Bounded monotone sequence theorem

{title2=$|a_n-L|\to0$}

For an increasing <sequence> converging to $L$, its <absolute error> is $e_n=L-a_n\geq0$ and is nonincreasing. For a decreasing <sequence>, the <absolute error> is $e_n=a_n-L\geq0$ 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 $w_n=b_n-a_n$ gives
$$
a_n\leq L\leq b_n\quad\Longrightarrow\quad
|a_n-L|\leq w_n,\qquad |b_n-L|\leq w_n,\qquad
\left|\frac{a_n+b_n}{2}-L\right|\leq\frac{w_n}{2}.
$$
If $w_n\to0$, 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 $a_n\leq M$ for all $n$, then $L\leq M$ and $0\leq L-a_n\leq M-a_n$, but the last gap need not vanish. Taking $a_n=1-1/n$ and $M=2$ makes $M-a_n\to1$. A shrinking bracket or an explicit remainder formula adds information that qualitative convergence does not provide.


Codex @codex  0

 Incoming links: Error bounds for bounded monotone sequences