I think of limit superior and inferior as the supremum and infimum, respectively, of the set of all limit/convergence points of the subsequences of $(x_n)$. Can I say that limsup is the supremum and lim inf is the infimum of the set of cluster points of the sequence $(x_n)$?

Moreover, can anyone please verify the answers to the following limSup and Liminf questions?

Find lim inf and lim sup of $(x_n)$ given by:

1)$ {{(-n)^n}}$ Since the sequence is unbounded, liminf and limsup does not exist.

2) ${(1+1^n)}$ is a constant sequence with all the terms equaling to 2. Hence, it converges to 2 and therefore, liminf=limsup = 2

3) ${(n\sin(n\pi/2))}$ Since $-1 \leq \sin(n\pi/2) \leq 1$, $\implies -1n \leq n\sin(n\pi/2) \leq 1n$. Since both $ -n$ and $n$ are unbounded, this implies the sequence is unbounded so liminf and limsup do not exist.

4)${(0, 1, 0, 1, 2, 0, 1, 3, 0, 1, 4, ...)}$ We can create two constant subsequences that converge to $0$ and $1$. Then liminf $= 0$ and lim sup $=1$.

Please let me know if there are any errors in my answers, or in the way I am thinking about limSup and LimInf. I, for one, am not confident if I understand the concept of these properly so any help will be highly appreciated.

Thank you.

  • $\begingroup$ Your understanding isn’t perfect, I don’t think. Limit supremum/infimum is just the supremum/infimum of every possible tail of the sequence. So, in the last example, you’ll always have something like $0,1,n$ as part of the tail (no matter where you truncate the sequence), so the limit supremum does not exist. $\endgroup$
    – Clayton
    Mar 15, 2018 at 2:05
  • $\begingroup$ Oh, ok! So that means if I take infinitely many tails, and find the limit of each of them and from those sets of limits, find the supremum and infimum, I'll get the limsup and liminf respectively? Also, then, in the last question, all the tails will have 0 as their liminf, right? $\endgroup$
    – Alea
    Mar 15, 2018 at 2:09
  • $\begingroup$ Correct on both accounts! Yes :) $\endgroup$
    – Clayton
    Mar 15, 2018 at 2:17
  • $\begingroup$ Thank you! Also, since we are talking about tails, for the sequence $n$ of natural numbers, the liminf would be 2, right? and not 1? $\endgroup$
    – Alea
    Mar 15, 2018 at 2:24
  • $\begingroup$ The sequence of natural numbers...think of the tail again. If you start at $n$, then your tail is like $n,n+1,\ldots$. This infimum would be $n$ (which goes to infinity the further along you start the tail) and the supremum is $\infty$. $\endgroup$
    – Clayton
    Mar 15, 2018 at 2:44

1 Answer 1


Use the definition(s) of $\limsup$ and $\liminf$. For example, the limit superior of a sequence, is defined as $\limsup a_n = \sup_{k \geq 1} \inf_{n \geq k} a_n$, and for $\liminf$ the $\sup$ and $\inf$ are switched.

Here is a better way of understanding these concepts. If $a_n$ is any sequence, then $b_n = \sup_{k \geq n} a_k$(in the extended reals) is an decreasing sequence(because the set $\{k \geq n\}$ keeps shrinking as $n$ grows larger). Therefore, every decreasing sequence either has a limit(if it is bounded) or goes to $-\infty$. The limit superior is defined as the limit of the decreasing sequence, if it exists, else it is $\pm\infty$ depending upon where the sequence goes (Note : $\limsup$ is defined over the extended reals, whence it can possibly take the values $\pm \infty$).

Similarly, the limit inferior would be defined as follows : $c_n = \inf_{k \geq n} a_k$ is an increasing sequence. So either it has a limit, which we call the limit inferior, or it goes to $\pm\infty$, whence the limit inferior would also be $\pm \infty$.

With this outlook, let us look at the questions we are given.

  • $a_n = (-n)^n$.

What is $b_n$ here? $b_n = \sup_{k \geq n} a_k$. But then $a_k$ is an increasing sequence that increases to $\infty$. So, $b_n$ is $\infty$, because the supremum is greater than each element, but the elements are themselves arbitrarily large. In particular, the limit superior is also $+\infty$. Following a similar logic gives you $\liminf a_n = -\infty$.

  • $a_n = 1 + 1^n = 2$

Of course, $a_n$ is constant, so $b_n = \sup_{k \geq n} a_k = 2$ (the supremum of the set $\{2\}$ is just $2$!) and the limit of the constant sequence $2$ is $2$. So the limit superior (analogously, limit inferior) are both equal to $2$.

  • $n \sin (\frac{n \pi}{2})$.

