I want to write an article which explains the limit superior. I also want to present the intuition behind this concept. Currently I would describe the limit superior as the "least upper bound of a sequence at infinity" which therefore can be used to estimate the sequence "at infinity".

To explain my idea: Take a sequence $(a_n)$. The supremum $\sup_{n\in\mathbb N} a_n$ is the least upper bound of all sequence elements and thus can be used to estimate the sequence elements.

Similarly we can define an "upper bound of $(a_n)$ at infinity" as

A real number $c$ is an upper bound at infinity of the sequence $(a_n)$, iff for all $\epsilon > 0$ almost all sequence elements $a_n$ are less then or equal $c+\epsilon$.

Given this definition, $\limsup_{n\to\infty} a_n$ is the smallest of all upper bounds at infinity (analogously as the supremum is the least upper bound of a sequence).

The limit superior can also be used to estimate the sequence at infinity: Having $\limsup_{n\to\infty} a_n = c \in \mathbb R$ we know that almost all sequence elements are less than or nearly equal $c$ and that $c$ is the best of all those upper estimates. Here we have to use "nearly equal" instead of "equal" because $c$ must not be an upper bound for $(a_n)$. Take $a_n=1+\tfrac 1n$ as an counterexample.

My question: How would you describe the intuition behind the limit superior? What do you think about the above description of the limit superior as the "least upper bound at infinity"? Are there characteristics of the limit superior I have missed?

I know that the presented intuition still needs to be elaborated. But that's also the reason I am asking the above questions here... ;-)

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    $\begingroup$ I don't know about intuition, but if the target audience is comfortable with the usual limit and it's just the limsup part causing them to stumble, I found the definition quite natural to digest with the following phrasing: Consider a 'parallel' sequence of intervals $(I_n)_{n \in \mathbb{N}}$ where each $I$ is the smallest interval (possibly unbounded, but closedness/openness/neither doesn't matter) that contains all later $a$'s; then limsup of the original sequence is the limit of (the sequence of) the upper endpoints. $\endgroup$ Commented Jan 21, 2016 at 5:15
  • $\begingroup$ This also serves as my personal motivation for diameter, oscillation, and Cauchy sequences. (Actually, it is almost too natural and likely a reflection of how it's almost impossible to explain something unless they already got it, but I am having trouble thinking of any 'lossy' (as opposed to lossless or tautologously equivalent) yet easier explanation of liminf and limsup.) $\endgroup$ Commented Jan 21, 2016 at 5:21
  • $\begingroup$ I think of "lim sup" and "lim inf" as ways of describing convergence behavior in the absence of a limit. When the (finite or infinite) limit exists, these two values are equal, and when the limit doesn't exist, these two values give you the tightest two bounds (lower and upper) that the sequence values oscillate between as $n \rightarrow \infty.$ Fractal dimension doesn't exist? Then use upper and lower limit version. Lebesgue density doesn't exist? Then use upper and lower limit version. Decimal digits of a real number have no limiting frequency? Then use upper and lower limit versions. $\endgroup$ Commented Jan 22, 2016 at 20:22
  • $\begingroup$ @Vandermonde Thanks a lot for that explanation, it's exactly what I was looking for! $\endgroup$
    – Ovi
    Commented Mar 29, 2018 at 1:49

6 Answers 6


I have two intuitions to offer:

  1. A sequence $(a_n)$ may have cluster points (these are points such that every neighborhood contains infinitely many elements of the sequence, or, more precisely, for every neighborhood and every index $N$, there is an element $a_n$ with an index $n>N$ which is in this neighborhood). The $\limsup$ is the largest of these cluster points.

  2. Consider the tail of the sequence $(a_n)_{n\geq N}$. This has a supremum $\sup_{n\geq N} a_n$. This supremum gets smaller and smaller as $N$ grows since the set over which we take the supremum gets smaller. The $\limsup$ is the infimum of these suprema (equivalently, it's the limit of the suprema of the tails). You may visualize this by a kind of decaying straicase which lies on top of the sequence and the step number $N$ of the stair is as low as possible while still above all elements with index $\geq N$.

While both intuitions are really valid definitions of the $\limsup$ I find both of them intuitive (at least as intuitive as I find any definition of the limit).


The idea behind the limsup that you write is not simple and will not convey a concise intuition. The shortest description of the limsup of a sequence needs two steps:

(1) The audience has to know what a limit point (also called accumulation point or cluster point) of a sequence is, and that a sequence can have many limit points, by examples.

(2) Granting the technical term introduced in (1), the limsup of a sequence is its largest limit point. (Similarly, the liminf is the smallest limit point.)

That's it! This is the same idea as Dirk describes in his answer. He also offers another intuition, but I do not find anything other than (2) to be so immediately intuitive, cutting right to the heart of what the limsup of a sequence means in such a concise way. The other possibilities offered here illustrate this.

