Bounded Sequence¶

Definition¶

A sequence $(a_n)_{n=1}^\infty$ of real numbers is bounded if there exists a real number $M\geq0$ such that

|a_i|\leq M \qquad\text{for every }i\geq1.

Equivalently, the sequence is both bounded above, meaning that $a_i\leq U$ for every $i$ for some real number $U\text{,}$ and bounded below, meaning that $a_i\geq L$ for every $i$ for some real number $L\text{.}$ A sequence is unbounded if no such $M$ exists.

Interpretation¶

A bounded sequence is one whose terms all stay between $-M$ and $M$ for one fixed number $M$ . The terms may move around inside this range, but none can escape it.

The sequence (-1) to the n times (1 + 1/n), alternating in sign with shrinking size, with all terms inside a shaded band between -M and M.

The terms of $a_n=(-1)^n\bigl(1+\tfrac{1}{n}\bigr)$ alternate in sign: $a_1=-2$ $- 2$ , $a_2=\tfrac{3}{2}$ $\frac{3}{2}$ , $a_3=-\tfrac{4}{3}$ $- \frac{4}{3}$ , and so on. All of them lie in the band between $-M$ and $M$ with $M=2$ $2$ , and $a_1=-2$ shows that no smaller $M$ works. The terms approach $1$ and $-1$ alternately, so this sequence is bounded but not convergent.

The absolute value measures each term's distance from zero, and the inequality $|a_i|\le M$ must hold with one common bound for every index. The order of the quantifiers matters: "there is an $M$ such that $|a_i|\le M$ for every $i$ " is the definition, whereas "for every $i$ there is an $M$ " holds for every sequence, with $M=|a_i|$ $∣ a_{i} ∣$ . Any larger $M$ also works, and a sequence bounded above by $U$ and below by $L$ satisfies $|a_i|\le\max\{|U|,|L|\}$ $max {∣ U ∣, ∣ L ∣}$ .

Boundedness is a common hypothesis in convergence results. Every convergent sequence is bounded, so an unbounded sequence diverges. The monotone convergence theorem adds monotonicity to obtain convergence. In the figure, the even-indexed terms form a convergent subsequence with limit $1$ .

A common slip is to equate "unbounded" with "tends to infinity". The sequence with $a_n=n$ for even $n$ and $a_n=0$ for odd $n$ is unbounded, yet it does not tend to $\infty$ , because it keeps returning to $0$ .

Math III shows that every bounded sequence has a convergent subsequence: the Bolzano–Weierstrass theorem.

Key Results¶

Every convergent sequence is bounded.
A bounded sequence need not converge.
Every finite sequence is bounded.

Examples¶

$a_n=1/n$ is bounded.
$a_n=(-1)^n$ is bounded but not convergent.

Prerequisites¶

Sources¶

Math I: Calculus · Lectures 2–3

Symbols: infinity

Last updated: 20 October 2026 03:02