Question

Sequências.

Prove que se $\displaystyle\lim_{n \to \infty} a_n \ \textgreater \ \displaystyle\lim_{n \to \infty} b_n$, existe $n_0$ $\in \mathbb{N}$ tal que para todo $n \ \textgreater \ n_0$ temos $a_n \ \textgreater \ b_n.$

Answer 1

Olá!

Assumindo que vale tal desigualdade, os limites existem.

Sejam

$\displaystyle \lim_{n\to\infty}a_n=x,\;\text{e}\;\lim_{n\to\infty}b_n=y.$

Da definição, temos

$\bullet\;\forall\;\epsilon>0,\exists \;n_1\in\mathbb{N}:\;\forall n>n_1,\;|a_n-x|<\dfrac{\epsilon}{2}\\ \\ \\ \bullet\;\forall\;\epsilon>0,\exists \;n_2\in\mathbb{N}:\;\forall n>n_2,\;|b_n-y|<\dfrac{\epsilon}{2}$

Tome $n_0=\text{max}\{n_1,n_2\}$ e tem que

$n>n_0\Rightarrow \left\{\begin{array}{lcr}|a_n-x|&<&\dfrac{\epsilon}{2}\\ \;&\;&\\ |b_n-y|&<&\dfrac{\epsilon}{2}\end{array}\right.\Rightarrow \left\{\begin{array}{lcr}x-\dfrac{\epsilon}{2}&<a_n<&x+\dfrac{\epsilon}{2}\\ \;&\;&\\ y-\dfrac{\epsilon}{2}&<b_n<&y+\dfrac{\epsilon}{2}\end{array}\right. \Rightarrow \\ \\ \\ \Rightarrow \left\{\begin{array}{lcr}x-\dfrac{\epsilon}{2}&<a_n<&x+\dfrac{\epsilon}{2}\\ \;&\;&\\ -\dfrac{\epsilon}{2}-y&<-b_n<&\dfrac{\epsilon}{2}-y\end{array}\right. \Rightarrow$

$\Rightarrow a_n-b_n> x-(\epsilon+y)\;\overset{x>y}{>}\;0-\epsilon\Rightarrow a_n-b_n>-\epsilon\Rightarrow \\ \\ \Rightarrow a_n>b_n-\epsilon$

Como vale para todo $\epsilon > 0$, segue que $a_n>b_n$ se tivermos $n>n_0.$

Bons estudos!