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

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?

• You say that $(x_n)$ is “obviously” increasing, but you have trouble to prove that $x_{n+1} / x_n \ge 1$? Feb 20, 2021 at 4:06
• Well I just mean I could not prove it is increasing in an elegant manner. It is easy to notice, but what you notice is not a proof. Feb 20, 2021 at 4:13

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 $$f(x) \leq 0$$ , $$x \in [\frac{3 - \sqrt{5}}{2} , \frac{3 + \sqrt{5}}{2}]$$ .

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.

• I just came up with the simplest proof of {$x_n$} being increasing sequence: note if $x_n \le x_{n+1} => \sqrt{x_n} \le \sqrt{x_{n+1}} => 1+ \sqrt{x_n} \le 1 + \sqrt{x_{n+1}} => x_{n+1} \le x_{n+2}$ so by induction {$x_n$} is increasing Feb 20, 2021 at 4:15
• @Gerald I understood form your question that your problem was the boundedness from above. Feb 20, 2021 at 4:17
• That the sequence is increasing also follows by induction quickly just considering $$x_{n+1}-x_n = \left(1+\sqrt{x_n}\right)-\left(1+\sqrt{x_{n-1}}\right) = \sqrt{x_n}-\sqrt{x_{n-1}}$$ Feb 20, 2021 at 4:20

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 $$x_1, we infer by induction that $$x_n for all $$n \ge 1$$. The induction step is simply an application of $$f$$ to both sides.