real-analysis
In many sites including wikipedia, limsup and liminf are defined using the pictures of oscillating sequences.
So, is there only this type of sequence which can have limsup and liminf?
(Ok, this is reasonable that a sequence jumping up & down can have limsup and liminf,but . . .)
I will write only about the first inequality $$\limsup\limits_{n \rightarrow \infty} s_n + \liminf\limits_{n \rightarrow \infty} t_n \leq \limsup\limits_{n \rightarrow \infty} (s_n + t_n).$$ (You wrote that you can prove the rest.)
By definition, if $\liminf t_n=t$ then for each $\varepsilon>0$ the inequality $t_n>t-\varepsilon$ holds for all but finitely many $n$'s. For such $n$'s we also have $s_n+t_n>s_n+t-\varepsilon$ and $$\limsup(s_n+t_n) \ge \limsup (s_n+t-\varepsilon) = t-\varepsilon+ \limsup (s_n).$$ (We have used monotonicity of $\limsup$ and that $\limsup (C+x_n)=C+\limsup x_n$ for any constant $C$.) Since the above inequality is true for each $\varepsilon>0$, we get that $$\limsup(s_n+t_n) \ge t + \limsup (s_n)=\liminf t_n+\limsup s_n.$$
EDIT: Note that the above proof does not work for $t=-\infty$ (it does not make sense to write $-\infty-\varepsilon$), but in this case the inequality is clear. (Of course, we have to omit indeterminate case $\infty-\infty$, i.e., in this case we assume that $\limsup s_n$ is finite.)
Or if you use $\liminf\limits_{n\to\infty} x_n= \lim\limits_{n\to\infty} \inf\limits_{k\ge n} x_k$ and $\limsup\limits_{n\to\infty} x_n= \lim\limits_{n\to\infty} \sup\limits_{k\ge n} x_k$ as the definition of limit inferior/superior then you can use $$\sup_{k\ge n} (x_k+y_k) \ge \sup_{k\ge n} x_k + \inf_{k\ge n} y_k$$ to get $$\lim_{n\to\infty}\sup_{k\ge n} (x_k+y_k) \ge \lim_{n\to\infty}\sup_{k\ge n} x_k + \lim_{n\to\infty}\inf_{k\ge n} y_k\\ \limsup_{n\to\infty} (x_n+y_n) \ge \limsup_{n\to\infty} x_n + \liminf_{n\to\infty} y_n.$$
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.
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 $\liminf a_n = -\infty$.
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 $2$.
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 $-\infty$.
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.
Best Answer
A sequence has a finite limsup and liminf if and only if it is bounded.
On the other hand if it is unbounded upwards, its limsup will be $\infty$. (And similarly for downwards unbounded sequences).