A sequence $b_n$ is decreasing and bounded. Prove it it convergent.
Proof:
Since $b_n$ is bounded, $b_n > L$ where L is the greatest lower bound as per the completeness Axiom.
Consider some $l>0$ We know $b_n > L + l$ is FALSE
Thus $b_n < L + l$ is TRUE for $n>N$, where N is considered a cut-off value
Rearranging $=>$ $|b_n-L|<l$ for $n>N$ $=>$ lim$b_n$ = L as $n -> infinity
$\endgroup$ 11 Answer
$\begingroup$Since $\{b_n\}$ is bounded, it has an infimum call it $S_*$. Now $b_n \ge S_*$ for all $n \in N$. Now since $S_*$ is an infimum, given $\epsilon \gt 0$, there exists atleast one $n_0 \in N$ such that $S_*+ \epsilon \gt a_{n_0}$. Now for all $n \ge n_0$, $a_n \le a_{n_0} \lt S_*+ \epsilon$ . Moreover, $$S_*- \epsilon \lt s_* \le a_n \le a_{n_0} \lt S_* + \epsilon$$ for all $ n \ge n_0$. Hence given $ \epsilon \gt 0$ there exists an $n_0 \in N$ such that for all $n \ge n_0$ we have $$|a_n-S_*| \lt \epsilon$$
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