Proof verification: sequence $ a_n = (-1)^n $ does not converge Theorem: Show that the sequence $ a_n = (-1)^n $ for all $ n \in \mathbb{N}, $ does not converge.
Proof: Suppose that there exists a limit $L$ such that $ a_n \rightarrow L  $. Specifically, for $ \epsilon = 1 $ there exists $ n_0 $ s.t. for all $ n > n_0$ then $|(-1)^n-L|<1$, from this we can also infer that $|(-1)^{n+1}-L|<1$ for all $ n > n_0 $.
Let $ n > n_0 $ be arbitrary, then $ 2 = | (-1)^n - (-1)^{n+1} | \leq | (-1)^n - L | + | L - (-1)^{n+1} | < 1 + 1 = 2  $, so since $ n > n_0 $ was arbitrary , hence for all $ n>n_0 $ we have 2<2 which is false, hence we get a contradiction.
My question: Regarding to when I wrote " $ n>n_0 $, we have 2<2 which is false, hence we get a contradiction. "
I'm trying to understand with respect to what there is a contradiction; we get that any statement of the form $ \forall n>n_0 (P(n) )$ is false ( since we got $ \forall n>n_0 . 2<2 ) $ [ and $ P(n) $ is a statement that depends on $ n $ ], and since we already have in the proof the statement " for all $ n > n_0$ then $|(-1)^n-L|<1$ " , then we get a contradiction. Is this correct?
Note: I know there are answers on this sequence's divergence, but they didn't really help me find an answer to my question.
 A: I say your statement $P(n)$ does not depend on $n$.
With respect to the known fact that $2=2$, you have deduced a contradiction. Hence your proof is good. (Although stop at the $2<2$ bit. You go on saying "so since $n$ was arbitrary", but that is not relevant.)
$a_n$ converges means that there is an $L$, so that for every positive epsilon, we can find a (big) index beyond which $a_n$ is epsilon-close to L.
$a_n$ does not converge means that for any $L$, there is a positive epsilon, so that for any index, we can find an even further index $t$, at which $a_t$ is not epsilon-close to L.
Maybe it will help you to see your proof from this second viewpoint.
Do you see how the "for anys" trade places with the "there exists at least one", when we contradict a complicated definition?
Try to step through your proof using the definition of "doesn't converge". It is in a way running your argument backwards.
So the task is: faced with any real $L$ candidate, to answer with a suitable epsilon.
A: Your statement $P(n):= 2< 2$ does not depend on $n$.
Although if $\lim (-1)^n = L$ then there would be an $n_0$ so that $n > n_0\implies 2< 2$.
But no such $n_0$ can exists because $P(n)$ is false independent of $n$.
.....
Alternatively (although this is a bit abstract for my taste) $n > n_0\implies 2< 2$ and $2< 2$ is false; implies $n > n_0$ is false and we are (in theory) claiming there is an $n_0$ that is is an upper bound for all integers.
