$\lim \sup s_n = \sup\{s_n : n > N\}$ when $N \rightarrow +\infty$
To me, this definition, at the first glance, is wired and caused some difficulties to understand it. Here is the understanding I finally got.
For any given $N$, $\sup \{s_n : n > N\} \geq t$ and for all $\epsilon > 0$, there exits a $N_1$ s.t. $\sup \{s_n : n > N_1\} < t + \epsilon$. Because $\lim \sup s_n$ is defined by the limit of sequence $(u_n)$ where $u_k = \sup {s_n : n > N_k}$ and $(N_k)$ is a strictly increasing sequence, i.e. $n < m \implies N_n < N_m$. Exercise 12.1 would improve the understanding of $\lim \sup s_n$ and $\lim \inf s_n$.
Besides, $\{s_n: s_n < \lim \sup \{s_n\} \}$ could be an empty set.
Recall the sequence $s_n = 1/n$, $\lim s_n = 0$, therefore $\lim \sup s_n = \lim \inf s_n = \lim s_n = 0$ and all $s_n > 0$. See page 61, $\lim \sup s_n$ is the largest value that infinitely many $s_n$'s can approach to.
Extra tip
If $(s_{n_k})$ is a subsequence of $(s_n)$, we have
and therefore
For any given $N$, $\sup \{s_n : n > N\} \geq t$ and for all $\epsilon > 0$, there exits a $N_1$ s.t. $\sup \{s_n : n > N_1\} < t + \epsilon$. Because $\lim \sup s_n$ is defined by the limit of sequence $(u_n)$ where $u_k = \sup {s_n : n > N_k}$ and $(N_k)$ is a strictly increasing sequence, i.e. $n < m \implies N_n < N_m$. Exercise 12.1 would improve the understanding of $\lim \sup s_n$ and $\lim \inf s_n$.
Besides, $\{s_n: s_n < \lim \sup \{s_n\} \}$ could be an empty set.
Recall the sequence $s_n = 1/n$, $\lim s_n = 0$, therefore $\lim \sup s_n = \lim \inf s_n = \lim s_n = 0$ and all $s_n > 0$. See page 61, $\lim \sup s_n$ is the largest value that infinitely many $s_n$'s can approach to.
Extra tip
If $(s_{n_k})$ is a subsequence of $(s_n)$, we have
$\{s_{n_k}: k > N\} \subseteq \{s_n : n > N\}$
$\lim \inf s_n \leq \lim \inf s_{n_k}$ and $\lim \sup s_{n_k} \leq \lim \sup s_n$
See page 74. Besides, $n_k \geq k$ because when $k=1$, $n_1 \geq 1$ and $n_{k} > n_{k-1}$.
Exercise 11.8
Exercise 12.4 and Exercise 12.5
If $(t_n)$ and $(s_n)$ are bounded sequences, then
Exercise 11.8
$\lim \inf s_n = - \lim \sup (-s_n)$
If $(t_n)$ and $(s_n)$ are bounded sequences, then
$\lim \inf s_n + \lim \inf t_n \leq \lim \inf (s_n + t_n) \leq \lim \sup (s_n + t_n) \leq \lim \sup s_n + \lim \sup t_n$
Exercise 12.6
Let $(s_n)$ be a bounded sequence, and let $k$ be a nonnegative real number.
1. $\lim \sup (ks_n) = k \cdot \lim \sup s_n$.
2. $\lim \inf (ks_n) = k \lim \inf (s_n)$.
If $k < 0$, then $k \lim \sup s_n = \lim \inf (ks_n) $ and $\lim \sup (ks_n) = k \lim \inf (s_n)$
A conclusion used in Exercise 12.9 but not directly given by the text
A conclusion used in Exercise 12.9 but not directly given by the text
$ \inf \{s_n : n \in \mathbb{N}\} \leq \lim \inf s_n \leq \lim \sup s_n \leq \sup \{ s_n: n \in \mathbb{N} \}$
Def. of $\lim \sup s_n$ and $\lim \inf s_n$ on Page 60
If $(s_n)$ is not bounded above, $\lim \sup (s_n) = + \infty$. If $(s_n)$ is not bounded below, $\lim \inf (s_n) = - \infty$.
Combining Theorem 11.8 and the first paragraph of section 12, we can conclude that, for any sequence $(s_n)$, $\lim \inf s_n$ and $\lim \sup s_n$ always make sense, regardless of whether $\lim (s_n)$ exists or not.
Combining Theorem 11.8 and the first paragraph of section 12, we can conclude that, for any sequence $(s_n)$, $\lim \inf s_n$ and $\lim \sup s_n$ always make sense, regardless of whether $\lim (s_n)$ exists or not.
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