LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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A nondecreasing sequence that is not bounded above diverges to $+ \infty$

Statement

Let $(x_{k})$ be a nondecreasing sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) whose range $S = {x_{k} : k \in N}$ is not bounded above (Lower bound, bounded below, bounded set). Then $(x_{k})$ diverges to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ): for every $M \in R$ there is $K \in N$ with $x_{k} > M$ for all $k \geq K$ .

Read together with the monotone convergence theorem this says that a nondecreasing sequence has exactly two possible behaviours, with nothing in between: it converges to the supremum of its range, or it runs away to $+ \infty$ .

Facts & Assumptions

Given: A nondecreasing sequence $(x_{k})$ of reals whose range $S = {x_{k} : k \in N}$ is not bounded above.

[L1]

Monotonicity: $x_{j} \leq x_{k}$ whenever $j \leq k$ (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L2]

Bounded above: $S$ is bounded above exactly when some $M \in R$ satisfies $s \leq M$ for every $s \in S$ (Lower bound, bounded below, bounded set).

[L3]

Trichotomy: for reals $s$ and $M$ , exactly one of $s < M$ , $s = M$ , $s > M$ holds, so the failure of $s \leq M$ is $s > M$ (Complete ordered field (least-upper-bound property), Ordered field).

[L4]

Divergence to $+ \infty$ : $x_{k} \to + \infty$ when for every $M \in R$ there is $K \in N$ such that $x_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[L5]

Every element of $S$ is a term of the sequence, and conversely (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Proof

technique · direct

1.1givenL2

Let $M \in R$ be arbitrary. Since $S$ is not bounded above, $M$ is not an upper bound of $S$ , so some $s \in S$ fails $s \leq M$ .

2.1step 1.1L3L5choose

By trichotomy that $s$ satisfies $s > M$ , and being an element of $S$ it is a term: fix $K \in N$ with $s = x_{K}$ , so $x_{K} > M$ .

3.1step 2.1L1

For every $k \geq K$ monotonicity gives $x_{K} \leq x_{k}$ , hence $x_{k} \geq x_{K} > M$ and so $x_{k} > M$ .

4.1step 3.1L4∎

For every real $M$ an index $K$ has been produced with $x_{k} > M$ for all $k \geq K$ , which is exactly divergence to $+ \infty$ .

Remarks

Only "not bounded above" is used, not unboundedness of the sequence. For a nondecreasing sequence the two coincide, since such a sequence is bounded below by $x_{0}$ (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences), so a nondecreasing sequence is unbounded exactly when its range is not bounded above. The hypothesis is stated in the one-sided form because that is the form the proof consumes.
The dual statement holds with the same proof: a nonincreasing sequence whose range is not bounded below diverges to $- \infty$ . Reflecting through the origin turns one into the other.
$+ \infty$ is not a limit. Divergence to $+ \infty$ and to $- \infty$ is deliberately not a case of Limits and Cauchy sequences of reals: a sequence diverging to $+ \infty$ is unbounded, hence not convergent (Every convergent sequence is bounded), and the arrow in $x_{k} \to + \infty$ is an abbreviation for the displayed quantifier statement and never an equation.
The companion statement is A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum: between them, a nondecreasing sequence converges to the supremum of its range or diverges to $+ \infty$ , with no third possibility.

A nondecreasing sequence that is not bounded above diverges to $+ \infty$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A nondecreasing sequence that is not bounded above diverges to +∞+∞

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A nondecreasing sequence that is not bounded above diverges to $+ \infty$