The answer to your more restrictive question is still yes with a reasonable definition of computable sequence (and I'll use your Busy Beaver example in the proof). Specifically, I will provide you with a computable $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $\vec{a} = \langle a_m| m \in \mathbb{N}\rangle$ defined by:

$a_m = \begin{cases} 1/f(m) & \text{if } f(m) \neq 0 \\\\ 0 & \text{otherwise.} \end{cases}$

converges to $0$, but its modulus of convergence is noncomputable.

Fix a computable Cantor pairing function $\langle \cdot, \cdot\rangle$ where $\langle e, n\rangle \geq n$ for all $n$ and some computable enumeration of Turing machines $\langle T_e| e \in \mathbb{N}\rangle$ with $T_0$ some trivial Turing machine halting in $0$ steps on every input. Then define $f$ as follows:

$f(\langle e, n\rangle) = \begin{cases} e & \text{if } T_e \text{ halts in exactly } n \text{ steps on input } 0 \\\\ 0 & \text{otherwise.} \end{cases}$

$f$ is clearly computable because we can use an appropriate Universal Turing machine to run program $e$ on input $0$ for $n$ steps to determine whether it halts in exactly $n$ steps or not. Furthermore, each positive Natural number $e$ is assumed at *most* once by $f$, mainly at $\langle e, n\rangle$ if $T_e$ halts in exactly $n$ steps on input $0$. Consequently, for any positive Natural number $e$, we'll have $|a_m - 0| < 1/e$ beyond the at most $e$ many places where $f$ assumes a value from {$1, 2, \ldots, e$}. Therefore, $\vec{a}$ converges to $0$.

But if the modulus of convergence $M$ for $\vec{a}$ were computable, where $M(k)$ is understood to satisfy $|a_m - 0| < 1/k$ for all $m > M(k)$, then the Busy Beaver problem would also be computable. To see this, first note that if the program with positive index $e$ halts on input $0$ in exactly $n$ steps and $e \leq k$, then we have $a_s = 1/f(s) = 1/e \geq 1/k$ where $s = \langle e, n\rangle \geq n$, so that $M(k) \geq n$. Consequently, $M(k)$ will give us an upper bound on the number of steps it takes for all programs halting on input $0$ with index at most $k$ to do so. Therefore, by taking the maximum code $e_s$ of all of the $s$-state Turing machines, $M(e_s)$ will in particular provide us with an upper bound on the number of steps that it takes for all $s$-state Turing Machines halting on input 0 to do so. Then we simply run all of the $s$-state programs for this many steps and take the maximum output to compute $BB(s)$.

If on the other hand, you wanted $\langle a_s| s < \mathbb{N}\rangle$ to be a convergent computable sequence of *Natural numbers*, then $a_s$ must be constant beyond some fixed Natural number $N$, and so we will always have a simple computable modulus of convergence given by the constant function assuming $N$ at every value.

DISCLAIMER: I know almost nothing about logic and computability theory), but how about the following: $a_n = 0$ if all even integers $k$ with $4 \leq k \leq n+4$ can be written as the sum of two primes, and $a_n = 1$ otherwise. Then $(a_n)$ is increasing, and converges to either $0$ or $1$ (classically, anyway). Unfortunately, we'll never know which until someone solves the Goldbach Conjecture; and moreover, even knowing which one it is might not be enough to calculate $N(\epsilon)$ explicitly...(to be continued...) – Zen Harper Jan 12 '11 at 0:34bothcases (although maybe constructivists wouldn't allow excluded middle here?!!!!), this isn't really an answer to your precise question, sorry. But I thought it might be interesting. – Zen Harper Jan 12 '11 at 0:35