Evaluating $\sqrt{6+\sqrt{6+\cdots}}$ Tough as introduction to analysis for beginners (Dutch handbook - I'm Belgian). Again
($n$) means index $n$, $x_1 = \sqrt6$, $x_{n+1} = \sqrt{6+x_n}$


*

*Question:


$$|x_{n+1} - 3| \le 1/5 \cdot |x_n - 3|$$
For me this means that $3$ as a 'limit', we need to find that the distance between $x_{n+1}$ and the 'limit' is $1/5$ the distance between the $x_n$ and the limit.
Where does the $1/5$ come from?


*

*Prove that $|x_n - 3|\le (1/5)^{n-1}$

*prove that the sequence converges to $3$.
ps: 
When I studied maths in 1980. we went quickly towards metric spaces, so these calculus minded times are nothing compared to those times. But still, as I didn't pass then, I'd like to restart on a new basis.
Thanks for all the help. 
If you know where maths can be studied in community on the net, always welcome.  
 A: For the inequality, by the definition of $x_n$ we have
$$x_n-3=\sqrt{6+x_{n-1}}-3.$$ 
Multiply by $\dfrac{\sqrt{6+x_{n-1}}+3}{\sqrt{6+x_{n-1}}+3}$. So we are multiplying by $1$ in a fancy way. We get 
$$x_n-3=\frac{x_{n-1}-3}{\sqrt{6+x_{n-1}}+3}.\tag{$\ast$}$$ 
The bottom is clearly $>5$, since the $x_i$ start and stay positive.  One can do better than $5$ here, for example we can without thought replace $5$ by $\sqrt{6}+3$, and with not much more by $\sqrt{6+\sqrt{6}}+3$.  But it doesn't matter, $5$ is good enough for a proof of convergence.  It would even be enough to observe that the denominator is $>3$. 
Taking absolute values, we find that
$$|x_n-3|=\frac{|x_{n-1}-3|}{\sqrt{6+x_{n-1}}+3}&lt\frac{|x_{n-1}-3|}{5}.$$ 
Iterate. The distance to $3$ gets divided by at least $5$ with each iteration, so after a (short) while $x_n$ is awfully close to $3$. Thus our sequence has limit $3$.
A: Let's define the auxiliary sequence $a_{(n)}$, $n\ge1$, n in${\mathbb N}$ as follow:
$$a_{n}=\frac{|x_{n+1} - 3|}{|x_n - 3|}$$
i). Taking into account that $x_{n}$ is positive, one sees that $|\sqrt{6+x_n}+3|>5$. Hence, our first inequality may be proved as follows:  
$$a_{n}=\frac{|\sqrt{6+x_n}-3|}{|x_n - 3|}=\frac{1}{|\sqrt{6+x_n}+3|}\le \frac{1}{5} \to \space a_{n}\le \frac{1}{5}.
$$
ii). Proving the second inequality:
$$a_{1}\cdot a_{2} \cdot a_{3}\cdots a_{n-1}=\frac{|x_{n} - 3|}{|\sqrt6 - 3|}\le \left({\frac{1}{5}}\right)^{n-1} \to \space |x_{n} - 3|\le {|\sqrt6 - 3|}\left({\frac{1}{5}}\right)^{n-1} \le \left({\frac{1}{5}}\right)^{n-1}.$$
iii). Using the inequality from the previous point we get immediately that:
$$\lim_{n\to\infty} |x_{n} - 3|\le 0 \to \lim_{n\to\infty} x_{n}=3.$$
The proof is complete.
A: We will also assume that $x_n$ lies between 0 and 3.
$3-x_{n+1} = 3 - \sqrt{6 + x_n} = 3 - \sqrt{9-(3-x_n}$.
Call $3-x_n$ as $a_n$.
So $a_{n+1} = 3 - \sqrt{9-a_n} = 3(1 - \sqrt{1-\frac{a_n}{9}}) \le 3\frac{a_n}{18} = \frac{a_n}{6}$ (by the Taylor series expansion).
Also, $a_{n+1}$ does not become negative like this, and as we will show its absolute value keeps decreasing. So we justified in assuming that $x_n$ lies between 0 and 3.
So $a_{n+1} \le \frac{a_n}{5}$ also
As $a_{n+1} \le \frac{a_n}{5}$, and $a_1$ = $3 - \sqrt{6} \lt 1$, $a_n \le \frac{1}{5}e^{n-1}$ in general.
So the series converges to 3.
