you could always work with the sequence {(n^2-n-3)/(1+2n)}, and show that it diverges to +∞

(if M is negative, than -M is positive, saying (-M) < a is the same as saying -a < M). the same N should work.

what i would do is re-write the numerator as:

3+n-n^2 = (1/2)(6 + 2n - 2n^2) = (1/2)(6+3n) - (n/2)(1+2n) = (9/4) + (3/4)(1+2n) - (n/2)(1+2n).