I'm trying to show that if the following result is true $(a_{n}) \to - \infty \Leftrightarrow (-a_{n}) \to \infty$
Attempt:
$(a_{n})\to - \infty \Leftrightarrow \forall C < 0, \exists N$ s.t. $a_{n} < C, \forall n > N$ $\tag{1}\label{1}$
Now I obviously need to introduce a negative sign somehow to get to the definition of converging to infinity.
Multiplying $\eqref{1}$ by a negative I get:
$-(a_{n}) = (-a_{n})$ on the RHS and $\forall C > 0, \exists N \text{s.t. } a_{n} > C, \forall n < N$. But now I the last inequality is incorrect...
I'll show one direction. You should show the other. Assume $a_n \to -\infty$. Therefore, for all $C < 0,$ there is some $N$ such that $a_n < C$ for all $n \geq N.$
To show $-a_n \to \infty,$ we want to show that for all $C > 0,$ there is some $N$ such that $-a_n > C$ for all $n \geq N.$
Therefore, let $C > 0$ be arbitrary. Therefore $-C < 0,$ so by the assumption that $a_n \to -\infty,$ there is some $N$ (which we now fix) such that $a_n < -C$ for all $n \geq N.$
Let $n \geq N.$ It follows that $a_n < -C$ so $-a_n > C.$
As an important point that seemed (at least to me) to be tripping you up: Note that when we apply the assumption of $a_n \to -\infty,$ we don't need $'C'$ explicitly. We just need a negative number. When we assumed $C > 0,$ we obviously get $-C < 0$ so we can then apply the limit definition.