# Math Help - Question about bounded sequences and limits

1. ## Question about bounded sequences and limits

Hi everyone,

I've got a problem on a practice test that I can't quite get my head around.

Let the sequence a subscript n, going from 1 to infinity, be bounded above by M. If the sequence goes to L as n goes to infinity, show (using an epsilon - N argument) that L <= M.

I'm given the hint: Assume to the contrary that L > M, then (L-M)>0 so you may take epsilon = (L-M)/2 in your epsilon - N argument.

I've gotten to the stage where I have (L+M)/2 < a subscript n < (3L-M)/2 and am not sure how to proceed.

2. Originally Posted by pineapple89
Hi everyone,

I've got a problem on a practice test that I can't quite get my head around.

Let the sequence a subscript n, going from 1 to infinity, be bounded above by M. If the sequence goes to L as n goes to infinity, show (using an epsilon - N argument) that L <= M.

I'm given the hint: Assume to the contrary that L > M, then (L-M)>0 so you may take epsilon = (L-M)/2 in your epsilon - N argument.

I've gotten to the stage where I have (L+M)/2 < a subscript n < (3L-M)/2 and am not sure how to proceed.

So you took $\epsilon=\frac{L-M}{2}$ ,and thus there exists $N_\epsilon\in\mathbb{N}\,\,\,s.t.\,\,\,n>N_\epsilo n\Longrightarrow |a_n-L|<\epsilon$ $\Longleftrightarrow -\epsilon . But $-\epsilon+L=\frac{L+M}{2}>\frac{M+M}{2}=M$ $\Longrightarrow a_n>-\epsilon+L>M$ , and this contradicts the definition of M...

Tonio