How to find the limit of $a_1=1\:,\:a_{n+1}=\frac{a_n+2}{a_n+1}$ How to find the limit of $a_1=1\:,\:a_{n+1}=\frac{a_n+2}{a_n+1}$?
I think to find the limit, first I need to prove that the sequence is convergent then it is easy to find the limit : 
$L = \frac{L+2}{L+1}  \to  L = \sqrt{2}$ because the sequence $>0$ 
any hint to prove that the sequence is convergent? 
thanks 
 A: Hint:
Compute $a_{n+1}-a_n$ in function of $a_n$ and $a_{n-1}$, and show that
$$|a_{n+1}-a_n|<k\,|a_n-a_{n-1}| \quad (0<k<1),$$
then  rewrite $a_n-a_0$ as
$$a_n=(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+\dots +(a_1-a_0).$$
A: First, suppose the sequence converges to some $L$. We have
$$a_{n+1}=\frac{a_n+2}{a_n+1}=1+\frac{1}{1+a_n},$$
and we can take the limit of both sides. On the left,
$$\lim_{n\to\infty}(a_{n+1}-L+L)=\lim_{n\to\infty}(a_{n+1}-L)+L=L.$$
Now since $a_n>0$ (or even $a_n>1$) for all $n$, we know that $\lim\limits_{n\to\infty}a_n\neq -1$. Thus the limit of the right hand side is
$$\lim_{n\to\infty}\frac{a_n-L+L+2}{a_n-L+L+1}=\frac{\lim\limits_{n\to\infty}(a_n-L+L+2)}{\lim\limits_{n\to\infty}(a_n-L+L+1)}=\frac{L+2}{L+1}.$$
As you noted, this gives $L^2=2$. Of course $L\ge 0$, so $L=\sqrt 2$.
As you also noted, this is not a proof, since we began by assuming $L$ exists. We can show this sequence is Cauchy as follows. Write
\begin{align}
a_{n}-a_{n-1}&=1+\frac{1}{1+a_{n-1}}-1-\frac{1}{1+a_{n-2}}\\
&=\frac{a_{n-2}-a_{n-1}}{(1+a_{n-1})(1+a_{n-2})}.
\end{align}
In particular, $a_n>1$ for all $n$, and so
$$|a_{n}-a_{n-1}|=\left|\frac{a_{n-2}-a_{n-1}}{(1+a_{n-1})(1+a_{n-2})}\right|<\frac12|a_{n-1}-a_{n-2}|.$$
By induction,
\begin{align}
|a_n-a_{n-1}|&<\frac12|a_{n-1}-a_{n-2}|\\
&<\frac{1}{2^2}|a_{n-2}-a_{n-3}|\\
&\hspace{3.5pt}\vdots\\
&<\frac{1}{2^{n-2}}|a_2-a_1|\\
&=\frac{1}{2^{n-1}}
\end{align}
which shows the sequence is Cauchy.
A: A standard trick is to take the fixpoint $L= \sqrt{2}$ of the equation and to consider 


*

*$a_n - L = a_n - \sqrt{2}$
Using 


*

*$a_{n+1} = 1+\frac{1}{a_n + 1} \stackrel{a_1 = 1}{\geq} 1 \Rightarrow a_n \geq 1$ for all $n\in\mathbb N$
you get immediately:
$$0\leq \left|a_{n+1} - \sqrt{2} \right| = \left|1+\frac{1}{a_n + 1} - \left(1+\frac{1}{\sqrt 2 + 1}\right) \right|$$
$$= \left|\frac{\sqrt 2 - a_n}{(a_n + 1)(\sqrt 2 + 1)}\right|\stackrel{a_n \geq 1}{\leq}\underbrace{\frac 1{2(\sqrt 2 + 1)}}_{= q < 1}\left|a_n - \sqrt 2\right|$$
$$\leq q^n\left|a_1 - \sqrt 2\right|= q^n(\sqrt 2-1)\stackrel{n\to\infty}{\longrightarrow}0$$
