Why is $\tan^{-1}(\infty)$ here equal to $\dfrac{\pi}{2}$? In the question below in the last step of the solution, they seem to be claiming that $\displaystyle \lim_{n \to \infty+}\tan^{-1} \left (\dfrac{n+1}{d} \right) = \dfrac{\pi}{2}$, which I agree with but what about $\dfrac{5\pi}{2}$ etc.? How can they say it must be $\dfrac{\pi}{2}$?
Also if they aren't claiming what I said above it seems that they are claiming that $\displaystyle \lim_{n \to \infty+}\left[ \tan^{-1} \left (\dfrac{n+1}{d} \right)+\tan^{-1} \dfrac{n}{d}\right] = \pi $, which isn't necessarily true.
Problem and solution


The inverse tangent function can be extended to the extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty,\infty\}$, so
$$ \arctan \infty = \frac{\pi}{2}, \arctan(-\infty) = -\frac{\pi}{2}
$$
The definition here comes from the limiting behaviour
$$ \lim_{x \to \infty} \arctan x = \frac{\pi}{2}
$$
