Bounded monotone sequence theorem

A monotone sequence of real numbers has a finite limit of a sequence exactly when it is a bounded sequence. In the increasing case, only an upper bound needs to be supplied, because the first term already gives a lower bound; in the decreasing case, only a lower bound needs to be supplied. The limit of a sequence is the exact edge of the set of terms:The upper bound $M$ or lower bound $m$ need not itself be the limit of a sequence. A supremum or infimum identifies the sharp bound. ^[ref]

For example, $a_n=1-1/n$ is increasing and bounded above by one, and $b_n=1+1/n$ is decreasing and bounded below by one. Both have limit of a sequence one, although neither ever equals one. Their absolute errors are exactly $1/n$. They also bracket any sequence $x_n$ satisfying $a_n\le x_n\le b_n$: the squeeze theorem gives $x_n\to 1$, even if $(x_n)$ itself is not a monotone sequence.

The theorem is useful when a recurrence relation gives successive improvements and a bound, before the answer has been identified. It proves existence of a limit of a sequence; further algebra, often using continuity, identifies the answer. A finite plot illustrates the formulas but does not establish the infinite claim. Error bounds for bounded monotone sequences require additional quantitative information.

A real increasing sequence bounded above converges to its supremum; a decreasing sequence bounded below converges to its infimum. For the increasing case, the definition of supremum provides a term above $L-\epsilon$, and every later term remains between that term and $L$. Negation reduces the decreasing case to the increasing case. This elementary result is distinct from the measure-theoretic monotone convergence theorem for integrals.

The existence of the real supremum in this short proof uses the least-upper-bound property, equivalently completeness of the real numbers. The approximation property of a supremum supplies $a_N>L-\epsilon$, and monotonicity keeps every later term within the same tolerance:The corresponding decreasing statement follows from the approximation property of an infimum. The converse uses the fact that every convergent sequence is a bounded sequence. ^[ref]

For the two sequences in bounded monotone sequence theorem,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 $\epsilon >0$, it suffices to choose an integer $N>1/\epsilon$, and then every $n\ge N$ works. For $\epsilon =0.01$, 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 $\epsilon =0.01$, $n=7$ 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.

The supremum of an increasing bounded sequence need not be a maximum of its set of terms. For $a_n=1-1/n$, every term is strictly less than one; the supremum and limit of a sequence are one, but there is no maximum. Every candidate term is beaten by the next. The decreasing sequence $1+1/n$ similarly has infimum one without a minimum.

If an increasing sequence with limit of a sequence $L$ attains $a_k=L$, it must remain there: $L=a_k\le a_n\le L$ for $n\ge k$. The decreasing case is identical with inequalities reversed. Thus attainment of the limit of a sequence is equivalent to an eventually constant tail of a sequence for a convergent monotone sequence. The increasing sequence $0,1/2,1,1,\dots$ has both supremum and maximum one. ^[ref]

Dropping either condition destroys the implication. The increasing sequence $a_n=n$ has a lower bound one but no upper bound, and no finite limit of a sequence. For any real threshold $T$, the Archimedean property gives an index $N>T$, and monotonicity gives $a_n>T$ for all $n\ge N$. Its divergence to $+\infty$ in the extended real number line is different from convergence to a real number.

Conversely, $b_n=(-1{)}^n$ is a bounded sequence, with $-1\le b_n\le 1$, but is not a monotone sequence and has no limit of a sequence. Its subsequence at even integers is constantly one, and its subsequence at odd integers is constantly minus one. These two subsequential limits cannot both equal a single limit of a sequence. ^[ref]

The ambient real numbers also matter. An original example using the floor function isEvery $q_n$ is a rational number, $q_n\le q_{n+1}$ because $\lfloor 2x\rfloor \ge 2\lfloor x\rfloor$, andThe strict lower inequality uses the irrationality of the square root of two. Hence this bounded sequence converges to $\sqrt{2}$ in the real numbers, but has no limit of a sequence in the rational numbers. Convergence to a rational candidate would also be convergence in the real numbers, contradicting uniqueness of the limit of a sequence. The formulas give $q_1=1$, $q_2=5/4$, $q_3=11/8$, $q_4=11/8$: this counterexample even has a plateau. This illustrates why completeness of the real numbers matters: the rational numbers lack the least-upper-bound property needed to contain the required supremum. ^[ref]

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.