Disproving the proposition that $a_n = (-1)^nn$ has a limit. Is the following Proof Correct?

- Show that the sequence $a_n = (-1)^nn,\forall n\in\{1,2,3,...\}$ does not have a limit.

Proof. Assume on the contrary that $\lim a_n = L$ consequently there exists an $N\in\mathbf{R^+}$ such that given any $n\in\mathbf{Z^+}$ $|(-1)^nn-L|&lt;|L|$ whenever $n&gt;N$ the archimedian property guarantees that such an $n$ exists consequently for some $n&gt;N$ we have $|(-1)^nn-L|&lt;|L|$  but $|(-1)^nn-L| = |(-1)^{n+1}(n+L)| = |(-1)^{n+1}|\cdot|n+L|$ implying that $|n+L|&lt;|L|$ resulting in a contradiction.

$\blacksquare$
The proof is missing $\varepsilon$ if you intend to use the $\varepsilon-N$ definition of the limit for sequences. If $(a_n)$ had a limit $L&lt;+\infty$ then so would $(b_n)$ where $b_n:=|a_n|$. This follows from triangle inequality
$$|b_n-|L||=||a_n|-|L||\leqslant |a_n-L|&lt;\varepsilon$$
whenever $n\geqslant N$ for some large enough $N$. But $\lim_nb_n=\lim_n |a_n|=\lim_nn=+\infty$.
if $n$ is even we get $$a_n=n$$ and this tends to infinity for $n$ tends to infinityif $n$ is odd, then we get $$a_n=-n$$ and this tends to $-\infty$ if $n$ tends to infinity, so no Limit exists