Question 3.T.7: A monotonic sequence is convergent if, and only if, it is bo...
A monotonic sequence is convergent if, and only if, it is bounded. More specifically,
(i) if (x_{n}) is increasing and bounded above, then
lim x_{n} = sup \left\{x_{n} : n ∈ \mathbb{N}\right\};
(ii) if (x_{n}) is decreasing and bounded below, then
lim x_{n} = inf \left\{x_{n} : n ∈ \mathbb{N}\right\}.
If (x_{n}) is convergent then, by Theorem 3.2, it must be bounded.
(i) Assume (x_{n}) is increasing and bounded. The set A = \left\{x_{n} : n ∈ \mathbb{N}\right\} will then be non-empty and bounded, so, by the completeness axiom, it has a least upper bound. Let sup A = x. By Definition 2.6, given ε > 0 there is an x_{N} ∈ A such that
x_{N} > x − ε.
Since x_{n} is increasing,
x_{n} ≥ x_{N} for all n ≥ N. (3.6)
But x is an upper bound for A, hence
x ≥ x_{n} for all n ∈ \mathbb{N}. (3.7)
From (3.6) and (3.7) we arrive at
|x_{n} − x| = x − x_{n} ≤ x − x_{N} < ε for all n ≥ N,
which means x_{n} → x.
(ii) If (x_{n}) is decreasing and bounded, then (−x_{n}) is increasing and bounded, and, from (i), we have
lim(−x_{n}) = sup(−A).
But since sup(−A) = − inf A, it follows that lim x_{n} = inf A.