I had known real analysis for a couple of years in college years before I read (2), and until then I never felt fully comfortable with limsups and liminfs, or solving problems involving them. After reading (2) I immediately understood this notion and never found it hard to use again. Ironically, the place where I found (2) was in a book on $p$-adic analysis! (See the first page of Chapter IV in Koblitz's Springer GTM on $p$-adic numbers, where he explains Hadamard's limsup formula for the radius of convergence $r$ of a power series in terms of $1/r$ being the greatest accumulation point of a sequence.)


Every sequence $\langle u_n: n\in\mathbb{N}\rangle$ has a natural extension $\langle u_n: n\in{}^\ast\mathbb{N}\rangle$ where ${}^\ast\mathbb{N}$ are the hypernatural numbers (positive hyperintegers). The limsup of the sequence (when it is finite) is then exactly the sup of the shadows of $u_H$ for infinite indices $H$.

So I would suggest explaning this as follows: look at terms in the sequence for indices that are so huge that you can't express them even using a computer the size of the universe, in the total time allotted to our civilisation, so that for all practical purposes they are infinite. Evaluate the sequence at those indices, and take the maximal value among the ones you get. That's the limsup.

Then you can point out that a way of writing this precisely depends on which background formalisation of analysis you use.

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    $\begingroup$ Interesting view. However, I have a hard time imagining somebody who does not have an intuition about the $\limsup$ and finds this explanation helpful at the same time. How should one parse "evaluate the sequence at integers that are too large to comprehend them"? $\endgroup$
    – Dirk
    Commented Jan 22, 2016 at 10:08
  • $\begingroup$ @Dirk, I didn't say large integers are "too large to comprehend" and in fact the notion of infinity is a natural product of the human imagination. Rather, an infinite integer in, for example, a hyperreal framework can be thought of as an ordinary integer that's so huge that we have no way of accessing it even using the fastest growing functions we have, Ackermann, busy beaver, you name it. $\endgroup$ Commented Feb 2, 2016 at 13:54

I would probably avoid the phrase "at infinity", since it's kind of vague and I'm not sure if it'd help someone who doesn't already understand the concept. Instead, I would talk more explicitly about numbers that are an upper bound outside a bounded segment — in other words, an upper bound if we throw out some finite initial piece. This works for both sequences and functions, as long as we take "finite" in the sense of "finite length" for functions. ("Bounded" makes sense for both as is.)

The word "least" is also a bit problematic, since the limsup itself might not be an "upper bound at infinity" itself — it's just the greatest lower bound of the set of "upper bounds at infinity". This is genuinely different than for the least upper bound: by completeness of the reals, if a set has an upper bound, then the set of its upper bounds really does contain a least element. (That is, the least upper bound is an upper bound itself.)

This latter issue seems a bit harder to intuitively describe for the limsup. If your audience is already comfortable with the notion of supremum and infimum, you can just say it's the infimum of all the numbers that are upper bounds outside a bounded interval. If not, I'm not sure what's the best way to explain it... however, I suspect someone who has trouble understanding supremum and infimum will inevitably have a hard time with limsup and liminf no matter how it's presented, because it's a strictly harder concept.


For me, this picture from wikipedia helped the most in understanding limsup and liminf

enter image description here


Limit supremum (of a sequence of numbers)

I will try to give intuition only using words, without using mathematical symbols.

  • What is the smallest number which is greater than infinitely many members of a sequence? --> limit supremum

Story: To be a candidate for the limit supremum, a number has to be greater than infinitely many members of the sequence. For this, we need to consider tails of the sequence. Why tails? Why not heads? Because given any member, all the members to the left of it (head) are finite in number as the sequence starts on the left and the starting point is known ([avoiding bi-infinite sequences for now] a sequence is a mapping from the set of natural numbers which starts at 1, so the starting side is known. You can tell the first natural number, the first 2 natural numbers, etc. but cannot tell the last 2 natural numbers), only the right side is infinite. So, where there is talk of infinitely many members, you have to consider a tail. Any tail of the sequence is infinitely long. Hence, the supremum of any tail is greater than infinitely many members of the sequence. Thus, the supremum of any tail is a candidate for limit supremum. Further, as successive tails are subsets of the previous ones, the corresponding supremums form a monotonically decreasing sequence of candidates for the limit supremum. However, the infimum (= limit, in this case) of this sequence of candidates is the smallest number which already bounds infinitely many members of the original sequence from above--which is precisely the purpose. So, we reject all larger candidates, and this infimum, as the best/most parsimonious candidate, is elected as the limit supremum of the original sequence.

Limit infimum (of a sequence of numbers)

  • What is the largest number which is smaller than infinitely many elements of the sequence? --> limit infimum

You can write the story now.

  • $\begingroup$ Wow, thanks a lot. Really nice answer! :-) $\endgroup$ Commented Oct 3, 2022 at 22:42

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