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Just posting a slight elaboration on JeffE's answer.

We know that two functions/cases exist that can compute the function f(n):

A function that always returns true (for all n, there exist n number of consecutive 0's) A function that will return true if n is smaller than an integer N, where N is defined as the maximum length of consecutive 0's that exist in the given irrational number (otherwise it returns false).

One and only one of these functions can be correct. We do not know which, but we know for certain that an answer exists. Computability requires that a function exist that can determine the answer within a finite amount of steps.

The number of steps in case 1 is trivially bound to just returning 1.

In case 2 the number of steps are also finite. For every integer $N$ we can construct a Turing machine $T_N(n)$ that accepts if $n < N$ and otherwise rejects in finite time. So not knowing an upper bound on $N$ doesn't matter. For every $N$ there exists a Turing machine, namely $T_N(n)$, that computes correctly whether $n < N$ (we don't know which of these are correct, but it doesn't matter, one exists).

While it may not be possible to choose between the two cases (though one seems more likely than another), we know that exactly one of them must be correct.

As a side note: our solution supposes that while we can not determine which function will elicit a correct value the essence of computability does not rely on the constructability of the proof. Pure Existence is sufficient.