Finding $\sqrt {6+\sqrt{6+\sqrt{6+...+\sqrt 6}}}$

Question

For positive integer $n$, $$x_n=\sqrt {6+\sqrt{6+\sqrt{6+...+\sqrt 6}}}$$ where $6$ is written $n$ times.

How can we find the $x _\infty$ ?

I coded a program any found that $x _\infty$ would be equal to $3$. So

$$x _\infty =3$$

But I need to know how can we prove it. Or simply how can we find $x _\infty$?

Answer 1

HINT:

If $y=\sqrt{n+\sqrt{n+\cdots}},y=\sqrt{n+y}$

$\implies y^2=n+y$ and $y\ge0$

For convergence see Convergence of nested radicals and On Infinite Radicals

Answer 2

Take $n \to \infty$, then we can write your expression as $$y = \sqrt{6 + \sqrt{6 + \sqrt{6 \dots}}}$$ $$\implies y = \sqrt{6 + y}$$ $$\implies y^2 = 6 + y$$ $$\implies y^2 - y - 6 = 0$$ $$\implies y = \frac{1 \pm \sqrt{1 + 24}}{2}$$ $$\implies y = \frac{1 \pm 5}{2} $$ $$\implies y = 3, -2$$ But we know that $y \ge 0$ since the RHS in of the original expression is $ > 0$ $$\implies y = 3$$ $$\text{Q.E.D.}$$

Answer 3

By definition, it is clear all $x_n > 0$. Notice for any $n > 1$, $$|3 - x_{n}| = |3 - \sqrt{6+x_{n-1}}| = \left|\frac{3-x_{n-1}}{3+\sqrt{6+x_{n-1}}}\right| < \frac13 |3-x_{n-1}|$$ This leads to for all $n > 0$, $$|3-x_n| \le \frac{1}{3^{n-1}}|3 - x_1| = \frac{1}{3^{n-1}}|3-\sqrt{6}|$$ Since the RHS tends to $0$ as $n \to \infty$, $\lim_{n\to\infty} x_n$ exists and equals to $3$.