proving convergence for a sequence defined recursively The sequence $\left \{ a_{n} \right \}$ is defined by the following recurrence relation:
$$a_{0}=1$$ and $$a_{n}=1+\frac{1}{1+a_{n-1}}$$ for all $n\geq 1$
Part 1)- Prove that $a_{n}\geq 1$ for all $n\geq 0$
Part2)- Prove that the sequence $\left \{ a_{n} \right \}$ converges to some real number $x$, and then calculate $x$
For the first part, I could prove it using induction.
For the second part: The problem is how to prove that this sequence is convergent. Once the convergence is proved, then from the recurrence relation we can deduce that $x=\sqrt{2}$. In order to prove it is convergent, I tried to see how this sequence converges to $x$. I calculated the terms $a_{0}$, $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$. I can see that the sequence is neither decreasing nor increasing, so the monotone convergence theorem cannot be applied. I can see that the distance between two consecutive terms is getting smaller and smaller, so I tried to prove that this sequence is contractive. $\left | a_{n+1} -a_{n}\right |=\frac{1}{\left | 1-a_{n} \right |\left | 1+a_{n} \right |}\left | a_{n}-a_{n-1} \right |$, and obviously, $\frac{1}{\left | 1+a_{n} \right |}\leq \frac{1}{2}$. I need to prove that $\frac{1}{\left | 1-a_{n} \right |}\leq \alpha $ where $0< \frac{\alpha }{2}< 1$, and hence the sequence is contractive and therefore it is convergent. If you have any idea how to prove $\frac{1}{\left | 1-a_{n} \right |}\leq \alpha $ or any other idea please share...
 A: Hint: Find $c$ such that
$$
\frac{a_{n+1}-\sqrt2}{a_{n+1}+\sqrt2}=c\,\frac{a_{n}-\sqrt2}{a_{n}+\sqrt2}.
$$
A: The function $\displaystyle f(x) = 1 + \frac{1}{1+x}$ has derivative $\displaystyle f'(x) = - \frac{1}{(1+x)^2}.$ On $[1,\infty)$ we have $-1/4 \leq f'(x) <0$ so the sequence converges by the Contraction Mapping Theorem.
A: This answer shows that if $b$ and $d$ have the same sign, then
$$
\frac{a+c}{b+d}\text{ is between }\frac{a}{b}\text{ and }\frac{c}{d}
$$
Therefore,
$$
a_{n+1}=\frac{a_n+2}{a_n+1}\tag{1}
$$
is between $1$ and $2$.
Furthermore,
$$
\begin{align}
a_{n+1}^2-2
&=\left(\frac{a_n+2}{a_n+1}\right)^2-2\\
&=\frac{2-a_n^2}{(a_n+1)^2}\tag{2}
\end{align}
$$
Therefore, induction and $(2)$ show that
$$
0\le(-1)^n(2-a_n^2)\le\frac1{4^n}\tag{3}
$$
Thus, $(3)$ shows that
$$
\lim_{n\to\infty}a_n=\sqrt{2}\tag{4}
$$
A: You already have found out  that the limit point would have to be $\sqrt{2}$. We have to make use of this knowledge. The simplest way is to replace the given sequence $(a_n)_{n\geq0}$ by the new sequence
$$b_n:=a_n-\sqrt{2}\qquad(n\geq0)\ .$$
Now we have to prove $\lim_{n\to\infty}b_n=0$ which (if true) is certainly simpler than proving convergence to an unknown number $\xi$.
One has $b_0=1-\sqrt{2}\doteq-0.414$, and an easy computation shows that the $b_n$ obey the following recursion:
$$b_n={1-\sqrt{2}\over 1+\sqrt{2}+b_{n-1}}\ b_{n-1}\ .$$
Note the factor $b_{n-1}$ on the right. If we can show that the fraction in front of it is $<1$ in absolute value we are done. As the $b_{n-1}$ appears also in this fraction we have to be a little careful. I leave it to you to prove  the following assertion: 
For all $n\geq0$, if $|b_{n-1}|\leq1$, then $$|b_n|\leq{\sqrt{2}-1\over\sqrt{2}}\ |b_{n-1}|<1\ .$$
A: Also a solution would be that, $\lim_{n\to \infty}a_n = \lim_{n\to \infty}a_{n-1} = L$. By that logic $$L = 1+\frac{1}{1+L} = \frac{2+L}{1+L}$$. This leaves you with $L=\sqrt{2}$ or $L=-\sqrt{2}$. Since, $a_n>0 \forall n \in \mathbb{Z} $, hence $L=\sqrt{2}$. Thus it converges.
