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Suppose $a_n > 0$ for all $n$. How do I show that $\limsup a_n^{-1} = (\liminf a_n)^{-1}$?

My first thought was to consider when the sequence is bounded and unbounded. In the first case, it was easy

For unboundedness, I have an idea, i.e. $0 < a_n < \infty \implies 0< \frac{1}{a_n} < \infty$

But I don't think the transition is correct (it's very sloppy)

Could someone formally perfect that last step for me? Because I will know that $\limsup a_n^{-1} = \infty$ and $(\liminf a_n)^{-1} = \infty$

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A hint: Inversion, as a map from $[0,\infty]$ to itself, inverses the order on this set. Since $\limsup$ and $\liminf$ are defined just in terms of ordering, and you get one from the other by reversing the order, the result should be easy to prove just by chasing the definitions. No clever ideas required. –  Harald Hanche-Olsen Sep 26 '12 at 23:07
    
This can be found (together with many other basic facts on $\liminf$ and $\limsup$) in the book Kaczor, Nowak: Problems in Mathematical Analysis. This is Problem 2.4.22 The problem is stated on p.45 and a solution is given on p.203. –  Martin Sleziak Sep 28 '12 at 5:31
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2 Answers 2

up vote 2 down vote accepted

The statement is false for sequences $\langle a_n:n\in\Bbb N\rangle$ such that $\liminf_{n\in\Bbb N}a_n=0$: $0^{-1}$ isn’t defined. It should say that

  • $\limsup_{n\in\Bbb N}a_n^{-1}=(\liminf_{n\in\Bbb N}a_n)^{-1}$ when $\liminf_{n\in\Bbb N}a_n\ne 0$, and
  • $\limsup_{n\in\Bbb N}a_n^{-1}=\infty$ when $\liminf_{n\in\Bbb N}a_n=0$.

Thus, boundedness of the sequence isn’t really a consideration. I’ll deal instead with the case in which $\liminf_{n\in\Bbb N}a_n=0$.

In this case we know that for each $\epsilon>0$ there is an $n_\epsilon>0$ such that $\inf\{a_k:k\ge n\}<\epsilon$ whenever $n\ge n_\epsilon$. Thus, for each $n\ge n_\epsilon$ there is a $k\ge n$ such that $a_k<\epsilon$.

We want to show that $\limsup_{n\in\Bbb N}a_n^{-1}=\infty$, i.e., that for each positive $x$ there is an $m_x\in\Bbb N$ such that $\sup\{a_k^{-1}:k\ge n\}>x$ whenever $n\ge m_x$. Let $\epsilon=\frac1x$, and let $m_x=n_\epsilon$. Then for each $n\ge m_x$ there is a $k\ge n$ such that $a_k<\epsilon$ and hence such that $a_k^{-1}>\frac1{\epsilon}=x$. This implies that $\sup\{a_k^{-1}:k\ge n\}>x$ whenever $n\ge m_x$, which is exactly what we needed.

You should take another look at what you did in the bounded case: since you didn’t correctly identify the actual trouble spot, you probably don’t have a completely correct argument.

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You can also take this tack. If you have a sequence, its lim inf is the smallest limit point of the sequence and its lim sup is the largest. You can find the proof of this fact on this website.

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Of course, these must both be positive and finite. –  ncmathsadist Sep 26 '12 at 23:36
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