Assume that the terms of the sequence $(a_n)$ satisfy the conditions
$$ 0 \leq a_{n+m} \leq a_n + a_m \; \; \; \text{for} \; \; n,m \in \mathbb{N} $$
Now prove that $\lim_{n \to \infty} \dfrac{a_n}{n} $ exists.
attempt to the solution:
First of all, clearly, the sequence $(a_n)$ is bounded below by $0$. Let $b_n = \dfrac{a_n}{n}$. Since $\dfrac{1}{n} > 0$, then $b_n > 0$ and $(b_n)$ is bounded. If we can prove that $(b_n)$ is monotonic, then we will solve the problem.
Notice that if we put $m=n$ in the condition we get
$$ a_{2n} \leq 2 a_n \implies \dfrac{a_{2n} }{2n} \leq \dfrac{a_n}{n} \implies b_{2n} \leq b_n$$
Can we assume from here that $(b_n)$ is decreasing? Im not hundred percent on this step. Can someone tell me if I am on the right track to solve the problem?
Best Answer
Unfortunately, that approach does not work. It is correct that $b_{2n} \le b_n$, and even $b_{kn} \le b_n$ for all positive integers $n, k$. But that does not imply that $b_n = a_n/n$ is decreasing. A counterexample is $$ a_n = \left\lceil \frac n2 \right\rceil = 1, 1, 2, 2, 3, 3, 4, 4, \ldots $$ which satisfies $0 \leq a_{n+m} \leq a_n + a_m$, but $$ b_n = \frac{a_n}{n} = 1, \frac 12, \frac 23, \frac 12, \frac 35, \frac 12, \frac 47, \frac 12, \ldots $$ decreases and increases alternatingly.
For a working proof see for example Prove $\lim_{n\to\infty} \frac{a_n}{n}$ exists for positive sequence where $a_{n+m} \leq a_n + a_m$.