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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!

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  • $\begingroup$ Is this from Hata's problem book ? $\endgroup$ Sep 5 '18 at 13:06
  • $\begingroup$ Yes it is. What do you think of the problems in that book? I am trying to solve them but I find them hard. $\endgroup$ Sep 5 '18 at 13:07
  • $\begingroup$ See also math.stackexchange.com/questions/1808404/… $\endgroup$
    – Robert Z
    Sep 5 '18 at 13:12
  • $\begingroup$ @RobertZ It may be the same problem, but the questions seem to differ in my opinion. $\endgroup$
    – Arthur
    Sep 5 '18 at 13:13
  • $\begingroup$ Answer: that is not taking the limit. Actually you could use other alphabets, that does not affect the $\liminf a_\nu / \nu$. 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, hence $\inf_{k \geqslant 1} x_k \leqslant \inf_{k \geqslant n} x_k \to \liminf x_k [n \to \infty]. $ $\endgroup$
    – xbh
    Sep 5 '18 at 13:21
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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 $\mathbb R\cup \{-\infty\}$

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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).

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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.

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