Sequence such that $x_{n+m} \geq x_n+x_m$, show that $(\frac{n}{x_n})$ converges (or not)

Question

Let $(x_n)$ be a sequence such that $x_n \geq 0$ $\forall n \in \mathbb N$ and that $$x_{n+m} \geq x_n+x_m \quad \forall n,m \in \mathbb N^* $$ I have the feeling that $(\frac{n}{x_n})$ converges but I have a little problem with the proof's end. Can someone help me (the statement can also be wrong but I do not find counterexamples) ?

We have $x_{n+m} \geq x_n+x_m \quad \forall n,m \in \mathbb N^*$, we want to show that $(\frac{n}{x_n})$ converges :

Let's show that $ \lim_{n\to\infty} \frac{n}{x_n} = a$ where $a = inf\{\frac{n}{x_n} | n \in \mathbb N^*\}$ :

We have that $x_{pq+r} \geq x_{pq}+x_r \geq px_q+x_r$ with $p,q,r \in \mathbb N^*$ that gives $\frac{1}{x_{pq+r}} \leq \frac{1}{px_q+x_r}$.

Let $\epsilon > 0$, as $a$ is an inf, we have that $\exists q \in \mathbb N^*$ such that $ a \leq \frac{q}{x_q} \leq a + \frac{\epsilon}{2} $

Let $N>q$, so we have for $n > N$ and by Euclidean division on $n$, $$a \leq \frac{n}{x_n} \leq \frac{pq+r}{px_q+x_r} = \frac{pq}{px_q+x_r}+\frac{r}{px_q+x_r} \leq \frac{q}{x_q} + \frac{q}{px_q+x_r} \leq \frac{q}{x_q} + \frac{q}{px_q} = (1+\frac{1}{p})\frac{q}{x_q} \leq 2 \frac{q}{x_q} $$ Now I can write $a \leq \frac{n}{x_n}\leq 2a+\epsilon$ but we want $a \leq \frac{n}{x_n}\leq a+\epsilon$, is there a way to fix it ?

Answer 1

Instead of $N > q$, pick $N >q\left(2+\dfrac{2a}\epsilon\right)$. If $n > N$, then $$p > \dfrac nq - 1 > 1 + \dfrac {2a}\epsilon\\ p\dfrac \epsilon 2 > \dfrac \epsilon 2 + a\\\dfrac \epsilon 2 > \dfrac{a + \frac \epsilon 2}p$$

From your calculation,

$$\begin{align}a \le \dfrac n{x_n} &\le \left(1 + \dfrac1p \right)\dfrac q{x_q} \\&\le \left(1 + \dfrac1p\right)\left(a + \dfrac \epsilon 2\right)\\ &\le a + \dfrac \epsilon 2 + \dfrac{a + \frac \epsilon 2}p\\&\le a + \dfrac \epsilon 2 + \dfrac \epsilon 2\end{align}$$