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
$$t_k = s_{n_k} \tag{2}$$
Let $(s_n) $ be a sequence.
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.If the sequence $(s_n) $ is unbounded above, it has a subsequence with limit $+\infty $.
Similarly, if $(s_n) $ is unbounded below, a subsequence has limit $-\infty$.
In each case, the subsequence can be taken monotonic.
If the sequence $(s_n) $ converges, then every subsequence converges to the same limit.
Every sequence $(s_n)$ has a monotonic subsequence.
Every bounded sequence has a convergent subsequence.
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) $.
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 } $.
Let $(s_n) $ be any sequence in $\R $, and let $S $ denote the set of subsequential limits of $(s_n) $.
- $S $ is nonempty.
- $\supr{ S } = \lim \supr{ s_n }$
$\infi{ S } = \lim \infi{ s_n }$. - $\lim s_n$ exists $\iff $ $S = \set{\lim s_n}$.
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 $.
The closure of $S $ = $S \cup \set{ \text{subsequential limits of } S } $
A set is closed $\iff $ the set coincides with its closure.
A set is closed $\iff $ the set contains all its subsequential limits.
$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 $.