Proof that $ a_n = (-1)^{n} \cdot n $ is divergent I've tried to prove that the sequence $ a_n = (-1)^{n} \cdot n $ is divergent. Is my proof correct?
Proof:
Assume there exists an $ L \in \mathbb{R} $ such that $ L \gt 0 $ and $\lim_{n \to \infty} = L $. Therefore, for $ \epsilon = L $, There exists an $ N $ such that for all $ n \gt N: 0 < a_n < 2L $. Choose $ n = 2N + 1 \gt N. $ $ a_n = (-1)^{2N + 1} \cdot n = -(2N + 1) < 0$, which is a contradiction.
Assume there exists an $ L \in \mathbb{R} $ such that $ L \lt 0 $ and $\lim_{n \to \infty} = L $. Therefore, for $ \epsilon = -L $, there exists an $ N $ such that for all $ n \gt N: 2L < a_n < 0 $. Choose $ n = 4N > N $. $ a_n = (-1)^{4N} \cdot 4N = 4N > 0 $, and that is a contradiction.
Assume $\lim_{n \to \infty} = 0 $. Therefore, for $ \epsilon = \frac{1}{2} $, there exists an $ N $ such that for all $ n \gt N, \left|a_n\right| \lt \frac{1}{2} $. For all $ n \in \mathbb{N},a_n \in \mathbb{Z} $ and $a_n ≠ 0 $. Therefore, we get a contradiction.
Therefore, $ a_n $ is divergent.
 A: We do not need to break this problem into multiple cases. I find that simplification and explicitly breaking down the components of precisely what you are trying to prove when first dealing with so many quantifiers and negations can help with understanding as well. Intuitively, to show that some sequence (in a nice space like $\mathbb{R}$) is convergent, we want to show that eventually, the terms all become arbitrarily close to a fixed point in the space. What this means is, if $L$ were a limit, given any positive $\epsilon$, we can specify some $N\in\mathbb{N}$ such that for all $n\geq N$, $|a_n-L|<\epsilon$. Thus, to show that some sequence is not convergent, we need only show that no point $L$ satisfies this limit property. That is, for any given $L$, we can find some positive $\epsilon$ such that for any $N$, there is still some $n\geq N$ such that $|a_n-L|\geq \epsilon$. This is easy in this case as you can see that the absolute values of our terms keep growing.
Now for the actual proof. For any $L$, suppose say $\epsilon=1$ (any positive one will work here, but for future problems, you may have to choose a more suitable ones that may depend on each choice of $L$). Then, for any $N\in\mathbb{N}$, as long as say  $\mathbb{N}\ni n>N+1+|L|>N$, then $|a_n-L|\geq|a_n|-|L|\geq 1$, so this sequence does not converge to $L$, thus does not converge to any point, and thus diverges.
