Proving a recursive sequence is Cauchy

Question

I was trying to find a cauchy sequence in the metric space of rational numbers which does not converge in rationals. I end up with the sequence $x_{n+1}= \frac{2+x_n}{1+x_n}$ where $x_1=1$. I proved this sequence converges to $\sqrt{2}$ which is outside of rationals. But how should I prove this sequence is cauchy ?

Answer 1

A sequence of rational numbers is Cauchy in $\mathbb Q$ if and only if it is Cauchy in $\mathbb R$. There isn't really any difference between the Cauchy condition in $\mathbb R$ and the Cauchy condition in $\mathbb Q$.

Answer 2

For $n, m$ large enough, you have $$|x_n-l|<\epsilon/2,~|x_m-l|<\epsilon/2$$ Since the sequence is convergent. $l$ denotes the limit. Then,

$$|x_n-x_m| = |x_n-l+l-x_m|\leq|x_n-l|+|x_m-l|\leq \epsilon$$

You have now proved that it is a Cacuchy sequence.

Answer 3

OK, how about showing the sequence $(x_n)$ of rationals is Cauchy, without using any knowledge of numbers other than rational numbers?

Let $x_1 = 1$ and $x_{n+1}= \frac{2+x_n}{1+x_n}$ for $n \ge 1$. So $x_2 = \frac32$.

By induction, $1 \le x_n < 2$ for all $n \ge 1$.

Write $\Delta_n = x_{n+1}-x_n$. Then $\Delta_1 = \frac32 - 1 = \frac12$.
Compute $$ \left|\frac{\Delta_{n+1}}{\Delta_n}\right| = \frac{1}{2x_n+3} \le \frac{1}{5}. $$ By induction, $$ |\Delta_n| \le \Delta_1\left(\frac15\right)^{n-1} = \frac12\;\left(\frac15\right)^{n-1} . $$ Then for $1 \le m < n$, \begin{align} |x_n - x_m| &= \sum_{k=m}^{n-1} |\Delta_k| \le \sum_{k=m}^{n-1}\frac{1}{2}\left(\frac15\right)^{k-1} = \frac{1}{8}\left[\left(\frac15\right)^{m-1} - \left(\frac15\right)^{n-1}\right] < \frac{1}{8}\left(\frac15\right)^{m-1} \end{align}

Now: $|\frac15|<1$, so $\lim_{k\to\infty}\left(\frac15\right)^k = 0$.
So, given $\varepsilon > 0$, choose $N$ so that $\frac18\left(\frac15\right)^{N-1} < \varepsilon$. From this we get: for all $n,m$ with $N\le n < m$, $$ |x_n-x_m| < \varepsilon . $$ That is, $(x_n)$ is a Cauchy sequence.