Prove: $\liminf_{n \to \infty} s_n \le \limsup_{n \to \infty} s_n$

Question

I am looking over examples and the definitions for this section but I am still not familiar with all the tricks. I appreciate any help with proving this (from hints to maybe a solution. It is only a review so I just need something to look at and make sure I am doing it right). Thank you.

Prove $$\liminf_{n \to \infty} s_n \le \limsup_{n \to \infty} s_n$$

Answer 1

Hint: for every $n$ we have $$\inf_{m \ge n}s_m \le \sup_{m \ge n}s_m$$

Edit: if the sequence $\{s_n\}$ is bounded then just take the limit and you are done.

If the sequence is not bounded make sure you are defining things in the following way:

set $\alpha_n = \inf_{m \ge n}s_m$ then define $$\liminf_{n \to \infty}s_n = \begin{cases}-\infty & \text{if}\ \alpha_n = -\infty\ \text{for all}\ n\\ \lim_{n \to \infty}\alpha_n & \text{otherwise.}\ \end{cases}$$

Do the same for the limit superior and finally everything works. (this prevents you from meaningless things like taking the limit of $-\infty$) :D

Answer 2

Let $s_{n}$ a sequence, for all $n$ and $m$, we have $$C_{n}= \inf\limits_{k\ge n} s_{k} \le \sup\limits_{k\ge m} s_{k}=B_{m} $$ This is true, since if $p\ge \max\left\{n,m \right\}$, then $$s_{p}\in\left\{ a_{n},a_{n+1},\dots\right\} \bigcap \left\{ a_{m},a_{m+1},\dots\right\} $$ so $$\inf\limits_{k\ge n} s_{k} \le s_{p} \le \sup\limits_{k\ge m} s_{k}$$ as, $C_{n}\le B_{m}$ for all $m$, then we have $$C_{n}\le \lim\limits _{n\rightarrow \infty} B_{n}$$ and for all $n$, ve have $$\lim\limits _{n\rightarrow \infty}C_{n}\le \lim\limits _{n\rightarrow \infty} B_{n}$$

Answer 3

Just use the definitions of $\lim \sup s_n = L$and $\lim \inf s_n = l$ of the sequence $\{s_n\}$.

For any $\epsilon > 0$ $\exists$ $k_1 \in \mathbb{N}$ s.t. $s_n > l - \epsilon$ $\forall$ $n>k_1$ and $s_n < l + \epsilon $ for infinitely many $n$.

For any $\epsilon > 0$ $\exists$ $k_2 \in \mathbb{N}$ s.t. $s_n < L + \epsilon$ $\forall$ $n>k_2$ and $s_n < L - \epsilon $ for infinitely many $n$.

Thus for for any natural number $k > \max\{k_1, k_2\}$, $s_k > l -\epsilon$ and $s_k < L+\epsilon$.

So we shall get $l - \epsilon < s_n < L + \epsilon$ $\forall n > k$ Taking $\epsilon \rightarrow 0$ we shall get $L \ge l$, i.e. $\lim \sup \geq \lim \inf$.

The above proof is valid when $\{s_n\}$ is a bounded sequence. We say $\{s_n\}$ is unbounded above when $\lim \sup s_n = \infty$ and unbounded below when $\lim \inf s_n = - \infty$. Proof is trivial in this case.