# Prove that a sequence is convergent using Cauchy's principle of convergence

I am facing difficulty in employing Cauchy's general principle of convergence to prove that a sequence is convergent.

Let's take this example:

Let $$x_1 = 2$$, and $$x_{n+1} = \sqrt{x_n+20}$$, for $$n=1,2,3,\dots$$
Prove that the sequence {$$x_n$$} converges.

I solved this using monotone convergence theorem. Using induction, it can be proved that the sequence is increasing and is bounded above by $$5$$. Thus, it is convergent.

But how do I prove this using Cauchy's principle, which states that, for each $$\epsilon > 0$$ there exists a positive integer $$m$$ such that

$$|S_{n+p}-S_n|<\epsilon$$, $$\forall$$ $$n\geq m$$ and $$p\geq1$$, $$\{S_n\}$$ being the sequence.

let $$x_{n+1}>x_{n}$$ then

$$x_{n+1}+20 >x_n+20$$ $$\Rightarrow$$ $$\sqrt{x_{n+1}+20}=x_{n+2}>\sqrt{x_n+20}=x_{n+1}$$

and we know that $$x_2 > x_1$$ so $$x_n$$ is increasing

also, note that $$2 \leq x_n < 5$$ $$\Rightarrow$$ $$2\sqrt{22} \leq\sqrt{x_a+20}+\sqrt{x_b+20}<10$$ hence $$\frac{1}{\sqrt{x_a+20}+\sqrt{x_b+20}}>8$$

$$x_{n+p}-x_n = \sqrt{x_{n+p-1}+20}-\sqrt{x_{n-1}+20}= \frac{x_{n+p-1}-x_{n-1}}{\sqrt{x_{n+p-1}+20}+\sqrt{x_{n-1}+20}}=\frac{\sqrt{x_{n+p-2}+20}-\sqrt{x_{n-2}+20}}{\sqrt{x_{n+p-1}+20}+\sqrt{x_{n-1}+20}}$$

$$\frac{\sqrt{x_{n+p-2}+20}-\sqrt{x_{n-2}+20}}{\sqrt{x_{n+p-1}+20}+\sqrt{x_{n-1}+20}}=\frac{x_{p+1}-\sqrt{22}}{(\sqrt{x_{n+p-1}+20}+\sqrt{x_{n-1}+20})...(\sqrt{x_{p+1}+20}+\sqrt{22})}< \frac{x_{p+1}-\sqrt{22}}{8^{n-2}}$$

$$\frac{x_{p+1}-\sqrt{22}}{8^{n-2}}<\frac{5-\sqrt{22}}{8^{n-2}}< \epsilon$$ $$\Rightarrow$$ $$\frac{1}{8^{n-2}}<\frac{\epsilon}{5-\sqrt{22}}$$ and we know there exists and $$m$$ such that

$$m > n \Rightarrow |\frac{1}{8^{n-2}}-0|=\frac{1}{8^{n-2}}<\frac{\epsilon}{5-\sqrt{22}}$$

so $$x_n$$ is convergent