Negating the Definition of a Convergent Sequence to Find the Definition of a Divergent Sequence My task is to write a precise mathematical statement that "the sequence $(a_n)$ does not converge to a number $\mathscr l$"
So, I have my definition of a convergent sequence:
"$\forall\varepsilon>0$ $\exists N\in\Bbb R$ such that $|x_n -\mathscr l|<\varepsilon$
$\forall n \in \Bbb N$ with $n>N$"
Would the correct negation of this be "$\forall\varepsilon>0$ $\exists N\in\Bbb R$ such that $|x_n -\mathscr l|>\varepsilon$
$\forall n \in \Bbb N$ with $n>N$"?
It doesn't seem that this is the answer as the next part of my task is to prove that a sequence is divergent using my formed proof, but it'd be difficult to do since it's a general proof of divergence and not just a proof that $(a_n)$ doesn't converge a specific number $\mathscr l$
Perhaps I should find a prove that $(a_n)$ tends to $\pm\infty$? This is more simple but it does not include monotone sequences such as $x_n:=(-1)^n$.
Can someone assist me with this task? All comments and answers are appreciated. 
 A: 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.  
A: The statement would be something like:
$ \forall L, \exists \epsilon > 0 : \forall N, \exists n>N : |a(n)-L| \geq \epsilon$
A: This is not the correct negation. Consider $x_n = (-1)^n$ and $l = 1$. The correct negation can be expressed as
$$\exists\ \epsilon > 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon$$
A: I am just going to define it in words and then explain the formulation.
The definition of convegence say that for every value of $\epsilon > 0$, you will definitely get a natural number $N(\epsilon)$ such that $|x_n-l|<\epsilon$ for all $n \geq N(\epsilon)$.(where $l$ is the point of convergence).
Now, to negate this statement, you just have to find at least one $\epsilon >0 $ for which there exist no $N(\epsilon)$ s.t. $|x_n-l|<\epsilon$ for all $n \geq N(\epsilon)$ holds true. This is equivalent to saying that for all natural numbers $N$ you will get atleast one natural number $M$ s.t. $|x_M-l|>\epsilon$ where $M > N $. 
For e.g. You will start with $N=1$ and you will definitely get a natural number $M_1$ greater than 1 for which $|x_{M_1}-l|>\epsilon$. If you start with $N=2$ then you will definitely get a natural number $M_2$ greater than 2 for which $|x_{M_2}-l|>\epsilon$ and this will keep on going for all natural numbers $N$.
The logical statement of this, then, converts to 
$$
\exists\ \epsilon > 0,\ \forall\ N \in \mathbb N\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon
$$
