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$$\limsup \left(\frac 1{a_n} \right)=\frac 1{\liminf(a_n )} $$

I know this is true base on the definition of $\limsup$ and $\liminf$, but I don't know how to prove it formally.

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Is it really true? Note that $1/x$ is not a decreasing function. P.S. I am not hinting that its false, I'm just asking: is it true? – goblin Sep 29 '13 at 12:14
up vote 0 down vote accepted

Assume that a subsequence $a_{n_k} \to a$. Of course we need some assumptions, and I'll assume that $a_n>0$ for any $n$. The $$\frac{1}{a_{n_k}}\to \frac{1}{a},$$ and since $a=\liminf_{n \to +\infty} a_n$ is the smallest possible choice, we deduce at once that the largest possible limit point of $1/a_n$ is $1/a$. Hence $$\frac{1}{a} = \limsup_{n \to +\infty} \frac{1}{a_n}.$$

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$$\lim \sup \left(\frac {1}{a_n}\right) = \inf_{k=1}^{\infty}\sup_{n=k}^{\infty}\left(\frac {1}{a_n}\right) = \left(\frac {1}{\sup_{k=1}^{\infty}\inf_{n=k}^{\infty} a_n}\right)\dots$$

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You want $\inf,\sup$, not $\min,\max$. I don't see how this is a proof, though. – Pedro Tamaroff Sep 26 '13 at 19:48
If you put $a_n=n/(n-1)$, say, then your max is not defined. – Martin Argerami Sep 26 '13 at 19:54
Still waiting for a real proof. – Diane Vanderwaif Sep 27 '13 at 14:43

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