Show a sequence is bounded and equals to infimum of supremum Let $(s_n)$ be a convergent sequence. I have already proved that $\{ s_n : n \in \mathbb{N} \} $ is bounded set. I m curious to know why 
$$ \lim_{n \to \infty} s_n = \inf \{ \sup \{ s_n,s_{n+1},.... \} : n \in \mathbb{N} \} $$
 A: Let $a$ and $b$ be real numbers such that $a \le s_n \le b$ for all $n\in \Bbb N$. Let $a_n = \sup\{s_n,s_{n+1},\ldots\}$, for all $n\in \Bbb N$. Then $a_1 \ge a_2 \ge a_3 \ge \cdots \ge a$, so by the monotone convergence theorem, $\lim_{n\to \infty} a_n$ exists. Then $s_n \le a_n$ for all $n$, and hence $\lim_{n\to \infty} s_n \le \lim_{n\to \infty} a_n$. On the other hand, if $L = \lim_{n\to \infty} s_n$, then given $\epsilon > 0$, there exists $k \in \Bbb N$ such that $s_n < L + \epsilon/2$ for all $n \ge k$. So if $n\ge k$, by definition of $a_k$, there exists an $m \ge k$ such that $s_m > a_k - \epsilon/2$. Since $a_n$ is decreasing, for all $n \ge k$, $a_n \le a_k < s_m + \epsilon/2 < L + \epsilon$. Therefore $\lim_{n\to \infty} a_n \le L + \epsilon$. Since $\epsilon$ was arbitrary, $\lim_{n\to \infty} a_n \le L = \lim_{n\to \infty} s_n$. Therefore, $\lim_{n\to \infty} s_n = \lim_{n\to \infty} a_n$.
A: Because you can construct an extraction of the sequence $(s_n)_n$ converging to you right hand side term. ;-)
Your right hand side term is called (independently on if $(s_n)_n$ converges or not) the limit superior of the sequence $(s_n)_n$ and the inf can be replaced by a lim as your sup sequences decrease. If you switch inf and sup in the right-hand side, you get what it's called the limit inferior of $(s_n)_n$. It is well-known that $(s_n)_n$  is convergent if and only if its limit inferior and superior are equal, that the limit inferiori is smaller than the limit superior, and that the limit inferior is the smallest adherence value of $(s_n)_n$ and that the limit superior is the biggest adherence value of $(s_n)_n$.
You can have a look here :
http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior
and here :
Prove that limit inferior is same as limit superior for a convergent sequence
A: Define $t_N = \sup \{ s_N,s_{N+1},\dots \}$. We have two observations to make. First, $t_N$ is "almost" a subsequence of $s_n$. By this I mean that there is :


*

*a subsequence $s_{n_N}$ and

*a sequence $\{ r_N \}_{N=1}^\infty$ 


such that 


*

*$n_N \geq N$,

*$t_N = s_{n_N} + r_N$,

*$r_N \geq 0$, and

*$\lim_{N \to \infty} r_N = 0$.


This follows by the definition of supremum: for each $N$ we can pick a member of $\{ s_N,s_{N+1},\dots \}$, call it $s_{n_N}$, which is arbitrarily close to $t_N$, where we call the difference $r_N$. This means we can actually do a little better than the above, in that we can actually require something like $1/N \geq r_N \geq 0$, but this is not necessary for the proof.
The other observation is that $t_N$ is a decreasing sequence, which means that its infimum is its limit. So now you need to check that if one sequence is "almost" a subsequence of a convergent sequence, then both have the same limit.
