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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.

Hosein Rahnama
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eChung00
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3 Answers3

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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.

Michael Hardy
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  • Do you mean largest number L? – user123124 Apr 03 '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
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    @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 01:25
  • @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 ..... –  Feb 25 '16 at 20:20
  • @probabilitydoesntsuck : No. Look at the example. Suppose $\varepsilon=0.0001$. It is not true that for all $n$ beyond some point $a_n > 3+\varepsilon$, since many of the numbers in the sequence are near $5$. $\qquad$ – Michael Hardy Feb 26 '16 at 04:39
  • Oops. Thanks I got it now –  Mar 02 '16 at 22:27
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    Why couldn’t it exist a sequence approaching 3 from above and 5 from below, where therefore lim inf = inf (the smallest inf) and lim sup = sup (the largest sup)? – Mr Frog Nov 05 '21 at 08:38
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I understand $\limsup s_n$ and $\liminf s_n$ as the largest and smallest subsequential limits of $s_n$.

Quinn Culver
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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?

Jean-Sébastien
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  • 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
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    @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
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    @eChung00 That would work, but you can get more using only $\limsup$ and $\liminf$ – Jean-Sébastien Sep 14 '13 at 16:36
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    "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 02 '17 at 15:37
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    @eChung00 When N=1, sup is 9 and inf is -9, not 8 and -8, as you said. – WhySee Dec 17 '17 at 02:25