## 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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