Sequence $(\frac{a_n}{n})$ is convergent.

Question

Let $(a_n)$ be a real sequence such that $a_{n+p}\leq a_n+a_p$ for all $n$ and $p$. Prove, that sequence $(\frac{a_n}{n})$ is convergent.

I tried to show that $a_n$ is monotone decreasing. How do I solve ? Any help.

Answer 1

You have that $a_2=a_1+a_1=2a_1$. By induction you get $a_n=na_1$. Hence if you denote $b_n=\frac{a_n}{n}$ you get $b_n=a_1 \forall n \in \mathbb{N}$. Hence the sequence $(b_n)$ is constant and thus, convergent

Answer 2

You can show, that: $$\lim_{n \to \infty} \frac{a_n}{n} = \lim_{n \to \infty} \frac{a_1 + a_1 + ... + a_1}{n} = a_1 \lim_{n \to \infty} \frac{n}{n} = a_1$$ where $a_1$ is added $n$ times.

Answer 3

Also by Stolz-Cesaro

$$\lim_{n \to \infty} \frac{a_n}{n} =\lim_{n \to \infty} \frac{a_{n+1}-a_n}{n+1-n} =a_1$$

Answer 4

Firstly, the question asked is wrong. Let $a_n = - n^2$, then

$$a_{n+p} = -(n+p)^2 \leq -n^2 - p^2 = a_n + a_p$$ holds for any $n, p$, but $\displaystyle \lim_{n\to\infty} \frac{a_n}{n}$ doesn't exist.

Secondly, if the conditon $a_n \geq 0$ added, then $\lim \cfrac{a_n}{n}$ exist.

Proof: Fixed a arbitrary $m\in\mathbb{N}^+$. For any $n > m$, let $$n = m*d +r,\quad 0\leq d,\; 1\leq r\leq m$$ Then $x_n = x_{m*d+r} \leq d\cdot x_m + x_r$, hence $$\frac{x_n}{n}\leq \frac{d}{n}x_m + \frac{1}{n}x_r \leq \frac{x_m}{m} + \frac{x_r}{n}$$ Take supremum limit for both side with respect to $n$, we have $$\begin{align*} \lim_{n}\sup \frac{x_n}{n} &\leq \lim_n\sup \Bigl(\frac{d}{n}x_m + \frac{1}{n}x_r\Bigr) \\ &\leq \lim_n\sup \Bigl(\frac{1}{m}x_m + \frac{1}{n}x_r\Bigr)\\ &= \frac{x_m}{m} \end{align*} $$ Since the $m$ is arbitrary, take infimum limit for both side with respect to $m$, we get $$\lim_{n}\sup \frac{x_n}{n} \leq \lim_{m}\inf \frac{x_m}{m} $$ This all.