Show $x_{n+1} = 1 + \sqrt{x_n}$ is upper-bounded and increasing

Question

Given $x_1 = 1$, it is clear that $x_n \ge 1$ for any $n \in \Bbb{N}$. I know that {$x_n$} is obviously increasing and limit as $n \to \infty$ is $\frac{3 + \sqrt{5}}{2}$, so {$x_n$} has an upper bound also. I seem to have trouble proving $x_{n+1} / x_n \ge 1$ and don't even have any idea on how to prove the sequence is upper-bounded. Can you guys give me some hints?

Answer 1

From the recurrence , we can deduce that $$x_2 = 2$$ $$x_3 = 1 + \sqrt{2} $$ $$ x_4 = 1 + \sqrt{1+\sqrt{2}} $$ $$ x_5 = 1 + \sqrt{1+\sqrt{1+\sqrt{2}}} $$ You can continue . Let $z = x_n$ as $ n \to \infty $ you can write $$ z = 1 + \sqrt{z} $$ Solving you get that $z = \frac{3 + \sqrt{5}}{2}$ ($z$ is positive real) . Also to prove $$ \frac{x_n}{x_{n-1}} \geq 1 $$ You can write it as $$ 1 + \sqrt{x_n} \geq x_n $$ $$ 0 \geq x_n^2 -2x_n + 1 $$ Which is true from the limit as let $$f(x) = x^2 -3x +1$$ You can write it as $$f(x) = \left(x - \frac{3 + \sqrt{5}}{2}\right)\left(x - \frac{3 - \sqrt{5}}{2}\right) $$ and for $f(x) \leq 0$ , $x \in [\frac{3 - \sqrt{5}}{2} , \frac{3 + \sqrt{5}}{2}] $ .

Answer 2

For boundedness from above just note that

$$x_n \leq 4\Rightarrow x_{n+1} = 1+\sqrt{x_n} \leq 1+\sqrt 4 =3 \leq 4$$

Now, the claim follows immediately by induction.

Answer 3

Let $f(x)=1+\sqrt{x}$, so $f(\phi^2)=\phi^2$ where $\phi=(1+\sqrt{5})/2$. Since $f$ is strictly increasing and $x_1<\phi^2$, it follows by induction that $f(x_n) <\phi^2$ for all $n \ge 1$. Also, since $x_1<x_2$, we infer by induction that $x_n<x_{n+1}$ for all $n \ge 1$. The induction step is simply an application of $f$ to both sides.