formal definition prove for $\lim_{a \to \infty} \{a_n\} = a$ also applies to $\{-a_n\} \to -a$

Question

By definition of any sequence $\{a_n\}$: if $\lim_{n \to \infty} a_n$ = a, then for all values $\epsilon > 0$, there exists a value $N \in \mathbb{N}$ such that for all values $n > N$ then $|a_n - a| < \epsilon$.

I want to prove that this also applies to any sequence $\{-a_n\}$ such that $\lim_{x \to \infty} -a_n = -a$

However I am not sure how to do this. I'm thinking of multiplying each side of the equation by a factor -1 such that the following is now stated:

$-a_n < -\epsilon$ and then define another value $\epsilon_2 = -\epsilon$.

For all purposes the above would then state:

for all $\epsilon_2 < 0$ there exists an $N \in \mathbb{N}$ such that for all values $n > N$ then $|-a_n -(-a)| < \epsilon_2$ which is equal to $|a_n - a| < \epsilon$.

However I feel like I am missing something to formally prove that whenever I have a convergent sequence $\lim_{n \to \infty} a_n = a$ then the sequence $\lim_{n \to \infty} -a_n = -a$ is also convergent.

Can someone help me figure this out?

Answer 1

there exists a value $N \in \mathbb{N}$ such that for all values $n > N$ then $a_n > \epsilon$.

This is wrong. The correct definition ends with "... there exists a value of $N\in\mathbb N$ such that for all values of $n>N$, $|a_n-a|<\epsilon$"

Answer 2

Definition of limit: $a_n \to a$ if for every $\epsilon >0$ there exists $N$ such that $n>N$ implies $|a_n-a| <\epsilon$.

$|-a_n -(-a)|=|a_n-a|$. Hence the $N$ that works for $(a_n)$ works for $(-a_n)$ also.