From which n is this sequence monotone?

My professor asked this question in an exam and I'm not sure how to solve it.

Given the sequence $a_n=\frac{(3n+3)^2}{(2n-10)^2}$ from which $n$ is this sequence monotone.

I'm not sure I understand what he means. I can determine that the sequence is monotone like this:

$$a_n=(\frac{3n+3}{2n-10})^2=(\frac{3}{2}\frac{(n-5)+18}{n-5})^2=(\frac{3}{2}\frac{(n-5)}{(n-5)}+\frac{18}{n-5})^2$$ Looking at $\frac{18}{n-5}$ we can determine that the limit of the number we we add to the rest of the equation is $0$ so $a_n$ is monotonically increasing to a limit $(\frac{9}{4})$.

But how do I find the $n$ from which the the sequence is monotone? Is it from the vertical asymptote at $n=5$ where $(2n-10)^2$ is $0$?

I'd appreciate any help!

Solution 1:

The sequence only makes sense for $n>5$. Then as both factors of the fraction are positive, you can ignore the squares. Next,

$$\frac{3n+3}{2n-10}=\frac32\left(1+\frac{6}{n-5}\right)$$ is decreasing.