Limits Question: Proving Sn <= b for Finite n

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The discussion centers on proving that if a convergent sequence \( S_n \) is bounded above by \( b \) for all but finitely many \( n \), then the limit \( \lim S_n \) must also be less than or equal to \( b \). The proof approach involves contradiction, assuming \( s = \lim S_n \) and \( s > b \). The participants clarify that the condition \( |S_n - s| < \epsilon \) leads to a contradiction when \( \epsilon \) is set to \( s - b \), as it implies \( S_n \) cannot consistently be less than \( b \).

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phygiks
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Hello,
I'm having trouble with this question.
Let Sn be a sequence that converges. Show that if Sn <= b for all but finitely many n, then lim sn <= b.
This is what I'm trying to do, assume s = lim Sn and s > b. (Proof by contradiction) abs(Sn-s) < E, E > 0. Don't know what to do from there, but maybe set E = s -b. E is epsilon by the way. Probably to start using latex...

If anyone could help, that would be awesome.
 
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Your idea is fine so far. Now what does |s_n - s| &lt; \epsilon = s-b for all n > N imply?
 
s is a upper bound, so the Sn-s is negative. So abs(Sn-s) < s -b doesn't hold true all n. I'm not sure though
 

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