Proof/Answer Verification: LimSup And LimInf 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. 
 A: 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.
