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Prove that the given sequence ${a_n}$ diverges to infinity.

$a_n=\frac{n^3+5}{-n^2+8n}$

I believe that the sequence diverges to -infinity. And I have this for my proof so far:

Let $M>0$ and let $N=$ ?. Then $n>N$ implies... I am confused on how to solve for the N. I believe I have to make the numerator larger and denominator smaller, but I have a hard time visualizing this.

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We have $$a_n=-\left(\frac{n^3+5}{n^2-8n}\right).$$ Informally, we now want to show that that if $n$ is "large" then $\dfrac{n^3+5}{n^2-8n}$ is large. We do not want to worry about negative values of $n^2-8n$, so suppose that $n\gt 8$, making $n^2-8n\gt 0$. For $n \gt 8$, we have $$\frac{n^3+5}{n^2-8n} \gt \frac{n^3}{n^2-8n},\tag{$1$}$$ for on the right we have made the numerator smaller than on the left.

Also, for $n\gt 8$, we have $$\frac{n^3}{n^2-8n}\gt \frac{n^3}{n^2},\tag{$2$}$$ for on the right we have made the denominator bigger than on the left.

Combining Inequalities $(1)$ and $(2)$, we get that if $n\gt 8$ then $$\frac{n^3+5}{n^2-8}\gt \frac{n^3}{n^2}=n,$$ and therefore $$a_n=-\left(\frac{n^3+5}{n^2-8}\right)\lt -n.\tag{$3$}$$

Finally, we want to show that for any number $K$, so matter how large negative it is, we can find $N$ such that if $n\gt N$, then $a_n\lt K$. That will show that $\lim_{n\to\infty}a_n=-\infty$.

So let $K$ be negative, and let $N$ be any integer greater than the larger of $8$ and $|K|$. By Inequality $(3)$, if $n\gt N$, we have $$a_n\lt -n \lt -N\lt -|K|=K.$$

Remark: It is useful to work as long as possible with positive quantities, since there our intuition is likely to be more accurate.

There were additional items of unpleasantness introduced because of the fact that the limit is $-\infty$. It would be more comfortable to let $b_n=-a_n$, and show that $\lim_{n\to\infty} b_n=\infty$, and then refer to a little theorem somewhere to the effect that a sequence $(x_n)$ has limit $p$ (which may be a real number or one of $\pm\infty$) iff the sequence $(-x_n)$ has limit $-p$.

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Can you explain this in another way? I think I understand, but I'm not sure. I get confused with all the inequality signs. –  Skippydo Sep 26 '12 at 14:55
    
I have added what I hope is full enough detail, perhaps too full. –  André Nicolas Sep 26 '12 at 15:39
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Working directly from the definition, you're trying to show that for $n$ sufficiently large, you can make $a_n < -M$ for any $M \in \mathbb{N} .$ It helps to rewrite the general term as $\frac{n + 5/n^2}{-1 + 8/n}.$ The end behavior is suggested by the leading terms in the numerator and the denominator, so there are essentially two items you need to control first - the $\frac{5}{n^2} $ and the $\frac{8}{n}.$ It's easy to see that the first item is going to zero, so we know that at a cetain point ($n \ge 3$), the numerator is bounded by a certain natural number. To control the denominator, we want to make it smaller - just note that $\frac{8}{n}$ is always positive. Below is a concise proof.

Let $M > 1$ so that for $n > N = \max (2, M-1),$ we have $\frac{n^3+5}{-n^2+8n} = \frac{n + 5/n^2}{-1 + 8/n} < - (n+1) < -M .$

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