Proving the limsup=lim of a sequence

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SUMMARY

The discussion centers on proving that if the limit of a sequence \( \{b_n\} \) exists and equals \( b \), then the limit superior (limsup) of the sequence also equals \( b \). The participant establishes that for sufficiently large \( n \), the terms of the sequence are within \( \epsilon \) of \( b \), leading to the conclusion that the limsup must also converge to \( b \). The key argument hinges on the definition of limsup as the supremum of all subsequential limits, which simplifies to \( b \) when the sequence converges.

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Homework Statement


Show that the if lim bn = b exists that limsup bn=b.

The Attempt at a Solution



Let limsup = L and lim = b

We know for all n sufficiently large
|bn-b|<ε
|bn| < b+ε

Therefore L ≤ b+ε and
|bn| < L ≤ b+ε

I'm trying to get |bn-L|<ε or |L-b|<ε both of which I believe imply that b=L.
The problem is I can't get my absolute value signs to be correct.
 
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I'm not sure why you would do that. The "limsup" of a sequence is defined as the supremum of the set of all subsequential limits. If the sequence itself converges, then every subsequence converges to that limit. That is the "set of all subsequential limits" contains ony a single number.
 

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