# Boundedness = covergence for a monotonically decreasing sequence

A monotone decreasing sequence ${x_{n}}$ converges if and only if is bounded from below

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Anyone agrees Miguel's autogenerated gravatar is borderline inappropriate? =D –  Pedro Tamaroff May 15 '12 at 22:03
Hint for the reverse implication: Let $\alpha$ be the greatest lower bound of the sequence. Show that in fact the sequence converges to $\alpha$. –  David Mitra May 15 '12 at 22:06
@PeterTamaroff Why you you consider it inappropriate? –  David Mitra May 15 '12 at 22:06
@DavidMitra Please trust me that if you don't know, then you will regret finding out. –  MJD May 15 '12 at 22:09
Hint for the forward implication: contrapositive. –  Cameron Buie May 16 '12 at 0:51
If $x_n$ is bounded from below, there exist $I = \inf x_n$. Given $\epsilon > 0$. Chose $n_0$ such that $$I \le x_{n_0} < I+ \epsilon.$$ Hence as $(x_n)$ is decreasing $$n \ge n_0 \Rightarrow I \le x_n \le x_{n_0} < I + \epsilon.$$ Then $(x_n)$ converges. Reciprocally if $(x_n)$ converges to $x$, for $n > > 1$, $x_n > x -1$ and $x_n$ is bounded below.