Consider this example:
$$
3-\frac12,\quad 5+\frac13,\quad 3-\frac14,\quad 5+\frac15,\quad 3-\frac16,\quad 5+\frac17,\quad 3-\frac18,\quad 5+\frac19,\quad\ldots\ldots
$$
It alternates between something approaching $3$ from below and something approaching $5$ from above. The lim inf is $3$ and the lim sup is $5$.
The inf of the whole sequence is $3-\frac12$.
If you throw away the first term or the first two terms, the inf of what's left is $3-\frac14$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac16$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac18$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac1{10}$.
. . . and so on. You see that these infs are getting bigger.
If you look at the sequence of infs, their sup is $3$.
Thus the lim inf is the sup of the sequence of infs of all tail-ends of the sequence. In mathematical notation,
$$
\begin{align}
\liminf_{n\to\infty} a_n & = \sup_{n=1,2,3,\ldots} \inf_{m=n,n+1,n+2,\ldots} a_m \\[12pt]
& = \sup_{n=1,2,3,\ldots} \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} : n=1,2,3,\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\{ a_m : m\ge n\} : n=1,2,3,\ldots \right\}.
\end{align}
$$
Just as the lim inf is a sup of infs, so the lim sup is an inf of sups.
One can also say that $L=\liminf\limits_{n\to\infty} a_n$ precisely if for all $\varepsilon>0$, no matter how small, there exists an index $N$ so large that for all $n\ge N$, $a_n>L-\varepsilon$, and $L$ is the largest number for which this holds.
What you're trying to prove, if you want to use the definition of a limit directly (at least in the $+\infty$ case) is that
For any $M>0$: there is an $N$ such that for any $n > N$, $s_n > M$
From your knowledge of the liminf, you know that
For any $M >0$: there is an $N$ such that $\inf \{s_n:n>N\} > M$
Why does this second statement imply the first?
For $\pm \infty$, you will need only the liminf or the limsup. For finite limits, however, you will need both.
The conclusion we want to reach is
For any $\epsilon > 0$: there is an $N$ such that $n>N$ implies $|s_n - L| < \epsilon$.
Using the liminf and limsup, we have the statements:
For any $\epsilon > 0$: there is an $N$ such that $n > N$ implies $\inf \{s_n : n>N\} > L - \epsilon$
For any $\epsilon > 0$: there is an $N$ such that $n > N$ implies $\sup \{s_n : n > N\} < L + \epsilon$
How do these two statements let you deduce the first?
You could also use the squeeze theorem if you note that
$$
\inf \{s_n: n \geq N\} \leq s_N \leq \sup\{s_n: n \geq N\}
$$
and take the limit as $N \to \infty$.
Best Answer
Consider the sequence $(x_n)_{n=1}^{\infty}$ defined $$ x_n = \begin{cases} 1 + n^{-1} & \text{if $n$ is even,} \\ -1 - n^{-1} & \text{if $n$ is odd,} \end{cases} = (-1)^{n}(1+n^{-1}). $$ Some casual observations about this sequence include (a) the odd part and the even part are identical, just mirrored, (b) the sequence has no limit, (c) the sequence has two unique subsequence limits: $-1$ and $1$. The point of this sequence is to illustrate the definitions of the limits superior and inferior. I'll focus on the $\limsup$ and you can think through the details for the $\liminf$.
Using the definition $\limsup_{n \to \infty} x_n = \inf_{n \in \mathbb{N}} \sup_{m \geq n} x_m$, let's figure out what the "sup sequence" looks like. In particular, for a given $n \in \mathbb{N}$, we have $$ \sup_{m \geq n} x_m = \sup_{m \geq n} (-1)^{m} (1 + n^{-1}) = 1+n^{-1}. $$ This follows from two observations. First, we can discard the lower subsequence because it is always bounded above by the upper sequence. Second, the upper subsequence is decreasing, which means its largest value is its first value. Whence $1 + n^{-1}$. It therefore follows that \begin{align} \limsup_{n \to \infty} x_n = \inf_{n \in \mathbb{N}} 1 + n^{-1} = 1. \end{align} Now, if we had instead defined $\limsup_{n \to \infty} x_n$ as $\lim_{n \to \infty} \inf_{m \geq n} x_m$, we would follow the same steps and compute \begin{align} \limsup_{n \to \infty} x_n = \lim_{n \to \infty} 1 + n^{-1} = 1. \end{align} The point is that the sequence of suprema are necessarily nonincreasing. In that case, the infimum and limit coincide (this is the monotone convergence theorem for sequences).
Aside. Why is the sequence of suprema nonincreasing? The simple fact that $A \subset B$ implies $\sup A \leq \sup B$ allows us to say that, for example, $ \sup \{x_m : m \geq 2\} \leq \sup \{x_m : m \geq 1\} $ and so forth.
Finally, there is the alternate characterization of the limit superior: $$ \limsup_{n \to \infty} x_n = \sup \{\bar{x} \in \mathbb{R} : \exists x_{n_k} \to \bar{x} \}. $$ The set on the right hand side is also called the "cluster points" of $x_n$, composed of the limits of subsequences of $x_n$. It is a worthwhile exercise to prove that this characterization works. In the meantime, we can verify it for our toy sequence by observing that $\{-1,1\}$ are the cluster points of $x_n$, and $\limsup_{n \to \infty} x_n = 1$.