Let $(x_n)$ be a sequence of positive real numbers.
Prove that:
$$\limsup_{n\to\infty} \frac1{x_n} = \frac1{\liminf_{n\to\infty} x_n}$$
I don´t know how to start.
[Math] lim sup of a sequence of postive numbers is equal to lim inf of the inverses of the numbers
real-analysis
Related Solutions
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 $x$.
A proof using context of subsequence for your reference:
($\Leftarrow$)Let $x_{n_k}$ be a convergent subsequence of $x_n$, then
$$\inf \{x_n:n\ge n_j\}\le x_{n_j}\le \sup\{x_{n}:n\ge n_j\}\text{ , }\forall j$$
$$\Rightarrow \lim\inf x_n\le\lim x_{n_k}\le \lim\sup x_n$$
So all convergent subsequence of $x_n$ converge to $\lim \sup x_n=\lim \inf x_n=x^*$, therefore $x_n$ converge to $x^*$
($\Rightarrow$)Let $x_n$ be a convergent sequence with limit $x^*$, then all subsequence of $x_n$ would converge to $x^*$, and the set of all subsequential limit is a singleton $\{x^*\}$. Note $\lim \sup x_n$ denotes the supremum of the set of subsequential limit and $\lim \inf x_n$ denotes the infimum of the set of subsequential limit, it follows that $$\lim \inf x_n=\lim \sup x_n=x^*$$
Best Answer
HINT
$\frac1x$ is decreasing and continuous on your domain.
Use this to estimate the two sides $$LHS\le RHS\le LHS$$