Thread: real analysis proof of divergence to -infinity

I have the sequence (3+n-n^2)/(1+2n). I need to prove that this sequence diverges to negative infinity. So I have to find an N such that n>N implies that the sequence is less than any real number M. I understand how it works but the proofs for positive infinity seem to be easier. I've been trying to reduce it, making it bigger and bigger until it's less than M: sequence<...<...<M. I've even simplified it to numerous little expressions, but I either end up losing the negative, getting something like n squared, or not being able to find the N. So lost, I appreciate the help!

josette

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).

Thanks for the advice. I proved that -a > M and then by rules for inequalities said that a<M. Which is fine just like that cuz M is defined the same in both situations. Thanks again!!!

Josette