Convergence and divergence of $\frac{a_n}{n}$ as $n\to\infty$ 
Suppose that a sequence $\{a_n\}_{n\geqslant 1}$ satisfies:
$$a_{m+n}\leqslant a_m+a_n$$
for all integers $m,n\geqslant 1$. Show that $\frac{a_n}{n}$ either converges or diverges to $-\infty$ as $n\to \infty$.

Resolution:
For an arbitrarily fixed positive integer $k$ we put $n=qk+r$ with $0\leqslant r<k$. Applying the given inequality for $q$ times we get $a_n=a_{qk+r}\leqslant qa_k+a_r$; so,
$\frac{a_n}{n}\leqslant\frac{a_k}{k}+\frac{a_r}{n}$ 
Taking the limit as $n\to\infty$, we get 
$\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\frac{a_k}{k}$.
The sequence $\frac{a_n}{n}$ is therefore bounded above. Since $k$ is arbitrary, we conclude that:
$\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant \inf_{k\geqslant 1}\frac{a_k}{k}\leqslant\lim \inf_{k\to\infty}\frac{a_k}{k}$,
which concludes the proof.
Question:
1) First the author says that $k$ is a fixed positive integer. However in the last expression $\lim \inf_{k\to\infty}\frac{a_k}{k}$ the author makes $k$ vary. How is this supposed to prove that $\frac{a_n}{n}$ converges since the behaviour of $\lim \inf_{k\to\infty}\frac{a_k}{k}$ is not known? Why is $k$ allowed to vary?
2)On the question it is mentioned the divergence to $-\infty$. How was that point addressed in the answer or resolution?
Thanks in advance!
 A: Question 1)
He proves that $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\frac{a_k}{k}$ is true for any and all natural numbers $k$. In proving that inequality, $k$ is fixed (so it's only proven for "one $k$ at a time"), but once it's proven, you're allowed to take it for granted for any value of $k$ you'd like, and let $k$ vary.
Question 2)
Note that the $\limsup$ on the left side of $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\lim \inf_{k\to\infty}\frac{a_k}{k}$ and the $\liminf$ on the right side are actually of the same sequence. So the $\limsup$ of the sequence is less than or equal to the $\liminf$ of that sequence. This implies that they must be equal, and either they're finite (in which case the sequence has a limit), or they're not (in which case the sequence diverges to $-\infty$, since it's bounded above).
A: The author first considers a fixed positive $k$ and proves $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\frac{a_k}{k}$. Consequently, $\sup_n \frac{a_n}n < \infty$, hence $\left(\frac{a_n}{n}\right)_n$ is bounded above.
Since $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\frac{a_k}{k}$ is valid for every $k$, by definition of the infinimum, $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\inf_k\frac{a_k}{k}$. By definition of $\liminf$, one also has $\inf_k\frac{a_k}{k}\leq \liminf_k \frac{a_k}{k}$.
Thus $\liminf_k \frac{a_k}{k} = \limsup_k \frac{a_k}{k} = \inf_k \frac{a_k}{k}$, thus   $\frac{a_n}{n}$ converges to a limit in $\mathbb R\cup \{-\infty, \infty\}$.
Since $\frac{a_n}{n}$ is bounded above,  $\inf_k \frac{a_k}{k}<\infty$, but it could very well be equal to $-\infty$. Thus the limit belongs to $\mathbb R\cup \{-\infty\}$
A: Answer: that is not taking the limit $k \to \infty$. Actually you could use other alphabets, that does not affect the quantity $\liminf a_\nu / \nu$. By definition, 
$$
 \liminf \frac {a_n} n = \lim_n \inf_{k \geqslant n} \frac {a_k}k, 
$$ 
and you could prove that for any sequence $x_n$, $\inf_{k \geqslant n} x_k$ is increasing in $n$, hence for all $n \in \mathbb N^*$, $\inf_{k \geqslant 1} x_k \leqslant \inf_{k \geqslant n} x_k$
Since $\inf_{k \geqslant n}x_k \to \liminf x_k [n \to \infty]$, we have $\inf_{k \geqslant 1}x_k \leqslant \liminf x_n$. 
Additionally, $\liminf x_n$ always exists or equals $\infty$, so you can write down $\liminf x_n$ whenever you want. 
