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.


5 Answers 5


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.

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    $\begingroup$ Do you mean largest number L? $\endgroup$
    – user123124
    Commented Apr 3, 2015 at 11:26
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    $\begingroup$ @Johan : Yes. I've fixed that now. Thanks for pointing it out. ${}\qquad{}$ $\endgroup$ Commented Sep 10, 2015 at 13:30
  • $\begingroup$ dumb question: are inf and sup just fancy math names for min and max? $\endgroup$
    – Jason S
    Commented Dec 15, 2015 at 23:24
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    $\begingroup$ @JasonS : The set of all numbers less than $3$ does not have a maximum, but does have a supremum, which is $3$. ${}\qquad{}$ $\endgroup$ Commented Dec 16, 2015 at 1:25
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    $\begingroup$ 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)? $\endgroup$
    – Mr Frog
    Commented Nov 5, 2021 at 8:38

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

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    $\begingroup$ This take a little doing to show but it's a great little exercise and it sheds real insight. $\endgroup$ Commented Sep 23, 2014 at 1:14
  • $\begingroup$ To expand on this, the sequence $s_n$ itself may not converge, but it could be bounded. We still want to develop a concept of a "limit" for $s_n$. If $s_n$ is bounded, then by the Bolzano-Weierstrass theorem, there exists at least one convergent subsequence of $s_n$. For each convergent subsequence of $s_n$, take its limit and put it into a set $S$. We then have that $\limsup s_n = \sup S$ and $\liminf s_n = \inf S$. Moreover, if $\limsup s_n = \liminf s_n = L$, then $s_n \to L$. $\endgroup$
    – mhdadk
    Commented Mar 17, 2023 at 19:55
  • $\begingroup$ For future readers: see my answer below for details on my explanation above. $\endgroup$
    – mhdadk
    Commented Mar 17, 2023 at 20:20

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?

  • $\begingroup$ 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. $\endgroup$
    – eChung00
    Commented Sep 14, 2013 at 16:29
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    $\begingroup$ @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. $\endgroup$ Commented Sep 14, 2013 at 16:31
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    $\begingroup$ @eChung00 That would work, but you can get more using only $\limsup$ and $\liminf$ $\endgroup$ Commented Sep 14, 2013 at 16:36
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    $\begingroup$ "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! $\endgroup$
    – Konstantin
    Commented May 2, 2017 at 15:37
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    $\begingroup$ @eChung00 When N=1, sup is 9 and inf is -9, not 8 and -8, as you said. $\endgroup$
    – WhySee
    Commented Dec 17, 2017 at 2:25

Since this old question was reopened by a new answer, I will give my input as I believe an important property is missing from all the answers provided so far.

Let $\{a_n\}\subseteq \mathbb{R}$ be a sequence of real numbers. We often study the limit of $a_n$ when $n$ goes to infinity in order to deduce more properties about the sequence itself. This value is denoted with $\lim_{n\to\infty} a_n$. The problem with the limit is that, sometimes, it might not exist.

A simple example is the sequence $$ a_n=(-1)^{n}, $$ which oscillates between $-1$ and $1$. It is easy to show, with the epsilon delta (or any other) definition of the limit, that this sequence does not have one. However, it does have two subsequences that converge, the sequence of even-indexed elements which converges to $1$, and the sequence of odd-indexed elements which converges to $-1$. This is not a coincidence.

The $\limsup$ and $\liminf$, unlike the limit itself, always exist, and this is perhaps their main and most important property. There are two main things that we need to show this.

  1. You might know that for any set $S\subseteq \mathbb{R}$, there always exists a supremum of $S$, which is either a real number or $\pm\infty$, or similarly the infimum. In other words, $$ \text{for every $S\subseteq \mathbb{R}$ we have $\sup(S),\inf(S) \in \mathbb{R}\cup \{\pm\infty\}$}. $$
  2. The second important property to know is that if any $\{b_n\}\subseteq\mathbb{R}$ is a monotonic sequence (might be increasing or decreasing), then the limit of $b_n$ exists, and is a member of $\mathbb{R}\cup\{\pm \infty\}$. (I consider the limits going to $\infty$ or $-\infty$ as well defined "existing" limits. The sequence may not converge to a real number, but its divergence towards $\infty$ (or $-\infty$) is a well defined behaviour which gives us a lot of information.)

