The limit of inverse trigonometry function

Question

What is the value of $\displaystyle \lim_{n \to \infty} (\arctan (n+1) - \arctan (n))? $

I have opened my old notes about manipulating trigonometry function in limit form, but don't have much experience in its inverses like this. This is a real analysis problem and maybe need some theorems to emphasize the understandings.

Answer 1

Hint $$ \lim_{x\to\infty} \arctan x=\pi/2. $$

Answer 2

Hint:

You could try it the other way around.

$$\arctan (n+1)-\arctan n=\arctan \left(\frac {1}{n^2+n+1}\right) $$

Hence $$\lim_{x\to\infty} \arctan (n+1)-\arctan n=\lim_{x\to\infty} \arctan \left(\frac {1}{n^2+n+1}\right) =0$$

Answer 3

Using MVT

$$\arctan (n+1)-\arctan (n)=\frac {1}{1+c_n^2 } $$ with $$n <c_n <n+1$$

When $n \to +\infty , \; c_n\to +\infty $