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The question is: prove that $x_n=n-3n^2$ diverges to $-\infty$ as $n\to\infty$ using only the definition of divergence.

I know that $x_n$ diverges to $-\infty$ iff for each $M\in\mathbb{R}$ there is an $N\in\mathbb{N}$ such that $n\geq N$ implies $x_n<M$ but I don't know how to go about rigorously showing that $x_n$ diverges. Is there a general way of proving divergence?

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If $n \geq 1$, then $n-3n^{2} = n(1-3n) \leq -2n$. Let $M < 0$; then $-2n < M$ if $n > |M|/2$; so for all $n \geq \lceil \frac{|M|}{2} \rceil + 1$ we have $n-3n^{2} < M$.

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  • $\begingroup$ Just curious, what is so special about -2n? Why not just -n? $\endgroup$ – Matt G Oct 9 '15 at 1:25
  • $\begingroup$ Because $n \geq 1$ iff $1 - 3n \leq -2$. $\endgroup$ – Megadeth Oct 9 '15 at 1:41
  • $\begingroup$ So that is just natural $\endgroup$ – Megadeth Oct 9 '15 at 2:41

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