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 $(\liminf a_n)^{-1} = \infty$
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
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.