# Thread: a simple limit-series problem

1. ## a simple limit-sequence problem

$\displaystyle lim(a_n)=a, lim(b_n)=b$

Prove:

if $\displaystyle a<b$, then there is an index $\displaystyle n_0$ so that for every $\displaystyle n>=n_0$ : $\displaystyle a_n$ $\displaystyle < b_n$

This one should be really easy, I just have no idea how to look at it... Any tips for calculus will be absolutely great!

$\displaystyle lim(a_n)=a, lim(b_n)=b$

Prove:

if $\displaystyle a<b$, then there is an index $\displaystyle n_0$ so that for every $\displaystyle n>=n_0$ : $\displaystyle a_n$ $\displaystyle < b_n$

This one should be really easy, I just have no idea how to look at it... Any tips for calculus will be absolutely great!

Take $\displaystyle \epsilon:=\frac{b -a}{2}\,\Longrightarrow\,\,\exists N_1\,,\,N_2\in\mathbb{N}\,\,s.t.\,\,|a_n-a|<\epsilon\,\,\forall n>N_1\,,\,\,|b_n-b|<e\,\,\forall n>N_1$ $\displaystyle \Longleftrightarrow -\epsilon<a_n-a<\epsilon\,,\,-\epsilon<b_n-b<\epsilon$

$\displaystyle \Longrightarrow\,\forall n>max(N_1\,,\,N_2)\,,\,b_n-a_n=b_n-b+b-a-(a_n-a)>-2\epsilon+b-a>0$ ...

Tonio

3. Thank you very much !!