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Let $$(u_n) : \begin{cases} u_1=0\\u_2=1 \\ u_{n+1}=\dfrac{3u_{n-1}+2}{10u_n+2u_{n-1}+2}, \forall n \in \mathbb{N}\end{cases} $$

Find the limit of $(u_n)$ when $n \rightarrow +\infty $

I can prove that $0 <u_n<1 ,\forall n \in \mathbb{N}$ but it 's all i can do, $(u_n)$ is not the decreasing or inscreasing sequence by calculate by computer i can see it.

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You probably mean $u_0=0$ and $u_1=1$. – Git Gud Dec 27 '13 at 16:29
The fixpoint (when $u_{n-1}=u_n=u_{n+1}$) is at $\frac{\sqrt{97}-1}{24}$. I assume that will be the limit – Ragnar Dec 27 '13 at 16:29
I assume you mean $$u_n = \begin{cases} 0, &\text{if }n=0\\ 1, &\text{if }n=1 \\ \dfrac{u_{n-2}+2}{10u_{n-1}+2u_{n-2}+2}, &\text{if }n\ge 2\end{cases}$$ for all $n\in \mathbb N_0$. The sequence $(u_{2n})_{n\in \mathbb N_0}$ is increasing and bounded above. The sequence $(u_{2n+1})_{n\in \mathbb N_0}$ is decreasing and bounded below. – Git Gud Dec 27 '13 at 16:40
how can you know it and prove it? – Haruboy15 Dec 28 '13 at 1:02
I'm very sorry I typed wrong my problem, please check again thanks you so much – Haruboy15 Dec 30 '13 at 4:10
up vote 1 down vote accepted

Hint: For the monotonicity part, it might be useful to write the recurrence as $$ u_{n+1}=\dfrac{3u_{n-1}+2}{10u_n+2u_{n-1}+2}=\dfrac{1}{\dfrac{10u_n+\frac23}{3u_{n-1}+2}+\dfrac23}.$$ Now, assume that $u_{n}>u_{n-2}$ and $u_{n-1}< u_{n-3}$ and check what it gives for $u_{n+1}$.

Then you know that both the even and the odd subsequence converge, say towards $a$ and $b$. Use this with the defining recurrence to get two equations for $a$ and $b$. Subtract them to get


If $a$ and $b$ are different, you can use this to calculate $a$ and $b$ and find complex values, so you conclude that they are the same and calculate the final answer.

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I'm very sorry I typed wrong my problem, please check again thanks you so much – Haruboy15 Dec 30 '13 at 4:10
@Haruboy15 I changed the expression, it is actually easier now. – Phira Dec 30 '13 at 10:26

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