Given-n-arc-tan-n-lim-x-n-1-n-

Question Number 31706 by gunawan last updated on 13/Mar/18

Given θ_{n} = a r c \tan n, \lim_{x \to \infty} θ_{n + 1} - θ_{n} =

Commented by abdo imad last updated on 18/Mar/18

i f t h e q u e s t i o n i s f i n d {l i m}_{n \to \infty} θ_{n + 1} - θ_{n} w e h a v e

l i m θ_{n + 1} - θ_{n} = {l i m}_{n \to \infty} a r c t a n (n + 1) - a r c t a n n

\frac{π}{2} - \frac{π}{2} = 0