Real analysis: limit of sequences question

[mjjoga]

ok so,
a) If s sub n→0, then for every ε>0 there exists N∈ℝ such that n>N implies s sub n<ε.
This a true or false problem. Now this looks like a basic definition of a limit because
s sub n -0=s sub n which is less than epsilon. n is in the natural numbers. But, I thought there should be an absolute value around s sub n. so does that make it false? or does the absolute value of a sequence equal the sequence and it's true?

 

[CompuChip]

In this case, the statement is true. For example, whether you look at s_n = 1/n or s_n = -1/n doesn't matter.
Note however, that the converse is not true: if for all ε>0 there exists N∈ℝ such that n>N implies s_n<ε that does not mean that s_n→0. The -1/n I mentioned is a counterexample.

So you are right that there is an absolute value sign in the definition of limit, and you could say:
s_n→0 <==> for all ε>0 there exists N∈ℝ such that n>N implies |s_n|<ε ==> for all ε>0 there exists N∈ℝ such that n>N implies s_n<ε
Note that the <= is missing in the second step.

 

[mjjoga]

Thank you! that makes a lot more sense to me now.
mjjoga