This was false with the reverse inequality. Take $x_n=n^2$, for instance. It would still have been true with $\sup$ instead of $\inf$, as it follows from the following.
Now it is true: if $x_n$ is subadditive, i.e.
$$
x_{n+m}\leq x_n+x_m\qquad\forall n,m\geq 1
$$
then $\lim \frac{x_n}{n}=\inf \frac{x_n}{n}$. Note that moreover this limit/inf can take any value in $[-\infty,+\infty)$. To see why the case $-\infty$ can occur, take $x_n=-n^2$. For real values, take $x_n=an$.
This is called Fekete's lemma. See here for a proof by one of our colleagues. The strategy is to prove that
$$
\inf \frac{x_n}{n}\leq \liminf \frac{x_n}{n}\leq\limsup\frac{x_n}{n}\leq\frac{x_k}{k}\qquad\forall k\geq 1.
$$
The first two inequalities are trivial. The last one follows from a neat application of the Euclidean division. Then take the inf on the rhs to see that all these quantities are equal. So $\frac{x_n}{n}$ tends to $\inf \frac{x_n}{n}$. This is in $[-\infty,\infty)$. As observed earlier, all these cases can occur. Only the case $+\infty$ must be excluded, as the inf of a nonempty set is $<+\infty$.
Note: taking the exponential, you get a similar statement for submultiplicative positive sequences. A famous application of this is the fact that $\|a^n\|^\frac{1}{n}$ converges for every element $a$ in a Banach algebra. With more work, one can show that the limit is the spectral radius of $a$. That's called the spectral radius formula.
I’ll get you started. For one direction, suppose that $\lim\limits_{n\to\infty}x_n=x$; we want to show that $$\limsup_{n\to\infty}x_n=\liminf_{n\to\infty}x_n\;.$$ The most natural guess is that this is true because both are equal to $x$, so let’s try to prove that.
In order to show that $\limsup\limits_{n\to\infty}x_n=x$, we must show that $\lim\limits_{n\to\infty}\sup_{k\ge n}x_k=x$. To do this, we must show that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that
$$\left|x-\sup_{k\ge n}x_k\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon\;.$$
Since $\lim\limits_{n\to\infty}x_n=x$, what we actually know is that for each $\epsilon>0$ there is an $m_\epsilon'\in\Bbb N$ such that $|x-x_n|<\epsilon$ whenever $n\ge m_\epsilon'$.
Show that if $|x-x_n|<\epsilon$ for all $n\ge m_\epsilon'$, then $\left|x-\sup\limits_{k\ge n}x_k\right|\le\epsilon$. Conclude that if we set $m_\epsilon=m_{\epsilon/2}'$, say, then $$\left|x-\sup_{k\ge n}x_k\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon$$ and hence $\limsup\limits_{n\to\infty}x_n=x$.
Modify the argument to show that $\liminf\limits_{n\to\infty}x_n=x$.
For the other direction, suppose that $$\limsup_{n\to\infty}x_n=\liminf_{n\to\infty}x_n=x\;;$$ we want to show that $\langle x_n:n\in\Bbb N\rangle$ converges. The natural candidate for the limit of the sequence is $x$, so we should try to prove that $\lim\limits_{n\to\infty}x_n=x$, i.e., that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x-x_n|<\epsilon$ whenever $n\ge m_\epsilon$. What we know is that
$$\lim_{n\to\infty}\sup_{k\ge n}x_k=x=\lim_{n\to\infty}\inf_{k\ge n}x_k\;,$$
i.e., that for each $\epsilon>0$ there is an $m_\epsilon'\in\Bbb N$ such that
$$\left|x-\sup_{k\ge n}x_k\right|<\epsilon\quad\text{and}\quad\left|x-\inf_{k\ge n}x_k\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon'\;.$$
(Why can I use a single $m_\epsilon'$ instead of requiring separate ones for each of the two limits?)
- Show that if $\ell\ge n$, then $$|x-x_\ell|\le\max\left\{\left|x-\sup_{k\ge n}x_k\right|,\left|x-\inf_{k\ge n}x_k\right|\right\}\;,$$ and conclude that setting $m_\epsilon=m_\epsilon'$ will ensure that $|x-x_n|<\epsilon$ whenever $n\ge m_\epsilon$ and hence that the sequence converges to $x$.
Best Answer
Suppose $|x_n|/n \to L.$ If $L = 0,$ we're done, so assume $L > 0.$ For $n$ large enough, $x_n/n$ is very close to either $L$ or $-L.$ If past a certain $n$ they are close to the same, then $\frac{x_n}{n}$ converges. Otherwise, infinitely many terms are close to either. Pick $m, n$ large enough with $x_n \approx nL, x_m \approx -mL,$ then $$(n+m)L \approx |x_{n+m}| \le |x_n + x_m| \approx |n-m|L,$$ contradiction. You can turn this into a rigorous $\epsilon-N$ argument.
Upon Feng's request: Fix $\epsilon = L/2$ and pick $N \ge 3L$ such that $|\frac{|x_n|}{n} - L| < \epsilon$ for $n \ge N.$ Pick $m, n \ge N$ as before with $|m-n| = 1,$ which is possible.
Then $$3L \le (n+m)(L - \epsilon) \le |x_{n+m}| \le |x_n + x_m| \le |n-m|(L+\epsilon) \le 2L,$$ contradiction.