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Let $\{a_n\}_{n=1}^\infty \subset R_{\geq0}$ satisfy $a_{n+k} \leq a_n +a_k \quad$ for any $n,k$, then $$\lim_{n\rightarrow \infty} \frac{a_n}{n} = \inf_{n} \frac{a_n}{n}$$

I was thinking to prove that $\{\frac{a_n}{n}\}$ is decreasing. We have $a_n \leq (n-1)a_1$ so $$\frac{a_n}{n} - \frac{a_{n+1}}{n+1} \geq \frac{a_n}{n} - \frac{a_n}{n+1} - \frac{a_1}{n+1} = \frac{1}{n+1} \left( \frac{a_n}{n}-a_1 \right)$$ But the last thing is negative, so the estimate is not good enough. How can I make it better?

user280390
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1 Answers1

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The inequality $\liminf\frac{a_n}{n}\geq \inf\frac{a_n}{n}$ is clear.

For the converse, let $p\in\mathbb{N}$ be arbitrary, and take an increasing sequence $(N_k)_k$ such that $\frac{a_{N_k}}{N_k}\to\limsup\frac{a_n}{n}$. Write $N_k=q_kp+r_k$, where $0\leq r_k\leq p$. It follows that $q_k\to\infty$. Now we have $$\frac{a_{N_k}}{N_k}\leq\frac{a_{q_k p}+a_{r_k}}{q_kp+r_k}\leq\frac{q_k a_p+r_k a_1}{q_k p+r_k}=\frac{a_p}{p+\frac{r_k}{q_k}}+\frac{r_k}{q_kp+r_k}a_1$$ Since $0\leq r_k<p$ and $q_k\to\infty$, we have $r_k/q_k\to 0$ and $\frac{r_k}{q_kp+r_k}\to 0$, so taking the limit when$k\to\infty$ in the inequality above yields $$\limsup\frac{a_n}{n}=\lim\frac{a_{N_k}}{N_k}\leq\frac{a_p}{p}.$$ But $p$ was arbitrary, so we can take the infimum for all $p$ and conclude that $$\limsup_n\frac{a_n}{n}\leq\inf_p\frac{a_p}{p}\leq\liminf_n\frac{a_n}{n}$$ hence the result we wanted.

Luiz Cordeiro
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