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.


1 Answer 1


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


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