Checking monotonicity of a sequence with derivative of its function

Question

I have a sequence $a_n = \frac {3+n^2}{1-n}$

I assume this sequence is contained in the function $f$ such that $a_n=f(n)$

Now I take the derivative of $f(n)$, which gives me $\frac{-(n+1)(n-3)}{(1-n)^2}$

I find that $(1-n)^2$ is positive for all $n>1$

I also find that, for $n>3$, the numerator will always be negative.

So $\frac{negative}{positive} = negative$

Since $f'(n)$ is negative for all $n>3$, it means the function $f(n)$ is eventually decreasing.

Which should also imply that the sequence $a_n=\frac {3+n^2}{1-n}$ is eventually decreasing, meaning it is a monotonic sequence.

However, the textbook answer says this sequence is not monotone.

What am I doing wrong?

Answer 1

Looks like, based on your analysis, that the sequence is monotonic for $n\ge4: a_4, a_5, a_6, \cdots$, but the sequence $a_2,a_3,a_4 \cdots$ is not monotonic. The term $a_1$ is either infinite or undefined.

Answer 2

$a_2 = -7$.

$a_3 = \frac{12}{-2}=-6$

$a_4=\frac{19}{-3}$

Hence $a_2 < a_3$ but $a_3 > a_4$, hence it is not monotonoe. Monotonic function is monotone over all the indices, it's not about becoming monotonic eventually.

Answer 3

You are fully right, the sequence is decreasing for $n \geq 3$. However $a_2=-7$ and $a_3=-6$, it means it is not decreasing for all $n \geq 1$.