Evaluating $\sqrt{6+\sqrt{6+\cdots}}$

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

1. 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?

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

2. 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.

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 $3$.

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.