Trouble understanding negation of definition of convergent sequence.

Question

Definition of convergent sequence: $$\forall \varepsilon >0, \exists N \in \mathbb R: \forall n \in \mathbb N \ (n \ge N \implies d(x_n,x) < \varepsilon)$$

I found the negation to be: $$\exists \varepsilon > 0, \forall N \in \mathbb R: \exists n \in \mathbb N \ (n \ge N \land d(x_n, x) \ge \varepsilon)$$

However, I am having trouble understanding what is meant by $(n \ge N \land d(x_n, x) \ge \varepsilon)$. Specifically, the conjunction part. ($\land$)

What does it mean when we say ($n \ge N$ and $d(x_n, x) \ge \varepsilon$)?

Answer 1

Say it aloud.
"There is some n where n is greater than N and the distance from $x_n$ to $x$ is greater than epsilon."

Even though N may be large, there is some $x_n$ has moved away from the proposed limit point by more than a minimal amount.

Answer 2

The phrase

• for all $N \in \mathbb{R}$ there exists $n \in \mathbb{N}$ such that $n \ge \mathbb{R}$ and BLAH$(n)$

can be understood more colloquially as

• BLAH$(n)$ is true for arbitrarily large natural numbers $n \in \mathbb{N}$

Another equivalent way to express it is by saying

• There is a sequence of natural numbers $n_i \to +\infty$ such that BLAH$(n_i)$ is true for all $i$.

So, the negation of convergent sequence could be expressed in words as

• there exists $\epsilon>0$ such that $d(x_n,x) \ge \epsilon$ for arbitrarily large $n$

or as

• there exists $\epsilon>0$ and a sequence of natural numbers $n_i \to +\infty$ such that $d(x_{n_i},x) \ge \epsilon$ for all $i$.