This is a weird sequence. Look at its terms : $1,0,-3,0,5,0,-7,0,...$. In other words, this sequence can be rewritten like this : $$a_n = \begin{cases} 0 \quad n \equiv 0(2) \\ n \quad n \equiv 1(4) \\ -n \quad n \equiv 3(4) \end{cases}$$

where $n \equiv m(k)$ means that $n-m$ is a multiple of $k$. Now, what is $b_n = \sup_{k \geq n} a_k$? If you look at the set $\{a_k,a_{k+1},a_{k+2},...\}$ then this contains a strictly increasing sequence of positive numbers, that comes from the $n$ part of the definition above. Therefore, $b_n$ is actually infinite, and so there, is the limit superior. Similarly, the $-n$ part contributes to the limit inferior becoming $-\infty$.

  • $0,1,0,1,2,0,1,3,0,1,4...$

This is the trickiest of the lot. Look at $b_n = \sup_{k \geq n} a_k$. The sequence $\{a_k,a_{k+1},a_{k+2},...\}$ contains a strictly increasing sequence of positive numbers. Hence, its limit superior is $+\infty$.

On the other hand, the limit inferior? Observe the sequence $c_n = \inf_{k \geq n} a_k$. Well, observing the sequence $\{a_k,a_{k+1},...\}$, we see that $0$ must be there in this set. But then, $0$ is also the smallest element of this set! So, $c_n = 0$ for all $n$, so the limit inferior is $0$.

EDIT : I'll give you a bonus here.

The sequence $-1,-2,-3,...$ will have both $\liminf$ and $\limsup$ as $-\infty$. Try to show this from the definitions. This is a quirky point, and a nice example.

EDIT AGAIN : The new definition : the limit inferior of $x_n$ is the infimum of the set $\{v\}$ such that $v < x_n$ for at most finitely many $x_n$.

This is exactly the set $V$ of all numbers which are greater than or equal to all the numbers that appear "after some time" in the sequence i.e. there is $N$ such that for all $n > N$, $v > x_n$.

For example, lets see this for the last example. In $0,1,0,1,2,0,1,3,0,1,4,...$, what is the set $V$? Which is that element, which is greater than or equal to all elements "after some time"? I claim, that there is none of them. This is because, this sequence has a subsequence that is exactly the natural numbers, but no fixed number can be greater than all the natural numbers. So here, the limit superior is $+\infty$.

Similarly, the limit inferior is defined as the supremum over all $\{w\}$ such that $w > x_n$ for only finitely many $n$. So, what is that set here? Well, W definitely contains all the negative integers and $0$. But anything more? No, because there are infinitely many zeros in the sequence, so anything $ a > 0$ will be greater than all those terms of the sequence which are $0$, but there are infinitely many of them so $a \notin W$! The supremum of $(-\infty,0]$ is $0$, so there you have it.

  • $\begingroup$ The book I'm using (Bartle) has the following definition for LimSup and LimInf: (a) The limit superior of $(x_n)$ is the infimum of the set V of $v\in \mathbb{R}$ such that $v < x_n$ for at most a finite number of $n \in \mathbb{N}$. And the definition is switched for lim inf. Is this definition equivalent to the one you provided above? Also, what does it mean 'for at most a finite number of $n \in \mathbb{N}$? Thank you. $\endgroup$
    – Alea
    Mar 15, 2018 at 3:14
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    $\begingroup$ If you are on MSE, that is good enough : in four months, I guarantee you will look back here and wonder how you did not ever get this question in your life! For better tips, there are many textbooks for basic analysis, so look up all of them, i.e. don't stick to one, and furthermore always keep definitions in mind while solving problems. Keeping definitions first in mind while attacking a question is very important in introductory analysis. $\endgroup$ Mar 22, 2018 at 5:45
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    $\begingroup$ Have you tried (Baby) Rudin? It's Rudin, Principles of Mathematical Analysis, (third edition is good) and the solutions are online!(math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_exs.pdf) These are the best, from my perspective. I would naturally suggest many but this is the best text, really, if you want to stick to one. Others like Bartle, Lee etc. are all sort of lacking in something or the other : this fellow does it brilliantly, and if you need answers, and good ones at that, you have them there. $\endgroup$ Mar 26, 2018 at 13:38
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    $\begingroup$ @A.Asad Unfortunately my time is up for today (as in : dinner and then sleep) but I personally still think Rudin does the basics well. I mean, that is the recommended textbook for a first semester, from pretty much all professors I have met. Even if it is difficult, sit with friends and decipher it. Even half a day, two or three well chosen problems can do you the world of good. Get somebody good in class, or post regularly on MSE (you may inform me by catching me on this page, where we have conversed) and get things clarified quickly. But don't worry : you will do well! $\endgroup$ Mar 26, 2018 at 13:50
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    $\begingroup$ It is unfortunate if Rudin is not of help. If it is down to a way of "thought" to improve your analysis, then if you notice I went to great lengths in my answer to emphasise the definitions. Break them down into their smallest bits, and while writing your answers, remember that you should write down everything you know, from definitions to initial hunches. The book "How to think about analysis", by Lara Alcock, helps you understand how to register proofs in analysis and pick yourself up if you are demotivated. I think you should try this book for more general strategies. $\endgroup$ Mar 28, 2018 at 22:51

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