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Determine whether the following sequence is increasing or decreasing:

$$\frac{(n-8)^2}{(1-n)^2}, n\geq 2$$

So the first few terms are: $36,\frac{25}{4},\frac{16}{9},...$ so let's assume the sequence is decreasing.

$a_n\geq a_{n+1} \Leftrightarrow \frac{(n-8)^2}{(1-n)^2} \geq \frac{(n-7)^2}{n^2} \Leftrightarrow \frac{(n-8)^2n^2-(n-7)^2(1-n)^2}{(1-n)^2n^2}\geq 0$

The denominator is positive $\forall n\in\Bbb N$ so now I only have to prove that the numerator is non-negative. In other words, I have to prove that $(n-8)^2n^2\geq (n-7)^2(1-n)^2 $ What would be some smart move now to prove this?

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You can rewrite your expression using polynomial division:

$$\frac{(n-8)^2}{(n-1)^2} = \left(\frac{n-8}{n-1} \right)^2 = \left(1-\frac{7}{n-1} \right)^2$$

since $\frac{7}{n-1}$ is a positive but decreasing sequence, $1-\frac{7}{n-1}$ is a increasing sequence. But since $1 < \frac{7}{n-1}$ for $n<8$ we need to consider the sign of the first few terms: We see that $\frac{(n-8)^2}{(n-1)^2}$ is first decreasing, but then obviously increasing after $n\geq 8$.

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    We know that $x\mapsto x^2$ is decreasing for $x\leq 0$, but increasing for $x\geq 0$, so we just check the sign of $1-\frac 7{n-1}$. – 2017-01-06