Real Analysis: Proof of convergence

[slb123]

Problem:
Prove if {b[sub:tt4teg2b]n[/sub:tt4teg2b]} converges to B and B ? 0 and bn ? 0 for all n, then there is M>0 such that |b[sub:tt4teg2b]n[/sub:tt4teg2b]|?M for for all n.

What I have so far:
I know that if {bn} converges to B and B ? 0 then their is a positive real number M and a positive integer N such that if n?N, then |b[sub:tt4teg2b]n[/sub:tt4teg2b]|?M . (by lemma)
PROOF (of lemma)- (Note: let E be epsilon)
since B ? 0, (|B|)/2=E>0. There is N such that if n?N, then
|b[sub:tt4teg2b]n[/sub:tt4teg2b]-B|<E. Let M=[(|B|)/2]. thus for n?N,
|b[sub:tt4teg2b]n[/sub:tt4teg2b]|=|b[sub:tt4teg2b]n[/sub:tt4teg2b]-B+B|?|B|-|bn-B|?|B|-[(|B|)/2]=[(|B|)/2]=M

i know that this is not the entire proof i need but i dont know what changes need to be made. what variations do i need to make in the proof for the lemma? please help point me in the right direction!

 

[DrSteve]

You're almost done. You just need to worry about the finitely many terms \(\displaystyle b_n\) for \(\displaystyle n<N\). So just let \(\displaystyle K=min\{ b_1,...b_{N-1}, M\}\).

Remark: You may want to change the roles of M and K when you write down the argument.