Cauchy Sequence that Does Not Converge

Another one, same idea: $$ a_n = \left(1+\frac{1}{n}\right)^n $$ a sequence of rationals, but its limit $e$ is not rational.

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 $$\rho (x,y) = | \tan (\frac{\pi x}{2} ) - \tan (\frac{\pi y}{2}) |.$$

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.