If you are not married to using the rationals, I would suggest also using the open interval $(-1,1)$. Here you can take the sequence $( 1 - \frac{1}{n} )_{n=1}^\infty$, and note (quickly) that it is Cauchy and that it should converge to $1$, which of course is not in $(-1,1)$.

The punch line -- if it can be called that -- is that $(-1,1)$ is homeomorphic to the the entire real line $\mathbb{R}$, meaning that they have the same topological structure.

This tells us that it is the underlying metric which tells us whether a sequence is Cauchy or not, and it is not a property of the topology alone. And there are metrics on $(-1,1)$ compatible with the topology in which the aforementioned sequence is not Cauchy; an example would be $$\rho (x,y) = | \tan (\frac{\pi x}{2} ) - \tan (\frac{\pi y}{2}) |.$$

Such complicated examples! Here's a simple one: $\{1/n\}_{n=1}^\infty$ is a Cauchy sequence in the interval $(0,\infty)$ and does not converge within the interval $(0,\infty)$ (with the usual metric).

Of course you could tack $0$ onto the space and get $[0,\infty)$, and within that larger space it converges. Every metric space has a completion, within which every Cauchy sequence converges.

A fairly easy example that does not arise directly from the decimal expansion of an irrational number is given by $$a_n=\frac{F_{n+1}}{F_n}$$ for $n\ge 1$, where $F_n$ is the $n$-th Fibonacci number, defined as usual by $F_0=0$, $F_1=1$, and the recurrence $F_{n+1}=F_n+F_{n-1}$ for $n\ge 1$. It’s well known and not especially hard to prove that $\langle a_n:n\in\Bbb Z^+\rangle\to\varphi$, where $\varphi$ is the so-called golden ratio, $\frac12(1+\sqrt5)$.

Another is given by the following construction. Let $m_0=n_0=1$, and for $k\in\Bbb N$ let $m_{k+1}=m_k+2n_k$ and $n_{k+1}=m_k+n_k$. Then for $k\in\Bbb N$ let $$b_k=\frac{m_k}{n_k}$$ to get the sequence $$\left\langle 1,\frac32,\frac75,\frac{17}{12},\frac{41}{29},\dots\right\rangle\;;$$ it’s a nice exercise to show that this sequence converges to $\sqrt2$.

These are actually instances of a more general source of examples, the sequences of convergents of the continued fraction expansions of irrationals are another nice source of examples; the periodic ones, like this one, are probably easiest.

You take any irrational number, say $\sqrt2$, and you consider its decimal expansion, $$ \sqrt2=1.4142\ldots $$ Then you define $x_1=1$, $x_2=1.4$, $x_3=1.41$, $x_4=1.414$, etc.

Here's another, well-known, example: Let $b>0$. Take $a_1>0$ rational and define $a_{n+1}={1\over2}(a_n+{b\over a_n})$. One can show that this sequence is bounded below and eventually monotone decreasing. From this it follows that $(a_n)$ converges to $\sqrt b$.

Taking $b$ to be prime, for example, gives a sequence of rational numbers that converge to an irrational number.

It is pretty simple to see an example using the following:

1.) Take any sequence of points that converges to a limit, which is not one of the terms in the sequence.

2.) Delete the limit from the metric space

3.) We have a Cauchy Sequence which is not convergent.

For Example:

The sequence 1, 1/2, 1/3, 1/4, 1/5, ...

We know that this converges to 0. So, now take the Metric Space R (all real numbers) and delete the limit 0 from the metric space.

So the Cauchy sequence 1, 1/2, 1/3, 1/4, 1/5, ... is not convergent in the metric space R - {0}.

How about the Cauchy sequence $$x_1=0.1$$ $$x_2=0.12$$ $$x_3=0.123$$ $$x_4=0.1234$$ $$x_5=0.12345$$ $$x_6=0.123456$$ $$\vdots$$in $\mathbb{Q}.$

Here is another idea, generalising David Mitra's example. Let $P(x)$ be a polynomial with integer coefficients with an irrational real root $\xi$. Newton's method to find $\xi$ provides a sequence of rationals converging to $\xi$. Take $x_0\in\mathbb{Q}$ close enough to $\xi$. then the sequence defined recursively as $$ x_{n+1}=x_n-\frac{P(x_n)}{P'(x_n)} $$ converges to $\xi$ and $x_n\in\mathbb{Q}$ for all $n$. David's example is obtained taking $P(x)=x^2-b$.

If we have a subset $A$ of $\mathbb{Q}$ and a limit point $p$ of $A$ such that $p$ is not in $A$, then we can generate such a sequence. Just take any sequence converging to $p$ (which we know exists since $p$ is a limit point of $A$). Furthermore since the sequence is convergent it is also a Cauchy sequence. More precisely for every convergent sequence $p_n$ in the ambient metric space $X$, s.t. $p_n$ is in $A$ for all $n$, we have that $p_n$ is also a cauchy sequence in $A$. Note that the point is that this holds even though $A$ does not contain the limit of the sequence.

This agood example for acauhy seqeance but is not convergent.Take Xn€Q suhc that Xn^2<2 this imples Xn<2^2 then Xn convergent to 2^2 which dont belong to Q. So is not convergent.