A sequence in which liminf < limsup

Question

In this question we are looking for a sequence in which liminf is strictly less than limsup. A majority of the examples that I came up with and found were in which liminf<=limsup. As for the second part of this we are also looking for subsequences that converge to liminf and limsup. Any help is appreciated.

Answer 1

How about $$a_n= \begin{cases} 1+\frac1n, & n\text{ even}\\ -1-\frac1n & n\text{ odd}. \end{cases}$$

Answer 2

Take $a_n=(-1)^n$. You have $$\liminf_{n\rightarrow\infty} (-1)^n = -1 < 1 = \limsup_{n\rightarrow\infty}(-1)^n$$

Answer 3

Perhaps the easiest way to think of an example of a sequence whose $lim \ inf < lim \ sup$ is to choose a bounded but not convergent sequence.

If a sequence is not bounded, and let's say, diverges to (positive) infinity, then $lim \ sup = lim \ inf = +\infty.$

Similarly, recall that a sequence converges $\iff lim \ sup = lim \ inf = limit.$

So for a bounded but not convergent sequence, you can choose, say, $a_n := \{0 , 1, 0, 1, 0, 1, 0, 1, 0... \}. $ Then $lim \ inf = 0 < \ lim \ sup = 1.$ You do not necessarily have to pick a sequence that alternates between only two values, but such a sequence makes it easy to pick two subsequences that converge to $lim \ inf$ and $lim \ sup.$ Here, the subsequence ${a_n}_k$ for all $n$ odd converges to $0$, while the subsequence ${a_n}_l$ for all $n$ even converges to $1$.