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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?
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?