# 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.

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 and 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.

• Do you mean largest number L? – user1 Apr 3 '15 at 11:26
• @Johan : Yes. I've fixed that now. Thanks for pointing it out. ${}\qquad{}$ – Michael Hardy Sep 10 '15 at 13:30
• dumb question: are inf and sup just fancy math names for min and max? – Jason S Dec 15 '15 at 23:24
• @JasonS : The set of all numbers less than $3$ does not have a maximum, but does have a supremum, which is $3$. ${}\qquad{}$ – Michael Hardy Dec 16 '15 at 1:25
• @MichaelHardy Shouldn't it be $a_n < L+\epsilon$ in the definition of $\liminf\limits_{n \to \infty} a_n$ as the Largest L such that ..... – user286838 Feb 25 '16 at 20:20

I understand $\limsup s_n$ and $\liminf s_n$ as the largest and smallest subsequential limits of $s_n$.

• This take a little doing to show but it's a great little exercise and it sheds real insight. – ncmathsadist Sep 23 '14 at 1:14

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?

• For your exercise, I thisnk the sequence has to be bounded in order to converge. I think I saw one theorem in the book. And I have one more question!! Let $s_n=(8,-8, 4,-4,5,-5,7,-7,9 ,-9 ,1,-1,0,0,...)$. As you stated above, lim sup decreases as n gets bigger and lim inf gets smaller as n gets bigger. When N = 1, -8 is smallest and 8 is largest. But when N = 3, -9 is smallest and 9 is largest which implies that inf decreased and sup increased. – eChung00 Sep 14 '13 at 16:29
• @eChung00 Boundedness is not enough, for example $1,-1,1,-1,\ldots$ is bounded, yet does not converge. Try something in terms of $\liminf$ and $limsup$. For example, what would be the $\liminf$ and $\limsup$ of the sequence $1,-1,1,-1,\ldots$? Compare them with the sequence in my answer. – Jean-Sébastien Sep 14 '13 at 16:31
• a bounded monotonic sequence?? – eChung00 Sep 14 '13 at 16:35
• @eChung00 That would work, but you can get more using only $\limsup$ and $\liminf$ – Jean-Sébastien Sep 14 '13 at 16:36
• "As NN increases, there are "less and less" value to choose from, hence the lim suplim sup can only decrease (or stay constant)." Crucial insight @Jean-Sébastien, thank you! – Konstantin May 2 '17 at 15:37

## protected by Zev ChonolesNov 12 '15 at 6:49

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