18. Sequences

Proof of (1):
Let \(a_n\) be an increasing sequence which is bounded above and let \(L\) be its least upper bound. Then we need to show \(\displaystyle \lim_{n\to\infty}a_n=L\) which means: For all   \(\varepsilon \gt 0\),   there is a positive integer,   \(N\),   such that if   \(n \gt N\)   then   \(|a_n-L| \lt \varepsilon\). So given an arbitrary number \(\varepsilon \gt 0\), we claim there is an index \(N\) with \(a_N \gt L-\varepsilon\). If not, then every \(a_n\) satisfies \(a_n \lt L-\varepsilon\) and \(L-\varepsilon\) would be an upper bound. But this is impossible because \(L\) is the least upper bound. Consequently, there is an \(N\) for which \(a_N \gt L-\varepsilon\). Now suppose \(n \gt N\). Then \(a_n \gt a_N\) because the sequence is increasing. So: \[ a_n \gt a_N \gt L-\varepsilon \quad \text{or} \quad \varepsilon \gt L-a_n=|a_n-L| \]

Proof of (2):
The proof is totally analogous to that for (1), reversing some signs and some inequalities.

Proof of (3):
If the sequence if bounded and monotonic, then it is either increasing and bounded above or decreasing and bounded below. In either case it converges.

Proof of (4):
Let \(a_n\) be an increasing sequence which is unbounded. We need to show \(\displaystyle \lim_{n\to\infty}a_n=\infty\) which means: For all   \(M \gt 0\),   there is a positive integer,   \(N\),   such that if   \(n \gt N\)   then   \(a_n \gt M\). Since the sequence \(a_n\) is increasing it is bounded below by \(a_1\). Since it is unbounded, it must be unbounded above. That means it gets bigger than any given number. In particular, given an arbitrary number \(M \gt 0\), there is a number \(N\) for which \(a_N \gt M\). Then if \(n \gt N\) then \(a_n \gt a_N\) because the sequence is increasing and so \(a_n \gt a_N \gt M\).

Proof of (5):
The proof is totally analogous to that for (4), reversing some signs and some inequalities.