Now finally we get to the question of the $\limsup$ and the $\liminf$. By the definition that you provided, $$ \limsup_{n\to\infty} a_n = \lim_{N \to \infty} \sup \{a_k\:|\: k \geq N\}. $$ Let's consider the sequence ${s_N}$ where $s_N=\sup \{a_k\:|\: k \geq N\}$. \begin{align*} s_0&=\sup\{ a_0, a_1, a_2, a_3, \ldots \}, \\ s_1&=\sup\{\phantom{ a_0, }\, a_1, a_2, a_3, \ldots \}, \\ s_2&=\sup\{\phantom{ a_0, a_1 }\,\,\, a_2, a_3, \ldots \}, \\ s_2&=\sup\{\phantom{ a_0, a_1, a_2 }\,\,\, a_3, \ldots \}, \\ \,\,\,&\,\,\,\vdots \end{align*} Suppose, for the sake of simplicity, that for each of these sets $A_N=\{a_k\:|\: k \geq N\}$, their supremum is contained in the set. This is not true, far from it, but let's just assume it is. Then the supremum $s_N$ is the largest element of $A_N$.
What does this tell us about $s_{N+1}$? Well, since $s_N$ is the largest element of $A_N$, $s_{N+1}$ he largest element of $A_{N+1}$, and $A_{N+1}$ is smaller than $A_N$, then $s_{N+1}$ must be at most as large as $s_N$, but not larger. In other words $s_{N+1} \leq s_{N}$. Or, more generally, $s_0\geq s_1 \geq \cdots$, that is $$ \text{$s_N$ is a }\textit{monotone decreasing} \text{ sequence. } $$ Therefore, it has a limit. Therefore the $\limsup_{n\to\infty}a_n$ exists !

In full generality, even if $s_N\not\in A_N$, we can show that for any $S \subseteq T$ we have $\sup(S) \leq \sup(T)$ (I will leave this as an exercise). Then $s_N \geq s_{N+1}$ because $A_N \supseteq A_{N+1}$.

The argument is the same for $\liminf_{n\to\infty}a_n$, except that $S \subseteq T \implies \inf(S)\geq \inf(T)$, so the sequence of infimums is an increasing sequence, again monotonic, and thus still converging.

  • $\begingroup$ (+1) regarding the statement in bold, and I know you mentioned this in point 2 already, but I would add in bold that this is only true if the sequence is defined on the set of extended real numbers $\overline{\mathbb R} = \mathbb R \cup {\pm \infty}$. $\endgroup$
    – mhdadk
    Commented Mar 20, 2023 at 22:00
  • $\begingroup$ The question mentions real sequences, and I did say that all sequences are real. And the supremum doesn't really make sense outisde of (partially) ordered sets. $\endgroup$
    – Kolja
    Commented Mar 21, 2023 at 12:59
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    $\begingroup$ (+1) thanks for the insight $\endgroup$ Commented Aug 17, 2023 at 14:58

The concepts of $\limsup$ and $\liminf$ are useful if a sequence $a_n$ does not converge, but is bounded. In this case, we still want to develop a concept of a limit for $a_n$. Suppose that $a_n$ is bounded and does not necessarily converge. Then, by the Bolzano-Weierstrass theorem, there exists at least one subsequence of $a_n$ that does converge. Note that there may be more than one subsequence of $a_n$ that converges. More specifically, let $Q$ be the set of subsequences of $a_n$ that converge. Then, for each sequence in $Q$, take its limit and put it into the set $S$, such that $$S = \{\lim s_n \mid s_n \in Q\}$$ We then define $\limsup a_n = \sup S$ and $\liminf a_n = \inf S$. Therefore, $\limsup a_n$ can be thought of as "the largest limit of all the convergent subsequences of $a_n$" and $\liminf a_n$ can be thought of as "the smallest limit of all the convergent subsequences of $a_n$". When $a_n$ does converge to a limit $L$, then the set $S$ consists of only one element: $S = \{L\}$, and so $\limsup a_n = \sup S = \liminf a_n = \inf S = L$. This is why the concepts of $\limsup a_n$ and $\liminf a_n$ are not interesting if $a_n$ converges.

For more details on $\limsup$ and $\liminf$, see section 3.5 in the book An Introduction to Real Analysis by Cesar O. Aguilar, available for free online here or as a PDF here. For a proof that the definitions of $\limsup$ and $\liminf$ given here are equivalent to the definitions given in the question, see Theorem 3.5.5.

  • $\begingroup$ That $\limsup a_n = \text{sup} S$ is a rather non-trivial observation. And it's a stretch to call that the definition of $\limsup$. Usually it is defined in the way it is presented in the question. At least I have never seen it defined this way. It would take some work to show that the definition that you have given is the same as the one in the question. $\endgroup$
    – Kolja
    Commented Mar 17, 2023 at 21:24
  • $\begingroup$ @Kolja thanks for bringing this up. The book I linked to does just that: it defines $\limsup a_n$ and $\liminf a_n$ it as I wrote here, and then in Theorem 3.5.5, it proves that this definition is equivalent to the one given in the question. $\endgroup$
    – mhdadk
    Commented Mar 17, 2023 at 21:34

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