0
$\begingroup$

I am stuck on this question:

Suppose $a_n$ is a sequence with $\liminf a_n=s$ and $\limsup a_n=t$ where $s$ and $t$ are real number. Then there must exist some subsequence $a_{n_k}$ that converges to $c$ where $s<c<t$.

For example, if $\lim\inf a_n=1$ and $\lim\sup a_n=3$, then does there exist a subsequence that converges to 2?

I think it is yes, because every bounded sequence has a convergent subsequence. However, I am not fully sure how to argue it.

$\endgroup$
1
  • 1
    $\begingroup$ how about 1,3,1,3,1,... ? $\endgroup$
    – Alan
    Feb 27, 2021 at 0:32

1 Answer 1

2
$\begingroup$

No, take $a_n=(-1)^n$.

Then, $\liminf a_n =-1$, $\limsup a_n =1$.

However, any converging subsequence converges to either $-1$ or $1$.

$\endgroup$
2
  • $\begingroup$ What about lim inf and lim sub with both positive numbers, such as 1 and 3 respectively? $\endgroup$ Feb 27, 2021 at 0:36
  • $\begingroup$ See @Alan's comment. You could simply define $a_n=2+(-1)^n$ $\endgroup$ Feb 27, 2021 at 0:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .