Calculating Lim Sup and Lim Inf for Given Sequences | Homework Solution

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Homework Statement



Let Asub2n-1 =(0,n/2^n) and Asub2n = (0, 2^n/n), Bsub2n-1 =[0,n/2^n] and Bsub2n =[0, 2^n/n] for n = 1,2,... find lim sup An, lim inf An, lim sup Bn, and lim inf Bn.

Homework Equations





The Attempt at a Solution


I know lim n/2^n approaches 0 as n approaches inf and lim 2^n/n approaches inf as n approaches inf but then where do I go??

lim An = lim sup An = (0,∞) (under lim m→∞, under sup n≥m) and lim An = lim inf An = (0,0) (under lim m→∞, under inf n≥m)??
 
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what is your definition of lim sup, lim inf?
 
lanedance said:
what is your definition of lim sup, lim inf?

I've been working on this using my notes. I hope this translates it may not make sense

n/2^n approaches 0 as n → ∞ and 2^n/n approaches ∞ as n → ∞ , so

lim(n→∞)An = lim(n→∞)supAn = lim(m→∞)⋃(n≥m)An = ⋂_(m=1)^∞⋃_(n=m)^∞A_n = (0,∞)
And
lim(n→∞)An = lim(n→∞) inf An = ⋃_(m=1)^∞ ⋂_(n=m)^∞ An = (0, ½ )
 
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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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