Proof that the sequence $(-1)^n$ for $n\in\mathbb{N}$ diverges I need to prove that the sequence $(-1)^n$ for $n\in\mathbb{N}$ diverges. Is this proof correct? Is there a better way to prove it? Probably my proof is very poor, but here it is.
We prove this by contradiction. First of all, we know that $-1 \leq (-1)^n \leq 1$. Now suppose it converges to a value $a$. Then we have $$\forall \epsilon >0, \exists n_0\in\mathbb{N}: |(-1)^n-a| < \epsilon \,\,\,\, \forall n\in\mathbb{N}, n> n_0$$ in particular, take $\epsilon :=1$.
Then $\exists n_0 \in \mathbb{N}$ such  that the above holds for $\epsilon =1$. Take now $n > n_0$ then we have $$|(-1)^n-a| \leq |(-1)^n|+|a| = |-1|^n+|a|=1+|a| \geq 1$$ where I have used the triangle inquality and the fact that $|b^n| = |b|^n$ and that $|a| \geq 0$.
Hence we have a contradiction because this doesn't hold for any $\epsilon >0$.

Is this proof correct? Because in my book the solution is quite different. It involves diving into cases where $n$ is odd or even. I am pretty sure I can't do the triangle inequality part, but I have no other idea since I don't understand the proof in the book.

 A: Another way to see that is 
$$a_{n+1}=-a_n \to a_{n+1}+a_n=0$$
And if $a_n$ converge to L then we should have
$$L+L=0\to L=0$$
What is not true because $a_n=\pm 1$.
A: One way:  suposse $a_n\to L\in\mathbb{R}.$ Taking $\epsilon =1,$ there exists $n_0$ natural number such that if $n\ge n_0$ then, $\left | (-1)^n-L\right |<1.$ For $n=2n_0$ we get $\left | 1-L\right |<1,$ so,  $L>0.$
For $n=2n_0+1,$ we get  $\left | -1-L\right |=\left | 1+L\right |<1,$ which implies $L<0$ (contradiction).
A: If ${\{a_n\}}_{n = 1}^{\infty}$ converges to $l$, where $a_n = {(- 1)}^n$ for all $n = 1 , 2 , \ldots$, all the subsequences of ${\{a_n\}}_{n = 1}^{\infty}$ converge also to $l$. In particular ${\{a_{2 n + 1}\}}_{n = 1}^{\infty}$, given by $a_{2 n + 1} = {(- 1)}^{2 n + 1} = - 1$ for all $n = 1 , 2 , \ldots$, converges to $l = - 1$, but ${\{a_{2 n}\}}_{n = 1}^{\infty}$, given by $a_{2 n} = {(- 1)}^{2 n} = 1$ for all $n = 1 , 2 , \ldots$, should converge also to $l$ and it's false clearly because $\lim_{n \to \infty} a_{2 n} = \lim_{n \to \infty} 1 = 1 \neq - 1 = l$. This argument proves that ${\{a_n\}}_{n = 1}^{\infty}$ doesn't converge.
