taylor expansion of arctan(x) at infinity

$\displaystyle \displaystyle \arctan(x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} + ... \ , \ |x| \le 1 $

You can't simply say that the expansion at infinity equals the expansion of $\displaystyle \arctan(1/x) $ at $\displaystyle 0 $. That wouldn't make any sense.

But $\displaystyle \displaystyle \arctan(x) + \arctan \Big(\frac{1}{x}\Big) = \frac{\pi}{2} $

so $\displaystyle \displaystyle \arctan(x) = \frac{\pi}{2} - \arctan \Big(\frac{1}{x} \Big) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^{3}} - \frac{1}{5x^{5}} + ... $

Would that then be considered the expansion at infinity? And would it be valid for x >1?