On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$? I'm reading Ghorpade's Course in Calculus and Real Analysis. He points up one example of divergent sequence:

If $a_n:=(-1)^n$ for $n\in \mathbb{N}$, then $(a_n)$ is divergent. This can be seen as follows: Let $a\in \mathbb{R}$. If $|a|\neq 1$, let $\epsilon:=\min\{|a-1|,|a+1|\}$. Then $\epsilon>0$. Observe that $|a_n-a|\geq \epsilon$ for all $n$. If $|a|=1$, then let $\epsilon:=2$, and observe that $|a_n-1|\geq \epsilon$ for all off $n$ and $|a_n-(-1)|\geq \epsilon$ for all even $n$.

It's not very clear to me how he did this process of seeing that the sequence is divergent. The choice of $\epsilon$'s still seems a lot tricky to me. I am a little lost on why he assumed that $\epsilon:=\min\{|a-1|,|a+1|\}$ and then $\epsilon:=2$. I presume there is a reason for such values, but at the moment, they seem only arbitrary choices to me.

Could you expand a little on the thought process behind this?

 A: For the sequence to converge to $1$, we must have $\forall\epsilon\gt 0,\exists N\in\Bbb N :\forall n\ge N:|a_n-1|\lt\epsilon$. 
For the sequence to diverge, we must have the converse to this statement, i.e. there is an $\epsilon$ for which, $|a_n-1|\ge\epsilon,\forall n\in\Bbb N$.
I imagine, it would have taken deliberation for him to come up with such $\epsilon$.
But the choice is not arbitrary, since is $\epsilon=\min\{|a-1|,|a+1|\}$, then we see that:
$|a_n-a|=|1-a|=|a-1|$ if $n$ is even, and $|-1-a|=|1+a|$ if $n$ is odd.
thus $|a_n-a|\ge\epsilon$. 
This part rules out $a_n$ have a limit that is not $1$.
now let $\epsilon=2$.
now if $a=1$, $|a_n-1|=|1-1|=0$ if $n$ even, and $|a_n-1|=|-1-1|=2$ if $n$ is odd.
Thus $|a_n-1|\ge\epsilon$, $\forall n$ where $n$ is odd.
So the limit cannot be $1$ and it cannot be $\ne 1$, thus there is no limit at all.
A: The chosen value of $\varepsilon$ are either the distance between $a$ and $1$ or the distance between between $a$ and $-1$, whichever is smaller.
That makes then the distances between $a_n$ and $a$, for the various values of $n$.
The point is that you cannot make $a_n$ close to $a$ by making $n$ big enough.
That works if $a$ is neither $1$ nor $-1$.  If $a=1$, then the distance between $a_n$ and $a$ will be bigger than $\varepsilon$ if whenever $a_n=-1$ if $\varepsilon=\text{(for example) }1/10$.  A similar thing works if $a=-1$.
