To prove : $\lim_{n \to +\infty} a_{n} $ diverges, where $a_{n}= (-1)^n$ from $\varepsilon$ definition 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.
 A: Suppose on the contrary that $\lim_{n \to \infty}a_n = a $
Let $\epsilon = 1$. Note that $a_n \in \{1,-1\}$ and the sequence alternative between the two values. Hence if it converges, then we have $|1-a|<1$ and $|-1-a| < 1$.
It is equivalent to  $|a-1|<1$ and $|a+1| < 1$.
We can rewrite it as $-1 < a-1 < 1$ and $-1 < a+1 < 1$.
$0 < a < 2$ and $-2 < a < 0$.
Hence there is no such $a$ that satisfies the system of inequalities as $a$ can't be positive and negative at the same time. We conclude that $a_n$ diverges.

Remark:
Suppose you can't decide which $\epsilon$ to pick, you can also work with arbitrary $\epsilon>0$ first.
$|a-1| < \epsilon$ and $|a+1| < \epsilon$.
$1-\epsilon < a <  1+\epsilon$ and $-1 -\epsilon < a < -1+ \epsilon$.
To make it infeasible, choose $\epsilon >0$ such that $-1 + \epsilon \le  1-\epsilon$ and we conclude that we can pick $0<\epsilon \le 1$.
A: 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.
