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Definition

Suppose $(s_n)_{n \in \N} $ is a sequence. A subsequence of this sequence is a sequence of the form $(t_k)_{k \in \N} $ where for each $k $ there is a positive integer $n_k $ such that

$$n_1 < n_2 < … < n_k < n_{k+1} < … \tag{1}$$

and

$$t_k = s_{n_k} \tag{2}$$

Theorem

Let $(s_n) $ be a sequence.

  1. If $t $ is in $\R $, there is a subsequence of $(s_n)$ converging to $t$
    $\iff \forall \epsilon > 0$ the set $\set{n \in \N \mid \abs{ s_n - t } < \epsilon} $ is infinite.

  2. If the sequence $(s_n) $ is unbounded above, it has a subsequence with limit $+\infty $.

  3. Similarly, if $(s_n) $ is unbounded below, a subsequence has limit $-\infty$.

In each case, the subsequence can be taken monotonic.

Theorem

If the sequence $(s_n) $ converges, then every subsequence converges to the same limit.

Theorem

Every sequence $(s_n)$ has a monotonic subsequence.

Theorem : Bolzano-Weierstrass Theorem

Every bounded sequence has a convergent subsequence.

Definition

Let $S_n$ be a sequence in $\R$. A subsequential limit is any real number or symbol $+\infty$ or $-\infty$ that is the limit of some subsequence of $(s_n) $.

Theorem

Let $(s_n) $ be any sequence. There exists a monotonic subsequence whose limit is $\lim \supr{ s_n } $, and there exists a monotonic subsequence whose limit is $\lim \infi{ s_n } $.

Theorem

Let $(s_n) $ be any sequence in $\R $, and let $S $ denote the set of subsequential limits of $(s_n) $.

  1. $S $ is nonempty.
  2. $\supr{ S } = \lim \supr{ s_n }$
    $\infi{ S } = \lim \infi{ s_n }$.
  3. $\lim s_n$ exists $\iff $ $S = \set{\lim s_n}$.
Theorem

Let $S$ denote the set of subsequential limits of a sequence $(s_n) $. Suppose $(t_n) $ is a sequence in $S \cap \R$ (i.e. the sequence $t$ consists of elements from $S$) and that $t = \lim t_n$. Then $t$ belongs to $S $.

Proof [expand]

Suppose $t $ is finite.

There exist $t_n \in (t- \epsilon, t+ \epsilon)$.

Let $\delta = \min{ \set{ \pare{t + \epsilon} - t_n, t_n - \pare{t - \epsilon}}} $ s.t.

$$(t_n - \delta, t_n + \delta) \subseteq (t - \epsilon, t + \epsilon) $$

Since $t_n$ is a subsequential limit, the set $\set{ n \in \N : s_n \in (t_n - \delta, t_n + \delta)} $ is infinite, so the set $n \in \N : s_n \in (t - \epsilon, t + \epsilon) $ is also infinite. Thus $t $ it self is a subsequential limit of $(s_n) $.

If $t = +\infty $, then clearly the sequence $(s_n) $ is unbounded above, so a subsequence of $(s_n) $ has limit $+ \infty$. Thus $+ \infty $ is also in $S $. Similar for $t = -\infty $.

Definition

The closure of $S $ = $S \cup \set{ \text{subsequential limits of } S } $

Definition: Closed Set

A set is closed $\iff $ the set coincides with its closure.

A set is closed $\iff $ the set contains all its subsequential limits.

Definition: Partial Sums

$s_n = \sum_{ k = m }^{ n }a_k = a_m + a_{m+1} + … + a_n $

If $s_n$ converges to a value $s $, then we say $\sum_{ m }^{ \infty } a_n $ converges to $s $.

If no such $s$ exists, then $\sum_{ m }^{ \infty } a_n $ diverges.

If $\sum_{ m }^{ \infty } a_n $ diverges, then $\limu{ n }{ \infty } \sum_{ m }^{ n } \abs{ a_n } = +\infty $.