“real: chapter_03” — 2011/5/22 — 22:50 — page9—#9
Sequences and Series 3-9
Case 2: α =−∞. In this case a
n
→−∞as n →∞and hence
given any real number M we have a stage N such that a
n
< M for
n ≥ N . Thus α
n
≤ M for n ≥ N . (Note that {α
n
} is a decreasing
sequence). Thus α
n
→−∞as n →∞. However, by Theorem 3.3.7,
η = inf
n≥1
α
n
= lim
n→∞
α
n
=−∞=α.
Case 3: −∞ <α<∞. We merely prove that η satisfies both the
properties of α as described in (vi) so that η = α. Let γ>ηbe given.
Now γ is not a lower bound for {α
n
/n ≥ 1}. Thus we can get N ∈ N
such that α
N
<γ. Since {α
n
} is decreasing, we have α
n
≤ α
N
<γ
for n ≥ N . Hence the sequence {α
n
} is eventually in (−∞, γ). Now
let δ<ηbe given. Since η ≤ α
n
for n ≥ 1, we also ave δ<α
n
for
each n ≥ 1. From the definition of α
n
(δ is not