Math 2.0

Sequences, series, summability Proposition

Convergent sequences are bounded

Builds on Convergent sequenceBounded sequenceEpsilon-N definition of the limit of a sequence

Statement

Let (an)(a_n) be a convergent sequence of real numbers. Then (an)(a_n) is bounded: there is a real number K≥0K \ge 0 such that ∣an∣≤K|a_n| \le K for all n∈ℕn \in \mathbb{N}.

Remarks

Let LL be the limit. Take ε=1\varepsilon = 1 in the epsilon-N definition to get NN. Then K=max⁡(∣a0∣,…,∣aN−1∣,∣L∣+1)K = \max(|a_0|, \dots, |a_{N-1}|, |L| + 1) works. The converse is false. The sequence (−1)n(-1)^n is bounded and divergent. The statement holds in each metric space, where bounded means contained in some ball.

Reading path

To understand this page from the ground up, read these 7 pages in order. The foundations come first.

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  1. Sequence definition
  2. Real sequence definition
  3. Bounded sequence definition
  4. Limit of a sequence definition
  5. Convergent sequence definition
  6. Epsilon-N definition of the limit of a sequence statement
  7. Convergent sequences are bounded statement

Where this leads

18 pages build on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Limit_of_a_sequence, Properties
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Tag0S4JF619V7
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Kindstatement, proposition
WritingW0 stub
MSC40A05
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UsesConvergent sequence, Bounded sequence, Epsilon-N definition of the limit of a sequence

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