Show that if sn[itex]\leq[/itex]b for all but finitely many n, then lim sn[itex]\leq[/itex]b.
The Attempt at a Solution
My question is regarding the absolute value portion of the proof:
by contradiction: Call lim sn s. Suppose s>b. Then
l sn-s l < [itex]\epsilon[/itex].
Then l sn-s l < s-b.
sn>b, contradiction to problem statement.
My question is this: The proof only seems to work if we assume sn-s is negative. But why couldn't sn 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