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Limit Supremum and Infimum. Struggling the concept

Hey I'm trying to figure out what $\limsup{S_{n}}$ is compared to $\lim{S_{n}}$ as well as the difference of $\lim{S_{n}}$ and $\liminf{S_{n}}$

So for example (this is my current thinking process) if I have a monotone non increasing sequence $S_{n}:=1/n$ (where $n=1$ and goes to infinity). The $\limsup{S_{n}}$ is 1, and $\liminf{S_{n}}$ is 0. But we know the $\lim{S_{n}}$ is 0.

How does $\lim{S_{n}}=\liminf{S_{n}}=\limsup{S_{n}}?$

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    See my answer from a previous question: http://math.stackexchange.com/questions/205223/limit-supremum-and-infimum-struggling-the-concept/205235#2052352012-10-16

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One definition of $\limsup s_n$ is $\limsup s_n = \lim_{n \to \infty} \sup_{k \geq n} s_k$ The corresponding definition of $\liminf s_n$ is $\liminf s_n = \lim_{n \to \infty} \inf_{k \geq n} s_k$ In your case, where $s_n = \dfrac1n$, we have $\sup_{k \geq n} s_k = \sup_{k \geq n} \dfrac1k = \dfrac1n$ Similarly, for $\liminf$. Hence, $\limsup s_n = \lim_{n \to \infty} \sup_{k \geq n} s_k = \lim_{n \to \infty} \dfrac1n = 0$

In general, if $\displaystyle \lim_{n \to \infty} s_n$ exists, then $\limsup s_n = \lim s_n = \liminf s_n$

Another way to define $\limsup$ and $\liminf$ is to look at the limit points of the sequence $s_n$ i.e. if $S = \{\text{Limit points of the sequence }s_n\}$ then $\limsup s_n = \displaystyle \sup_{s \in S} S$ and $\liminf s_n = \displaystyle \inf_{s \in S} S$ If $s_n = \dfrac1n$, then $S = \{0 \}$. Hence, $\limsup s_n = 0 = \liminf s_n$

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$limsupS_n$ of $x_n$ is the largest cluster point of $x_n$ if sequence is bounded above. $liminfS_n$ is the smallest cluster point if it is a bounded below sequence.

If a sequence converges to some $x$ its every subsequence converges to $x$. This is (most simpy) how lim$S_n$ = limsup$S_n$. The sequence of 1/n converges to 0, its only cluster point is 0. Thus lim$S_n$ = liminf$S_n$ = limsup$S_n$ = 0.