A question about suggesting idea to give a formal proof to a theorem about sequence Theorem Every sequence {$s_n$} has a monotonic subsequence whose limit is equal to $\limsup  s_n$. I think to show that there exist a monotonic subsequence is kind of straight forward but I could show that there exist such subsequences whose limit is $\limsup s_n$.
 A: Let $L=\limsup_n s_n$. (Note that $L$ can be $\infty$.) Probably the simplest approach is to prove first that $\langle s_n:n\in\Bbb N\rangle$ has a subsequence $\langle s_{n_k}:k\in\Bbb N\rangle$ converging to $L$, and then show that $\langle s_{n_k}:k\in\Bbb N\rangle$ has a monotonic subsequence.
By definition $$L=\lim_{n\to\infty}\sup_{k\ge n}s_k\;,$$ so for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $$\left|L-\sup_{k\ge n}s_k\right|<\frac{\epsilon}2$$ for all $n\ge n_\epsilon$. By the definition of supremum there is a $k_\epsilon\ge n_\epsilon$ such that $$0\le\left(\sup_{k\ge n}s_k\right)-s_{k_\epsilon}<\frac{\epsilon}2\;,$$ and therefore 
$$|L-s_{k_\epsilon}|\le\left|L-\sup_{k\ge n}s_k\right|+\left|\left(\sup_{k\ge n}s_k\right)-s_{k_\epsilon}\right|<\epsilon\;.$$
For $i\in\Bbb Z^+$ let $n_i=k_{1/i}$, so that $|L-s_{n_i}|<\frac1i$; clearly the sequence $\langle s_{n_i}:i\in\Bbb Z^+\rangle$ converges to $L$. However, it may not be a subsequence of $\langle s_n:n\in\Bbb N\rangle$, because the indices $n_1,n_2,n_3,\dots$ may not be increasing. To finish the argument, you must do two things.


*

*Show that $\langle n_i:i\in\Bbb Z^+\rangle$ has an increasing subsequence $\langle n_{i_j}:j\in\Bbb Z^+\rangle$; $\langle s_{n_{i_j}}:j\in\Bbb Z^+\rangle$ is then a subsequence of $\langle s_n:n\in\Bbb N\rangle$ converging to $L$.

*Show that $\langle s_{n_{i_j}}:j\in\Bbb Z^+\rangle$ has a monotonic subsequence. That subsequence will necessarily have the same limit, $L$, as $\langle s_{n_{i_j}}:j\in\Bbb Z^+\rangle$ itself.
A: Let $S=\limsup\limits_{n\to\infty} s_n$.

  
*
  
*First we construct a subsequence $s_{n_k}$ such that $\lim\limits_{k\to\infty} s_{n_k}=S$ and
  $$k\le l \Rightarrow |S-s_{n_k}| \ge |S-s_{n_l}|.$$
  

In the other words, we construct a subsequence which converges to $S$ and has the property, that the every term of the subsequence is closer so $S$ than the previous one (or in the same distance from $S$).
To prove this we only use the fact that for every $\varepsilon>0$ there are infinitely many $n$'s such that 
$$|S-s_n|<\varepsilon.$$
(Maybe it might be simpler to treat the case that there are infinitely many $n$'s such that $s_n=S$ separately.)

  
*
  
*If we have sequence with the above property, then one of the sets $\{k\in\mathbb N; s_{n_k}\ge S\}$ and $\{k\in\mathbb N; s_{n_k}\le S\}$ is infinite.
  We take the one, which is infinite, and we obtain a monotone subsequence converging to $S$.
  

A: I shall prove a similar property of the infimum
Let $\left(a_n\right)$ be a bounded sequence of real numbers and $a=\limsup{a_n}$, $\epsilon>0$
The set $\left\{n\in \mathbb{N}|a-\epsilon<a_n\right\}$ is infinite while the set $\left\{n\in \mathbb{N}|a+\epsilon<a_n\right\}$ is finite (why?) and thus the sets
$\left\{n\in \mathbb{N}|a-\epsilon<a_n\right\}$ and $\left\{n\in \mathbb{N}|a_n\le a+\epsilon\right\}$ are both infinite subsets of $\mathbb{N}$
Therefore, the set
\begin{equation}\left\{n\in \mathbb{N}|a-\epsilon<a_n\le a+\epsilon\right\}=\left\{n\in \mathbb{N}|a-\epsilon<a_n\right\}\bigcap\left\{n\in \mathbb{N}|a_n\le a+\epsilon\right\} \text{ is infinite (why?)} \tag{1}\end{equation}
(the reader should try to prove that if $X$ is infinite and $A,B\subseteq X$ are infinite and $B^c$ is finite then $A\bigcap B$ is infinite. Hint: $A=A\bigcap (B\bigcup B^c)$)
For $\epsilon=1,\frac{1}{2},...,\frac{1}{n}$ we can construct by (1) a subsequence $\left({{a}_{k_n}}\right)$ such as that $a-\dfrac{1}{n}<a_{k_n}\le a+\dfrac{1}{n}$
Then, \begin{equation}\forall n\in \mathbb{N}\ \ \left|{{a}_{k_n}}-a\right|<\dfrac{1}{n}\end{equation}and thus $a_{k_n}\to a$. 
Next note that every sequence has a monotone subsequence, the proof can be found here: http://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem#Proof
