Is this proof correct convergence

Question

I'm trying to show that $(a_{n}) \to -\infty \Leftrightarrow (-a_{n}) \to \infty$

Attempt $(\Rightarrow)$: We have $(a_{n}) \to -\infty$. By definition this means, for all $C < 0$ there exists $N$ such that $a_{n} < C$ for all $n > N$. Therefore we have that $ a_{n} < C < 0$. Therefore for all $ -C > 0$ there exists $N$ such that $-a_{n} > -C > 0$ for all $n > N$

$(\Leftarrow$): We have $(-a_{n}) \to \infty$. By definition this means, for all $C > 0$ there exists $N$ such that $-a_{n} > C$ for all $n > N$. Therefore we have that $a_{n} < -C < 0$. Hence we have for all $-C < 0$ there exists $N$ such that $a_{n} < -C$ for all $n > N$.

Answer 1

Your proofs are correct. Well done.

Answer 2

Yes, $$\forall C>0:\exists N:n>N\implies a_n>C$$

is strictly equivalent to

$$\forall C>0:\exists N:n>N\implies -a_n<-C$$

and to

$$\forall C<0:\exists N:n>N\implies -a_n<C.$$

Note that the condition $C>0$ is not even required.