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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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    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](http://books.google.com/books?id=HrO6QzUHU-gC&pg=PA45) and a solution is given on [p.203](http://books.google.com/books?id=HrO6QzUHU-gC&pg=PA203).2012-09-28

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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.2012-09-26