= Increasing and decreasing approximations to a common limit
{title2=$1-1/n\uparrow1,\quad1+1/n\downarrow1$}
For the two <sequences> in <bounded monotone sequence theorem>,
$$
a_{n+1}-a_n=\frac1{n(n+1)}>0,\qquad b_{n+1}-b_n=-\frac1{n(n+1)}<0.
$$
Their directions and bounds are explicit. One is a succession of lower approximations and the other a succession of upper approximations to the same <limit of a sequence>. To obtain <absolute error> strictly below $\varepsilon>0$, it suffices to choose an <integer> $N>1/\varepsilon$, and then every $n\geq N$ works. For $\varepsilon=0.01$, the first such index is $101$; at $100$ the <absolute error> equals the tolerance.
A <geometric sequence> offers a faster example. The <recurrence relation> $c_1=1/2$, $c_{n+1}=(c_n+1)/2$ gives $c_n=1-2^{-n}$. Thus $(c_n)$ is an increasing <bounded sequence> with <limit of a sequence> one and <absolute error> $2^{-n}$. For $\varepsilon=0.01$, $n=7$ already works, since $2^{-7}=1/128<0.01$, whereas $2^{-6}=1/64>0.01$. The <bounded monotone sequence theorem> guarantees both examples converge; the formulas explain their different speeds.
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