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


  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.

  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.


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.


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.

  • $\begingroup$ You show $\limsup a_k = 1/\liminf a_k$. But then you say, using this, we have $\limsup (1/a_n) = 1/\liminf a_n$, but surely now there's a fraction which changes things? $\endgroup$ – Numbersandsoon Jun 29 '14 at 9:09
  • 1
    $\begingroup$ Thanks for pointing out my mistake @BoSchmidt. I have edited $\limsup a_k$ (which was incorrect) to $\limsup 1/a_k$. $\endgroup$ – Martin Sleziak Jun 29 '14 at 9:36

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