Proving the Reciprocal Relationship of Lim Sup and Lim Inf for Bounded Sequences

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Suppose a_n is a bounded sequence. Then prove that lim sup a_n = 1/lim inf (1/a_n).

This seems completely obvious to me, I don't know how to do this any simpler.
 
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lim sup{an}
= limn->oosupk>n{ak} ... (justify this)
= limn->oo[1/infk>n{1/ak}] ... (justify this)
= 1/[limn->ooinfk>n{1/ak}] ... (justify this)
= 1/lim inf{1/an} ... (justify this)
 
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Let S={limit points of a_n}. Since a_n is strictly positive and bounded, the limit points of 1/a_n are precisely 1/s for s in S. It follows from there, doesn't it?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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