Since there is a sequence of all Turing machines?, define a sequence $(S_n)_n$ as

$S_n = \sum_{i + j = n} 2^{-i} \{i\}_j ,$

where $\{i\}_j$ is the bit ($0$ or $1$) giving whether the $i$th Turing machine halts before $j$ steps. The theoretical limit of this sequence is

$\sum_i 2^{-i} \{i\} ,$

where $\{i\}$ is the bit giving whether the $i$th Turing machine halts at all, but this is uncomputable (on pain of solving the halting problem).

In constructive mathematics

In Russian constructivism, all real numbers are computable, so a Specker sequence has no (located) supremum, giving a counterexample to the classicalleast upper bound principle? ($LUP$).