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 $x$.
Notice that the inequality $\limsup x_n \le \inf\{x_n,\dots\}$ is not correct. Indeed for $x_n = (-1)^n$, it would read $1 \le -1$.
You can try the following: suppose that $\liminf x_n = a < \limsup x_n = b$ and extract two subsequences that realize these limits, i.e. pick $\{x_{n_k}\}, \{x_{n_j}\}$ such that $x_{n_k} \to a$ and $x_{n_j} \to b$. Let $\epsilon$ be such that $a + \epsilon < b - \epsilon$ and let $K, J$ be large enough so that $x_{n_k} \le a + \frac{\epsilon}{2}$ for every $k \ge K$ and $x_{n_j} \ge b - \frac{\epsilon}{2}$ for every $j \ge J$.
Then for every $k \ge K$ and $j \ge J$ $$|x_{n_j} - x_{n_k}| \ge b - \frac{\epsilon}{2} - a - \frac{\epsilon}{2} = b - a - \epsilon \ge \epsilon.$$
Do you see how this contradicts the assumption that $\{x_n\}$ is a Cauchy sequence?
Best Answer
Original Answer: Since $\inf_{n\in\mathbb{N}} n^{-1}x_n=-\infty$, there exists a sequence $n_j\to\infty$ such that $\frac1{n_j}x_{n_j}<-j$. Taking $j\to\infty$ gives $\lim_{j\to\infty}\frac1{n_j}x_{n_j}=-\infty$. Since $\lim_{n\to\infty}\frac1nx_n$ exists (in the extended reals) $$ -\infty=\lim_{j\to\infty}\frac{x_{n_j}}{n_j}=\lim_{n\to\infty}\frac{x_n}{n} $$ where the final equality follows from $\frac1{n_j}x_{n_j}$ being a subsequence of $\frac1nx_n$.
Modifed Anaswer: To prove that $\lim_{n\to\infty}\frac1n x_n$ exists, for every $m,n\in\mathbb{N}$ with $n>m>0$ we get $n=mq+r$, some $0\leq r<n$, so subadditivity gives $$ a_n\leq qa_m+a_r $$ and hence $$ \frac{a_n}{n}\leq \frac{q}{mq+r} a_m+\frac1{mq+r} a_r. $$ Taking limit superior as $n\to\infty$ gives $$ \limsup_{n\to\infty}\frac{a_n}n \leq\limsup_{q\to\infty}\max_{r=0,1,\dots,m-1}\left[\frac{q}{mq+r} a_m+\frac1{mq+r} a_r\right]=\frac{a_m}m $$ So we can take inf over $m$, $$ \limsup_{n\to\infty}\frac{a_n}n \leq\inf_m\frac{a_m}m $$ and so in this case we conclude $\limsup_n \frac1n a_n=-\infty$ and so also $\liminf_n\frac1n a_n=-\infty$.