Subsequences
Let (s_n)^\infty_{n=1} be a sequence and let (n_k)^\infty_{k=1} be any increasing sequence of natural numbers. The sequence (s_{n_k})^\infty_{k=1} is called a subsequence of (s_n).
A subsequence is any sequence formed by deleting and renumbering terms of some original sequence. A subsequence must preserve the order of the terms of the original sequence, and consist of an infinite amount of terms.
If a sequence (s_n) converges to a real number s, then every subsequence of (s_n) also converges to s.
Let (s_{n_k}) be any subsequence of (s_n). Given any \varepsilon > 0, there exists a natural number N such that n \geq N implies |s_n - s| < \varepsilon. When k \geq N, n_k \geq k \geq N, so that |s_{n_k} - s| < \varepsilon. Thus \lim s_{n_k} = s.
Claim: s_n = (-1)^n is not a convergent sequence.
Using the previous theorem, if (s_n) converged to s, then every subsequence of (s_n) would also converge to s. But (s_{2n}) converges to 1, and (s_{2n+1}) converges to -1. Because the subsequences converge to different values, the original sequence does not converge.
Every bounded sequence has a convergent subsequence.
Let (s_n) be a sequence whose range T = \{s_n \colon n \in \mathbb{N}\} is bounded.
Suppose that T is finite. Then there is some number x \in T that is equal to s_n for infinitely many values of n. That is there exists n_1 < n_2 < \cdots < n_k < \cdots such that s_{n_k} = x for all k \in \mathbb{N}. It follows that the subsequence (s_{n_k}) converges to x.
Next, suppose T is infinite. Then the Bolzano-Weierstrass theorem implies that T has an accumulation point, y, in \mathbb{R}. We can construct a subsequence (s_n) that converges to y. For each k \in \mathbb{N}, let A_k = (y - 1/k,y+ 1/k) be the neighborhood about y of radius 1/k. Since y is an accumulation point of T, given any k \in \mathbb{N}, there are infinitely many values of n such that s_n \in A_k. In general we choose s_{n_k} \in A_k with n_k > n_{k-1}. By doing so we obtain a subsequence (s_{n_k}) of (s_n) for which |s_{n_k} - y| < 1/k for all k \in \mathbb{N}. Theorem 1.1.1 implies \lim s_{n_k} = y.
And for unbounded sequences
Every unbounded sequence contains a monotone subsequence that diverges to \pm \infty.
Suppose (s_n) is unbounded above. Given any M \in \mathbb{R}, there must be infinitely many terms of (s_n) larger than M. There exists n_1 \in \mathbb{N} such that s_{n_1} > 1. There must also exists n_2 > n_1 such that s_{n_2} > \max \{2,s_{n_1}\}. In general, given n_1,\dots,n_k there exists n_{k+1} > n_{k} such that s_{n_{k+1}} > \max \{k,s_{n_k}\}. It follows that the subsequence (s_{n_k}) is unbounded and increasing. By Theorem 1.3.2, \lim s_{n_k} = + \infty.
Finally if (s_n) is not unbounded above, it must be unbounded below, and a similar argument produces an unbounded decreasing subsequence having \lim s_{n_k} = -\infty.
Limit Superior and Limit Inferior
Let (s_n) be a bounded sequence. A subsequential limit of (s_n) is any real number that is the limit of some subsequence of (s_n). If S is the set of all subsequential limits of (s_n), then we define the limit superior of (s_n) to be
\lim \sup s_n = \sup S
And the limit inferior of (s_n) to be
\lim \inf s_n = \inf S
This defintion requires (s_n) to be bounded. When (s_n) is a convergent sequence, \lim \inf s_n = \lim \sup s_n = \lim s_n. If it happens that \lim \inf s_n < \lim \sup s_n, we say that (s_n) oscillates.
Let (s_n) be a bounded sequence and let m = \lim \sup s_n. Then the following properties hold:
(a) For every \varepsilon > 0, there exists a natural number N such that n \geq N implies s_n < m + \varepsilon.
(b) For every \varepsilon > 0, and for every i \in \mathbb{N} there exists an integer k > i such that s_k > m - \varepsilon.
Furthermore, if m is a real number satisfying these properties, m = \lim \sup s_n.
Let (s_n) be a bounded sequence and let m = \lim \sup s_n. Then m \in S, where S is the set of subsequential limits of (s_n). That is, there exists a subsequence of (s_n) that converges to m.
Combining both parts of Theorem 1.3.4, we have the existence of some subsequence (s_{n_k}) of (s_n) such that
m - \frac{1}{k} < s_{n_k} < m + \frac{1}{k}
Clearly then (s_{n_k}) converges to m.
For a bounded sequence (s_n), Corollary 1.3.5 says the set S of subsequential limits of (s_n) contains its supremum m. That means m is actually the maximum of S. Likewise, \lim \inf s_n = \min S. While this defintion is more correct, it is not readily generalized to unbounded sequences.
Suppose that (r_n) converges to a positive number r and (s_n) is a bounded sequence. Then
\lim \sup r_n s_n = r\lim \sup s_n
Les s = \lim \sup s_n and t = \lim \sup r_ns_n. By the Corollary 1.3.5, there exists a subsequence (s_{n_k}) of (s_n) such that \lim s_{n_k} = s. We know \lim r_{n_k} = r by Theorem 1.4.1. Thus \lim r_{n_k}s_{n_k} = rs. Thus rs \leq \lim \sup r_n s_n = t (Why).
Similarly, let (r_{n_k}s_{n_k}) be a subsequence of (r_ns_n) that converges to t. Then since r > 0,
\lim s_{n_k} = \lim \frac{r_{n_k}s_{n_k}}{r_{n_k}} = \frac{t}{r},
so that t/r \leq s. Thus t \leq rs. Since rs \leq t and t \leq rs, we have t = rs.
Practice
Recap
If a sequence (s_n) converges to a real number s, then every subsequence of (s_n) also converges to s.
Every bounded sequence has a convergent subsequence.
Every unbounded sequence contains a monotone subsequence that diverges to \pm \infty.
Let (s_n) be a bounded sequence and let m = \lim \sup s_n. Then the following properties hold:
(a) For every \varepsilon > 0, there exists a natural number N such that n \geq N implies s_n < m + \varepsilon.
(b) For every \varepsilon > 0, and for every i \in \mathbb{N} there exists an integer k > i such that s_k > m - \varepsilon.
Furthermore, if m is a real number satisfying these properties, m = \lim \sup s_n.
Let (s_n) be a bounded sequence and let m = \lim \sup s_n. Then m \in S, where S is the set of subsequential limits of (s_n). That is, there exists a subsequence of (s_n) that converges to m.
Suppose that (r_n) converges to a positive number r and (s_n) is a bounded sequence. Then
\lim \sup r_n s_n = r\lim \sup s_n