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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-06 (claude-sonnet-5)
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A monotone sequence converges if and only if it is bounded

Statement

Let (xk)(x_k) be a monotone sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences). Then (xk)(x_k) converges if and only if it is bounded, that is if and only if there is MRM \in \mathbb{R} with xkM|x_k| \le M for every kNk \in \mathbb{N}.

The forward implication holds for every sequence and is Every convergent sequence is bounded. What monotonicity buys is the converse, which is false for sequences in general.

Facts & Assumptions

Given: A monotone sequence (xk)(x_k) of reals, with range S={xk:kN}S = \{x_k : k \in \mathbb{N}\}.

[L2]

Monotone convergence: a nondecreasing sequence whose range is bounded above converges, to the supremum of its range; a nonincreasing sequence whose range is bounded below converges, to the infimum (A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum).

[L3]

Every convergent sequence of reals is bounded (Every convergent sequence is bounded).

[L4]

A sequence is bounded when some MM satisfies xkM|x_k| \le M for every kk; its range is bounded above by uu when xkux_k \le u for every kk, and bounded below by \ell when xk\ell \le x_k for every kk (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Lower bound, bounded below, bounded set).

[L5]

Absolute value: tM|t| \le M exactly when MtM-M \le t \le M (Basic properties of the absolute value).

Proof

technique · direct
1.1

Suppose (xk)(x_k) converges. Then it is bounded by [L3], and this direction uses neither the monotonicity hypothesis nor any case distinction.

givenL3
1.2

Suppose instead that (xk)(x_k) is bounded, and fix MM with xkM|x_k| \le M for every kk; then MxkM-M \le x_k \le M for every kk, so the range SS is bounded above by MM and bounded below by M-M.

givenL4L5assume-hyp
2.1

If (xk)(x_k) is nondecreasing then its range is bounded above by step 1.2, so (xk)(x_k) converges, to supS\sup S.

step 1.2L2assume-case nondec
2.2

If (xk)(x_k) is nonincreasing then its range is bounded below by step 1.2, so (xk)(x_k) converges, to infS\inf S.

step 1.2L2assume-case noninc
3.1

A monotone sequence is nondecreasing or nonincreasing, so those two cases exhaust the hypothesis, and in both a bounded monotone sequence converges.

step 2.1step 2.2L1cases: nondecreasing or nonincreasingcases-exhaustive
4.1

Both directions are established: a monotone sequence converges if and only if it is bounded.

step 1.1step 3.1

Remarks

Depends on

Used by

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Sources