Error bounds for bounded monotone sequences ($|a_n-L|\to 0$)

For an increasing sequence converging to $L$, its absolute error is $e_n=L-a_n\ge 0$ and is nonincreasing. For a decreasing sequence, the absolute error is $e_n=a_n-L\ge 0$ and is also nonincreasing. This tells us that later terms cannot be worse, but does not supply a computable stopping index for a chosen tolerance.

Suppose an increasing lower approximation $a_n$ and a decreasing upper approximation $b_n$ enclose the same limit of a sequence $L$. Their observable gap $w_n=b_n-a_n$ givesIf $w_n\to 0$, the squeeze theorem certifies a common limit of a sequence; the midpoint has a certified absolute error at most half the gap. This is the principle behind the bisection method. In the symmetric example $a_n=1-1/n$, $b_n=1+1/n$, the gap is $2/n$, each endpoint's exact absolute error is $1/n$, and the midpoint happens to equal one exactly. The bound is valid even when such symmetry is absent.

A fixed upper bound by itself supplies less. If $a_n\uparrow L$ and $a_n\le M$ for all $n$, then $L\le M$ and $0\le L-a_n\le M-a_n$, but the last gap need not vanish. Taking $a_n=1-1/n$ and $M=2$ makes $M-a_n\to 1$. A shrinking bracket or an explicit remainder formula adds information that qualitative convergence does not provide.

Even restricting to increasing sequences in $[0,1]$ with limit of a sequence one gives no common stopping index. For any positive integer $K$, defineThis 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 $r_n\le 1$. SetThen $(a_n)$ 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 $s_n=\sup \{r_k:k\ge n\}$; 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 $\epsilon$ requires $n+2>\exp (1/\epsilon )$, using the exponential function. At $\epsilon =0.1$, 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.