= Bounded monotone sequences have no universal convergence rate
Even restricting to increasing <sequences> in $[0,1]$ with <limit of a sequence> one gives no common stopping index. For any positive <integer> $K$, define
$$
a_n=\begin{cases}0,&n\leq K,\\1-2^{-(n-K)},&n>K.\end{cases}
$$
This is an increasing <bounded sequence> converging to one, but its <absolute error> is one through index $K$. Since $K$ can exceed any proposed index, the <bounded monotone sequence theorem> alone cannot say when the <absolute error> will fall below $1/2$. A zero difference between two successive terms is also insufficient: the initial plateau eventually ends.
There is no universal eventual rate either. Let $(r_n)$ be any proposed positive nonincreasing <sequence> tending to zero, with $r_n\leq1$. Set
$$
a_n=1-\sqrt{r_n}.
$$
Then $(a_n)$ is increasing, lies in $[0,1]$, and converges to one, but for every sufficiently large $n$ its <absolute error> satisfies $1-a_n=\sqrt{r_n}>r_n$. Indeed, $\sqrt{r_n}/r_n=1/\sqrt{r_n}\to\infty$, so no fixed constant multiple of $r_n$ bounds the <absolute error> eventually. If a proposed rate is not nonincreasing, replace it, after a finite initial segment if necessary, by the decreasing envelope $s_n=\sup\{r_k:k\geq n\}$; this also tends to zero, and the same construction defeats the original rate. This is an original construction, rather than a quantitative claim attributed to the qualitative theorem.
For a concrete slow example, $a_n=1-1/\log(n+2)$ increases to one, where $\log$ is the <natural logarithm>. Its <absolute error> is $1/\log(n+2)$, so making it less than $\varepsilon$ requires $n+2>\exp(1/\varepsilon)$, using the <exponential function>. At $\varepsilon=0.1$, this first happens at $n=22025$. The reciprocal and geometric examples in <increasing and decreasing approximations to a common limit> meet the same tolerance at $n=11$ and $n=4$, respectively. All three obey the same <bounded monotone sequence theorem>.
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