Can someone clearly explain about the lim sup and lim inf? Can some explain the lim sup and lim inf?
In my text book the definition of these two is this.
Let $(s_n)$ be a sequence in $\mathbb{R}$. We define 
$$\lim \sup\ s_n = \lim_{N \rightarrow \infty} \sup\{s_n:n>N\}$$
and
$$\lim\inf\ s_n = \lim_{N\rightarrow \infty}\inf\{s_n:n>N\}$$
The right side of these two equality, can I think $\sup\{s_n:n>N\}$ and $\inf\{s_n:n>N\}$ as a sequence after $n>N$?
And how these two behave as $ n$ increases? My professor said that these two get smaller as $n$ increases.
 A: I understand $\limsup s_n$ and $\liminf s_n$ as the largest and smallest subsequential limits of $s_n$.
A: Think of it this way. In the $\limsup$ , you are taking the biggest value past a certain $N$. As $N$ increases, there are "less and less" value to choose from, hence the $\limsup$ can only decrease (or stay constant).
Same thing applies with $\liminf$ except that as you get "less and less value" you can only increase (or keep it the same) the value of your $\liminf$.
As a simple example, take a sequence to be
$$
s_n=(4,-4,3,-3,2,-2,1,-1,0,0,\ldots)
$$
Fix $N=1$ then the largest value past or at $N=1$ is $4$ and the smallest is $-4$. A few steps later, say $N=4$, the largest value past or at $N$ is $2$ and the smallest is now $-3$. Further away, at $N=10$ we have $\sup_{n\ge 10}=\inf_{n\ge 10}=0$.
From this you see that $\limsup$ decreases and $\liminf$ increases.
Exercise: What needs to happen for the sequence to converge?
A: Consider this example:
$$
3-\frac12,\quad 5+\frac13,\quad 3-\frac14,\quad 5+\frac15,\quad 3-\frac16,\quad 5+\frac17,\quad 3-\frac18,\quad 5+\frac19,\quad\ldots\ldots
$$
It alternates between something approaching $3$ from below and something approaching $5$ from above.  The lim inf is $3$ and the lim sup is $5$.
The inf of the whole sequence is $3-\frac12$.
If you throw away the first term or the first two terms, the inf of what's left is $3-\frac14$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac16$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac18$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac1{10}$.
. . . and so on.  You see that these infs are getting bigger.
If you look at the sequence of infs, their sup is $3$.
Thus the lim inf is the sup of the sequence of infs of all tail-ends of the sequence.  In mathematical notation,
$$
\begin{align}
\liminf_{n\to\infty} a_n & = \sup_{n=1,2,3,\ldots} \inf_{m=n,n+1,n+2,\ldots} a_m \\[12pt]
& = \sup_{n=1,2,3,\ldots} \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} : n=1,2,3,\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\{ a_m : m\ge n\} : n=1,2,3,\ldots \right\}.
\end{align}
$$
Just as the lim inf is a sup of infs, so the lim sup is an inf of sups.
One can also say that $L=\liminf\limits_{n\to\infty} a_n$ precisely if for all $\varepsilon>0$, no matter how small, there exists an index $N$ so large that for all $n\ge N$, $a_n>L-\varepsilon$, and $L$ is the largest number for which this holds.
