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.
Best Answer
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.