Let $a_n>0$ be bounded and: $\displaystyle\limsup_{n\to\infty}\frac 1 {a_n}=\frac 1 {\displaystyle\liminf_{n\to\infty}a_n}$ [duplicate]

Let $a_n$ be a sequence such that $\forall n\in \mathbb n: 0<a\le a_n\le b <\infty.$

Prove:

1. $\displaystyle\limsup_{n\to\infty}\frac 1 {a_n}=\frac 1 {\displaystyle\liminf_{n\to\infty}a_n}$

2. $\displaystyle\limsup_{n\to\infty}a_n\cdot \limsup_{n\to\infty}\frac 1 {a_n} \ge 1$ and there's an equality iff $a_n$ is converging.

1. Suppose there are two subsequences: $a_{n_l}, \ a_{n_k}$ such that $\lim a_{n_k} = k, \ \lim a_{n_l}=l$ and suppose $l\le k$, so $\lim \frac 1 {a_{n_k}}=\frac 1 k , \ \lim \frac 1 {a_{n_l}}=\frac 1 l$ so clearly: $\frac 1 k\le \frac 1 l\le l\le k$ so it's easy to see once the largest limit (supermum) is 'inverted' it has to become the smallest limit (infimum).

   I realize this doesn't show equality, I don't know how to do the other way and I'm not even sure if what I did is good.

2. If $a_n$ converges, suppose to $L$ as its limit then we have: $L\cdot \frac 1 L=1$.

   If it does not converge then $a_n$ may tend to infinity or won't have a limit. From 1 we can change it to $\displaystyle\limsup_{n\to\infty}a_n\cdot \frac 1 {\displaystyle\liminf_{n\to\infty}a_n} \ge 1$ and from BW, every sequence has a converging subsequence, and since for converging subsequences: $\liminf a_n\le \limsup a_n$ we have $\frac {\limsup a_n} {\liminf a_n}=\limsup a_n\cdot \limsup\frac 1 {a_n} \ge 1$.

   This should probably be in absolute value since one of those subsequnce limits can be negative, but it isn't in absolute value in the question.

Solution 1:

1. This is easy. Assume that $m>0$ is an eventual lower bound for $a_n$ (i.e. $a_n \geq m$ for $n \gg 1$). Then $1/a_n \leq 1/m$, so that $\limsup_{n \to +\infty} \frac{1}{a_n} \leq \frac{1}{m}$. By definition of $\liminf$, this means that $\limsup_{n \to +\infty} \frac{1}{a_n} \leq \liminf_{n \to +\infty} a_n$. Now take an eventual upper bound $M$ for $1/a_n$, and deduce similarly that $\liminf_{n \to +\infty} a_n \leq \limsup_{n \to +\infty} \frac{1}{a_n}$.
2. This is proved here.

By your assumption, all the quantities are actually positive.

Solution 2:

I think the following lemmas are particularly useful (In fact, some take these as the definitions of limsup's and liminf's).

1) $\limsup a_n \le a$ if and only if for every $\epsilon>0$, $a_n\le a+\epsilon$ for all but a finite number of $a_n$'s.

2) $\limsup a_n \ge a$ if and only if for every $\epsilon>0$, $a_n\ge a-\epsilon$ for infinitely many $a_n$'s.

Similar statements hold for the liminf.

Assuming $$\limsup \frac{1}{a_n}=a$$ we wish to show that $$\liminf a_n=\frac{1}{a}.$$ Let $\epsilon>0$ and choose $\delta>0$ so that $\frac{1}{a+\delta}=\frac{1}{a}+\epsilon$. Then by (1), $\frac{1}{a_n}\le a+\delta$ for all but a finite number of $a_n$'s, hence, $a_n\ge \frac{1}{a+\delta}=\frac{1}{a}+\epsilon$ for all but a finite number of $a_n$'s. Likewise by (2), $\frac{1}{a_n}\ge a+\delta$ for infinitely many $a_n$'s, hence, $a_n\le \frac{1}{a+\delta}=\frac{1}{a}+\epsilon$ for infinitely many $a_n$'s. We have just shown that, $$\liminf a_n = \frac{1}{a}.$$ This is what we wanted to show.

Similar arguments hold for the other cases.

Solution 3:

To get the first one, you just need to notice that $$\sup\{1/{a_k}; k\ge n\} = \frac1{\inf\{a_k; k\ge n\}}$$ and take limit for $n\to\infty$ to get $$\lim_{n\to\infty}\sup\{1/a_k; k\ge n\} = \lim_{n\to\infty}\frac1{\inf\{a_k; k\ge n\}} = \frac1{\lim_{n\to\infty}\inf\{a_k; k\ge n\}}.$$ (Although you should also check the cases when the LHS is $+\infty$ and when you have zero in the denominator.)

Once you have shown the first part, you get $$\limsup_{n\to\infty} \frac1{a_n} = \frac1{\liminf_{n\to\infty} a_n} \ge \frac1{\limsup_{n\to\infty} a_n}.$$ Then you can simply multiply this by $\limsup\limits_{n\to\infty} a_n$.

Again, you should check separately the cases where some of the above values is $+\infty$ or when you divide by zero.