Example 1

Consider the sequence $(u_n)$ on the rational line $\mathbf{Q}$ defined by $u_n = \sum_{p=0}^n 2^{-p(p+1)/2}$. If $m > n$ we have

and therefore $(u_n)$ is a Cauchy sequence. But this sequence has no limit in $\mathbf{Q}$; for if the rational number $a/b$ were a limit of $(u_n)$, then by (1) we should have for all $n$ where $h_n$ is an integer (depending on $n$); that is, for all $n$. Now the left-hand side of this inequality is an integer for all $n$, and must therefore be zero whenever $n$ is greater than an integer $n_0$ such that $b < 2^{n_0}$; we should therefore have $a/b = u_n$ for all $n > n_0$, which is absurd.