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 $x$.
Hint for (a): both sequences are monotone and bounded.
Hint for (b): you can explicitely find $\bar b_n$ and $\underline b_n$. Your intuition is correct, btw.
Hint for (c): $\bar b_n\ge\underline b_n$ always.
Hint for (d): $\bar b_n\ge a_n\ge\underline b_n$ to prove in one direction and use the fact that a convergent sequence satisfies the Cauchy criterion to prove in another direction.
Anothe way to understand $\limsup$ and $\liminf$ is to take all convergent subsequences; then find $\sup$ and $\inf$ of the set of their limits. This approach, for example, instantly gives the answer to (b).
Best Answer
Given a sequence of real numbers, we can consider the set of accumulation points in the extended real number line. The set is non-empty and it has a minimum. The minimum is called the limes inferior. A convergent sequence has precisely one accumulation point, namely its limit.
That being said, $\infty$ is the limes inferior if and only if the sequence converges to $\infty$ and the limes inferior equals $-\infty$ if and only if the sequence is not bounded from below.