I am trying to study the convergence of the series of functions
$$\sum\_{n=1}^{\infty}\frac{1}{\ln n} \arctan\frac{(x-2)^n}{n^2},$$
with $x\in\mathbb{R}$.

I would like to use the inequality $\arctan(x)\le\frac{\pi}{2},$ but I am not sure about that. Could someone please give me a hint? I'll post my (tenative) solution, but I need a hint for starting.

Thank you in advance!

${\bf EDIT:}$ (based on Tito Eliatron hint). I study the convergence of the power series
$$\sum\_{n=1}^{+\infty}\frac{(x-2)^n}{n^2}.$$
By applying the square criterion, it is clear that the series converges when $x<3$ and diverges for $x>3$. Thus, for $x<3$, it is
$$\sum\_{n=1}^{\infty}\frac{1}{\ln n} \arctan\frac{(x-2)^n}{n^2}\sim\sum\_{n=1}^{\infty}\frac{1}{\ln n}\frac{(x-2)^n}{n^2}.$$

Moreover, we have
$$\sum\_{n=1}^{\infty}\frac{1}{\ln n}\frac{(x-2)^n}{n^2}\le \sum\_{n=1}^{\infty}\frac{(x-2)^n}{n^2},$$
which converges. Thus our starting series converges by comparison test. We only have to study the case $x=2$. In this case
$$\sum\_{n=1}^{\infty}\frac{1}{\ln n} \arctan\frac{(x-2)^n}{n^2}=\sum\_{n=1}^{\infty} 0$$
which converges.

Could someone please tell me if am I wrong with something? Thank you in advance!