Negating the Definition of a Convergent Sequence to Find the Definition of a Divergent Sequence

Solution 1:

This is not the correct negation. Consider $x_n = (-1)^n$ and $l = 1$. The correct negation can be expressed as $$\exists\ \epsilon > 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon$$

Solution 2:

No, what you've written is not correct.

It looks like you need practice negating multiply-quantified statements. The key idea is that when you move a negation past a quantifier, it flips the quantifier from universal to existential or vice versa. Thus for instance

$\neg (\forall x$ $P(x))$

is logically equivalent to

$\exists x (\neg P(x))$.

Here $\neg$ means "not".

It would also be a good exercise to find an example to show that your proposed negation of "$a_n \rightarrow l$" need not be correct. As a hint: your condition implies that the sequence is unbounded.

I didn't really understand the second part of your question. In particular I don't follow "the next part of my task is to prove that a sequence is divergent using my formed proof". Maybe it's best to focus on one question at a time. Once you understand the negation question properly, you can ask the next part as a new question if you like.

Solution 3:

The statement would be something like:

$ \forall L, \exists \epsilon > 0 : \forall N, \exists n>N : |a(n)-L| \geq \epsilon$