[Math] Negation of sequence convergence

convergence-divergencereal-analysissequences-and-series

Just to verify that the negation of sequence convergence of ($a_n$) to a limit, a would be something like:

"There exists a $ \epsilon >0$ s.t. for all $n \in \mathbb{N}$ there exists an $n> \mathbb{N}$ s.t. $|a_n-a|\geq\epsilon$"

Is this right? My instructor wrote as a hint that we should start off by excluding all real numbers as limits. I don't quite understand what he means by that. Can someone give me a hint or clarify this?

Best Answer

The negation of the above statement would be $\exists \epsilon_o>0$ such that $\forall k \in \mathbb{N}~\exists p>k $ such that $$|a_p-a|\geq \epsilon_0$$

Related Question