TITLE: To prove : $\lim_{n \to +\infty} a_{n} $ diverges, where $a_{n}= (-1)^n$ from $\varepsilon$ definition
QUESTION [6 upvotes]: How to prove : $\lim_{n \to +\infty} a_{n}$ diverges, where $a_{n}= (-1)^n$ from $\varepsilon$ definition.
My way :
From definition :
$a = \lim_{n \to +\infty} a_{n} = \forall \varepsilon>0 \exists N(\varepsilon) \in \mathbb{N} : \forall n \geq N(\varepsilon) \implies \mid a_{n} -a \mid < \varepsilon$ ---(1)
Taking the negation of (1) statement we get :
$a \neq  \lim_{n \to +\infty} a_{n} = \exists \varepsilon>0 \forall N(\varepsilon) \in \mathbb{N} \exists n \geq N(\varepsilon) : \mid a_{n} -a \mid \geq \varepsilon$ ---(2)
We've to prove (2).
We've to divide the statement (2) into two parts :
(i) $\forall a \geq 0 \exists \varepsilon>0 \forall N(\varepsilon) \in \mathbb{N} \exists n = 2N+1\geq N(\varepsilon) : \mid a_{n} -a \mid \geq \varepsilon$ ---2(a)
(ii) $\forall a < 0 \exists \varepsilon>0 \forall N(\varepsilon) \in \mathbb{N} \exists n = 2N\geq N(\varepsilon) : \mid a_{n} -a \mid \geq \varepsilon$ ---2(b)
From 2(a) and 2(b) we get :
$\forall a \mathbb{R} \exists \varepsilon>0 \forall N(\varepsilon) \in \mathbb{N} \exists n \geq N(\varepsilon) : \mid a_{n} -a \mid \geq \varepsilon$ ---(3)
But, I'm confused to choose the proper $\varepsilon$ so that my work would be true. I'm stuck here. Please help me.
My previously asked the same question got downvoted due to lack of enough data, so I deleted the previous post and asked again with my detailed work.

REPLY [2 votes]: Convergent means:
There exists $a$ a number such that for every $\epsilon > 0$ there exists $N= N_{\epsilon}$ such that for every $n > N_{\epsilon}$ we have $|a- a_n| < \epsilon$.
The negation of the above statement: for every $a$ number there exists $\epsilon> 0$ such that for every $N_{\epsilon}$ there exists $n > N_{\epsilon}$ such that $|a-a_n| \ge \epsilon$.
Let's show that this negation is true: Take any $a$ real number. We'll show that $\epsilon =1$ works. Indeed, take any $N$. Take now any natural number $n > N$. Now let's show that at least one of the inequalities
$$|a_n - a| \ge 1$$
or
$$|a_{n+1} - a| \ge 1$$
holds. Indeed, we have
$$|a_n - a| + |a_{n+1} - a| \ge |a_{n} - a_{n+1}| = 2$$
so at least one of the inequalities is true.