How can I prove that $\limsup a_n= \sup\{a_n,a_{n+1},...\}$?
I also need to prove: for two sequences $a_n>0$ and $b_n \ge 0$, then $\limsup(a_n b_n) \le \limsup(a_n) \limsup(b_n)$.
I thought that instead I can prove $\limsup a_n=\sup\{s|s \text{ is a sub-limit of } a_n\}=\sup\{a_n,a_{n+1},...\}$
Quick approach, using the equivalent definition:
We have $$ \limsup_{n \to \infty} a_n b_n = \lim_{n \to \infty} \sup_{m \geq n} a_m b_m = \lim_{n \to \infty} \sup_{m=m',m \geq n} a_m b_{m'} \leq\\ \lim_{n \to \infty} \sup_{m,m' \geq n} a_m b_{m'} = \lim_{n \to \infty} \left(\sup_{m \geq n} a_m b_{m'}\right) \left(\sup_{m' \geq n} a_m b_{m'}\right) =\\ \left(\lim_{n \to \infty}\sup_{m \geq n} a_m b_{m'}\right) \left(\lim_{n \to \infty}\sup_{m' \geq n} a_m b_{m'}\right)=\\ \limsup a_n \cdot \limsup b_n $$
Using the sublimit definition:
Let $a = \limsup \{a_n\}$ and $b = \limsup \{b_n\}$. Consider an arbitrary convergent subsequence $a_{n_k}b_{n_k}$ of $\{a_nb_n\}$. If we can show that $$ \lim_{k \to \infty} a_{n_k} b_{n_k} \leq a \cdot b $$ then we'd be able to say that $a \cdot b$ is an upper-bound to the set of all sublimits of $a_n b_n$. So, by the definition of a supremum, we could deduce that $$ \limsup(a_n b_n) \le \limsup(a_n) \limsup(b_n) $$ Which is what we want.
To prove that this is the case, we could proceed as follows: let $\epsilon > 0$ be arbitrary. Let $L = \lim_{n \to \infty} a_{n_k}b_{n_k}$, noting that $L \leq \limsup(a_n b_n)$. There exists a $K$ such that for $k > K$, we have:
In particular, we have (for any such $k$) $$ L - \epsilon \leq a_k b_k \leq (a+\epsilon)(b+\epsilon) $$ Noting that $\epsilon > 0$ was arbitrary, we conclude that $L \leq ab$, as desired.
