Subsequential Limit of a Sequence

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The discussion centers on finding the subsequential limits, limit superior, and limit inferior for the sequence s_n = 4^{1/n}. It highlights that every sequence has subsequences, such as (s_{2n}), even if they converge to the same limit. The limit inferior (lim inf) is defined as the infimum of all subsequential limits, while the limit superior (lim sup) is the supremum of those limits. Since the sequence converges, all subsequences will converge to the same limit, making both lim inf and lim sup equal to that limit. The key takeaway is that the absence of distinct subsequential limits does not indicate a lack of subsequences.
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I have a problem that asks for the subsequential limits, the limit superior, and the limit inferior for the sequence

s_n = 4 ^{\frac {1} {n}}

I haven't had trouble with my other problems, but I don't see any subsequences in the sequence (other than the sequence itself). Am I missing somthing?
 
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bphys said:
I have a problem that asks for the subsequential limits, the limit superior, and the limit inferior for the sequence

s_n = 4 ^{\frac {1} {n}}

I haven't had trouble with my other problems, but I don't see any subsequences in the sequence (other than the sequence itself). Am I missing somthing?
All sequences have subsequences. For example, isn't ##(s_{2n})## a subsequence of ##(s_n)##?
 
The "lim inf" of a sequence is the "infimum" (lower bound) of the set of all subsequential limits and the "lim sup" is the "supremum" (upper bound) of that same set. When you say you "can't find any subsequences" what you really mean is that you cannot find any subsequences that converge to different limits. That's not a problem. Because this sequence itself converges, it follows that all subsequences coverge to the same limit. "Limit inferior" and "limit superior" ("lim inf" and "lim sup") are both equal to that limit.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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