need help, thanks!

[sheilaw]

It is a fact that every convergent sequence is bounded.
Recall that, by denition, a sequence fang has the limit L if for every
"E"> 0 there is a corresponding integer N such that
abs(an-L) < "E" whenever n > N:
Use this denition to prove that every sequence that converges to 0
is bounded, i.e., to prove that if lim(an)=0 (n is infinite)
then there is a positive
number M such that
abs(an)<and= M for all n are natural number


I had tried that, but i'm not sure in the right way

abs(an-L)=abs(an-0)<"E"
abs(an)<"E"
because "E">0
thus M="E" and abs(an)<M

 

[daon]

fang? de nition? Its important to copy questions exactly as they appear.

This is a good one for an indirect approach.

Assume \(\displaystyle \{a_n\}\) is not bounded. Then \(\displaystyle \{|a_n|\}\) is not bounded above.

Pick any epsilon. For concreteness' sake, say \(\displaystyle \epsilon = 1\).

By assumption, \(\displaystyle a_n \rightarrow 0\), so we are guarenteed there exists an \(\displaystyle N \in \mathbb{N}\) such that \(\displaystyle m > N \Rightarrow |a_m| < 1\)

Now, the set \(\displaystyle S = \{|a_m|, |a_{m+1}|, |a_{m+3} |, \dots\}\) is not bounded above. If it were, we'd have \(\displaystyle \{|a_n|\}\) being bounded by the maximum of \(\displaystyle a_1, \dots a_N\) and any bound on \(\displaystyle S\).

This means a countable subset of \(\displaystyle S\) must be greater than 1.

So we have a contradiction. See it?