idea, we set $\phi$ an extraction such that $a_{\phi (n)} \to L$, $a_{\phi (n)-1} \to \ell _1 \in [\ell , L]$, $a_{\phi (n)-2} \to \ell _2 \in [\ell , L]$, $a_{\phi (n)-3} \to \ell _3 \in [\ell , L]$. Then we have : $$\ell _1 + \ell _2 = \frac{2}{L}=2\ell $$ and $$\ell _2+ \ell _3 = \frac{2}{\ell _1}$$ Since $\ell _1, \ell_2 \geqslant \ell$ we obtain $\ell _1 = \ell _2 = \ell$ and then $\ell_2 + \ell _3= \frac{2}{\ell}=2L$ then $\ell_2=\ell_3 =L$ and we conclude $\ell =L$ : the sequence converges to $1$.

For the other questions, I don't know but here is a similar topic : How show that $|a_{n}-1|\le c\lambda ^n,\lambda\in (0,1)$

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  • $\begingroup$ Oh...I was told without detail that the third problem needs some "high-level" knowledge. $\endgroup$ – Yijun Yuan Oct 13 '15 at 10:49
  • $\begingroup$ @袁轶君 : Could you manage to provide details, even for the second one ? I also gived the answer for the second question on the related post. $\endgroup$ – M.LTA Oct 13 '15 at 16:23

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