Divergence of the sequence $n^2-100n$ as $n$ goes to $\infty$.

Question

State the definition of divergence to $+\infty$. Argue straight from this definition that $n^2-100n$ diverges to $+\infty$.

The definition is the negation of the definition of convergent sequence which says $\forall \epsilon>0 ~ \exists N \in \mathbb{R}$ s.t. $ |a_n-\ell| < \epsilon$ $\forall n > N$. The negation would then be $\exists \epsilon>0$ s.t.$ ~ \forall N \in \mathbb{R}$, $ |a_n-\ell| \ge \epsilon$ whenever $\exists n > N$. But I don't know how to prove divergence; so far I've only met with far more simple things.

Answer 1

Alternatively:

For $n > 100$, $$n^2 - 100n = n(n-100) \geq n \cdot 1 = n$$

Hence given $M > 0$, choose $N = \max(101,M)$. Then

$$n > N \ \Longrightarrow \ n^2 - 100n > M$$

Answer 2

The right definition of divergence to $\infty$ (supplied by Winther) is: for any $M>0$ there exists a $N$ such that if $n>N$ then $a_n>M$.

For $M>0$, take $N = \sqrt{M+50^2}+50$, then $n > N \implies n> \sqrt{M+50^2}+50$ which implies that $(n-50)^2-50^2> M$ which is the same as $n^2-100n>M$. Therefore $n^2-100n \to \infty$ as $n \to \infty$.