Proof of the Week for 20120806

Monotonic Sequence Convergence Theorem, Part 2: A bounded, monotonic decreasing sequence is convergent.

Proof: Let \left(a_n\right) be a monotonic decreasing sequence, a_n\in\mathbb{R} . \left(a_n\right) is a bounded set of real numbers, thus, via the completeness axiom of real numbers, it must have an infimum. We call the infimum A . Thus a_n\ge A for all n \in\mathbb{N} . We want to show that \lim_{n \to \infty}a_n = A

Let \epsilon > 0 . A + \epsilon > a_m . For n \ge m, a_n \le a_m and A + \epsilon > a_n \ge A. Thus, 0 \le a_n - A < \epsilon for all n \ge m , which implies that \left(a_n\right) converges to A .

Note, part 1 of the proof can be found here.

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