Is My Initial Approach to Proving a Function's Limit Correct?

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Prove that the limit when x--> infinite of [(-1 )^n-1]/n^2=0
So for ε > 0,exists N>0 so that n>N => |x -a|< ε
What I do is |[(-1 )^n-1]/n^2| < ε. Here I remove the absolute value and I have (1^n-1)/n^2 < ε

I know how to keep this going,but is it correct until now?
 
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Elaia06 said:
Prove that the limit when x--> infinite of [(-1 )^n-1]/n^2=0
So for ε > 0,exists N>0 so that n>N => |x -a|< ε
The right side should be |[(-1 )^(n-1)]/n^2| < ε
Elaia06 said:
What I do is |[(-1 )^n-1]/n^2| < ε. Here I remove the absolute value and I have (1^n-1)/n^2 < ε
|(-1)^(n - 1)| = 1, so all you need to do is find n so that 1/n^2 < ε.
Elaia06 said:
I know how to keep this going,but is it correct until now?

Put parentheses around your exponent. What you wrote, (-1)^n - 1 would be interpreted as
$$(-1)^n - 1 $$
 
Thank you so much! :)