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## Homework Statement

Show that if s

_{n}[itex]\leq[/itex]b for all but finitely many n, then lim s

_{n}[itex]\leq[/itex]b.

## Homework Equations

## The Attempt at a Solution

My question is regarding the absolute value portion of the proof:

by contradiction: Call lim s

_{n}s. Suppose s>b. Then

l s

_{n}-s l < [itex]\epsilon[/itex].

Choose [itex]\epsilon[/itex]=s-b.

Then l s

_{n}-s l < s-b.

-(s

_{n}-s)<s-b

s

_{n}>b, contradiction to problem statement.

My question is this: The proof only seems to work if we assume s

_{n}-s is negative. But why couldn't s

_{n}be greater than s? Am I missing something important in the wording of the problem? Perhaps the "all but finitely man n"? I actually don't quite grasp what that means.

If the sequence was 1/n

Choose