Let $a_1,a_2,\ldots$ be a sequence of positive real numbers. Suppose that $a_{m+n}\leq a_m+a_n$ for all $n\geq 1$. Does $\lim_{n\rightarrow\infty}\dfrac{a_n}{n}$ always exist?
From $a_{m+n}\leq a_m+a_n$ we know that $a_n\leq na_1$, so that $\dfrac{a_n}{n}\leq a_1$, which means the sequence $\dfrac{a_n}{n}$ is bounded both from above and below. But this is not enough to conclude that the limit exists.
Best Answer
Consider $b_n = e^{a_n}$ for all $n$. Fix $m\in\mathbf{N}^*$. The sequence $(b_n)_n$ satisfies $$b_{p+q}\leq b_p b_q$$ for all $p,q\in\mathbf{N}$. For each $n\in\mathbf{N}$ note $p(n)$ and $q(n)$ integers such that for all $n$ we have $n = p(n)m+q(n)$, with $0\leq q(n)<m$. (Euclidian division.) We have $$b_n^{1/n} \leq b_m^{p(n)/n} b_{q(n)}^{1/n}.$$ Making $n\to + \infty$ if what preceeds you get that $$\limsup_{n\to + \infty} b_n^{1/n} \leq b_m^{1/m}$$ and this is for all $m\in\mathbf{N}^*$. From this you get $$\limsup_{n\to + \infty} b_n^{1/n} \leq \inf_{m\in\mathbf{N}^*} b_m^{1/m} \leq \liminf_{n\to + \infty} b_n^{1/n}$$ and as you always have $\liminf \leq \limsup$, you conclude that $\liminf$ and $\limsup$ are equal (which implies that $(b_n^{1/n})_n$ converges) to $\inf_{m\in\mathbf{N}^*} b_m^{1/m}$ which implies that $(b_n^{1/n})_n$ converges towards $\inf_{m\in\mathbf{N}^*} b_m^{1/m}$. This implies that $(a_n)_n$ converges. From this, you can conclude I guess.