I am trying to review some ideas form the calculus course and came across the much hated (back then) topic of $\limsup$, $\liminf$, $\sup$ and $\inf$. I tried to solve the following problem, Find $\limsup$, $\liminf$, $\sup$ and $\inf$ of $a_n$ , where $$ a_n=(-1)^n + \frac{1}{n} + 2\sin{\frac{n\pi}{2}} $$ As far as I remember to solve such problems, one uses the fact that, if a subsequence $a_{n_k}$ of the sequence $a_n$ has an accumulation point, than this is also an accumulation point of $a_n$. It is trivial, how to find the subsequence and obtain that

$$ \sup a_n=\frac{3}{2}, \inf a_n =-1, \liminf a_n = -1 \limsup a_n = 1 $$

This method works fine, but according to my book the problem can be solved much faster using the following theorem, which does not require finding all the subsequences of $a_n$.

$$ \limsup a_n = \lim_{k\to \infty} \sup (a_k,a_{k+1},\dots)\\ \liminf a_n = \lim_{k\to \infty} \inf (a_k,a_{k+1},\dots) $$ Is it possible to apply this theorem to the sequence $a_n$ as defined above and solve the problem without using the sub-sequences? Somehow, I am not able to understand what $\sup (a_k,a_{k+1},\dots)$ will be in my case.

  • $\begingroup$ If you use the "right" $k$ (i.e. even and odd I think), you can use the two parts to get the desired results. I don't see this as any easier than looking at subsequences. $\endgroup$
    – vadim123
    Jan 9, 2016 at 18:28
  • $\begingroup$ It would be my guess that the theorems aren't there to help you solve the problem, they are there to help you prove the results. Finding subsequences is probably the best way to convince yourself that your answers are correct, but it can be hard to prove without these definitions. $\endgroup$ Jan 9, 2016 at 18:29
  • $\begingroup$ $\limsup a_n = \lim_{k\to \infty} \sup (a_k,a_{k+1},\dots)$ is not theorem, it's definition $\endgroup$
    – luka5z
    Jan 9, 2016 at 18:38
  • $\begingroup$ @luka5z they call it a theorem in the book I have. $\endgroup$ Jan 9, 2016 at 18:46

1 Answer 1


Your answer is partly wrong: $\liminf a_n=-3$, not $-1$. You can see that by taking $n=4k+3$ for $k\in\Bbb N$: $\sin\frac{n\pi}2=-1$ there. I did a numerical confirmation of my answer by taking $n$ out beyond $1000$.

That theorem of yours does have the advantage of not requiring you to find all the relevant subsequences of $a_n$. However, it has the great disadvantage of requiring you to find infinitely many suprema and infima. It is often easier to find the desired subsequences. Just consider how I got the right answer to $\liminf$ in your problem.

It is good to have a theorem to allow an alternate approach, but remember that there are reasons the primary approach is use. Use the right tool for the job!

Finding $\lim_{k\to \infty} \inf (a_k,a_{k+1},\dots)$ is not easy: it is, of course, $-3$. In fact, $\inf (a_k,a_{k+1},\dots)=-3$, so the limit is not needed. Proving that seems to require considering the subsequence $a_{4k+3}$ so we have not escaped subsequences. Your problem is definitely not easier with your theorem. The subsequence tool is much easier here.

Let me know if you really need a proof that $\inf (a_k,a_{k+1},\dots)=-3$.

  • $\begingroup$ could you please rovide the prof that $\inf(a_k,a_{k+1}..)=-3$. Thank you for your answer. $\endgroup$ Apr 19, 2016 at 11:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.