# Limit Equals Infinity for a Sequence

I have to solve the following question in my assignment:

Prove that: $\lim\limits_{n \to \infty} \frac{n^2-n}{n+2}$ = $\infty$.

I have to prove this with the following definition:

"A series $(a_n)^\infty_{n=1}$ approches infinity if for every real number ${M}$, a natural number ${N}$ such that for every ${n}>{N}$, ${a_n}>{M}$".

I know how to solve the question with limit arithmetic, however, with the definition I have to work with, it's harder.

Your help is appreciated, thank you.

*Sorry for the bad mathjax, I am a newbie with the system.

• A note on terminology. What you are giving us is a sequence. A series is a sum. – Tim Raczkowski Mar 21 '15 at 14:20
• Thanks, fixed the mistake. – Alan Mar 21 '15 at 14:21

$$\lim\limits_{n\to\infty}\frac{n^2-n}{n+2}\color{red}{>}\lim\limits_{n\to\infty}\frac{n^2-n\color{red}{-6}}{n+2}=\lim\limits_{n\to\infty}\frac{(n+2)(n-3)}{n+2}=\lim\limits_{n\to\infty}n-3=\infty$$
$$\frac{n^2-n}{n+2}\ge \frac{n^2-n}{2n}=\frac{1}{2}(n-1)$$ with $n\ge 2$
Can you show that, for every natural number $n\ge 2$, that $\frac{n-1}{n+2} \ge \frac 14$? This means that $\frac{n^2 - n}{n+2} = \frac{n(n-1)}{n+2} \ge \frac n 4$ if $n \ge 2$, so that $$n > 4N \implies \frac{n(n-1)}{n+2} > N$$ for any natural number $N$.
$$\frac{n^2-n}{n+2}\ge\frac{n^2-n}{2n}=\frac n2-\frac12>M\iff n>2M+1$$