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


it is small because the individual angles considered are arctangents from lines with positive slopes: the angle between the positive $x$ axis and a line of positive slope $m$ is $\arctan m$ which is between $0$ and $\pi / 2.$ 
You should have emphasized an earlier line
$$ \arctan \left( \frac{i+1}{d} \right) -  \arctan \left(  \frac{i-1}{d} \right) = \arctan \left( \frac{2d}{i^2 + d^2 - 1} \right) \approx  \frac{2d}{i^2 + d^2 - 1} $$ which becomes very small.
