Is the sequence $\frac{1}{\arctan(-n)}\cdot\frac{3n-2}{n^2+n+10}$ increasing or decreasing?

Question

Is the following sequence increasing or decreasing $$\frac{1}{\arctan(-n)}\cdot\frac{3n-2}{n^2+n+10}$$?

I managed to come to conclusion that $a_n=\frac{1}{\arctan(-n)}$ is strictly increasing and that for $b_n=\frac{3n-2}{n^2+n+10}, b_1<b_2<b_3>b_4>b_5...$. So $b_n$ is increasing for the first $3$ terms and after that it's decreasing. So what can I conclude about the product of theses $2$ sequences? I know that the product of $2$ increasing sequences doesn't have to be increasing.

Answer 1

Let $(a_n)$ be $\ge 0$ and decreasing, and $(b_n)$ be $\le 0$ and increasing. Then:

$$a_{n+1}b_{n+1} - a_n b_n = a_{n+1}(b_{n+1} - b_n) + (a_{n+1} - a_n)b_n \ge 0$$

So your sequence is increasing.

Answer 2

Your approach is correct and almost solves the problem, you just have to pay attention to the details.

Since you do not give a proof for the monotonicity of $(b_n)$, let us do it here. We suspect that from a certain $n$ onwards, $(b_n)$ will decrease. Indeed,

$$b_n \ge b_{n+1} \iff \frac {3n-2} {n^2+n+10} \ge \frac {3n+1} {n^2+3n+12} \iff 7n^2 + 30n -24 \ge 4n^2 + 31n + 10 \iff \\ 3n^2 -n -34 \ge 0 .$$

Given that $n \in \Bbb N$, the above is equivalent to $n \ge \dfrac {1 + \sqrt{409}} 6 \in (3,4)$ it follows that, indeed, $(b_n)_{n \ge 4}$ decreases. Clearly, it is also positive.

On the other hand, since $\arctan$ is an increasing function, it follows that $n \mapsto \dfrac 1 {\arctan n}$ is decreasing. It is also positive.

Now, while it's true that the product of two decreasing functions is not necessarily decreasing (think of $(-n) \cdot (-n) = n^2)$), it is nevertheless true that the product of two positive decreasing functions is decreasing: if $x < y$ then $f(x) \ge f(y)$ and $g(x) \ge g(y)$, so $f(x) g(x) \ge f(y) g(x) \ge f(y) g(y)$ (a similar argument can be given for increasing functions).

Remember that $\arctan(-n) = -\arctan n$. Since $n \mapsto \dfrac 1 {\arctan n}$ and $n \mapsto \dfrac {3n-2} {n^2+n+10}$ are both decreasing and positive, it follows that $n \mapsto \dfrac 1 {\arctan n} \dfrac {3n-2} {n^2+n+10}$ will also be so. Finally, putting the missing "$-$" in front of this product makes it negative and increasing.