calculate-n-0-arctan-1-n-2-n-1-

Question Number 85164 by mathmax by abdo last updated on 19/Mar/20

c a l c u l a t e \sum_{n = 0}^{\infty} a r c t a n (\frac{1}{n^{2} + n + 1})

Commented by Cmr 237 last updated on 19/Mar/20

a r c t a n (\frac{1}{n^{2} + n + 1}) = a r c t a n (n + 1) - a r c t a n (n)

S = - a r c t a n (0) = 0

Commented by mind is power last updated on 19/Mar/20

= \lim_{n \to \infty} {a r c t s n (n + 1) - a r c t a n (0)}

Commented by mathmax by abdo last updated on 19/Mar/20

l e t S_{n} = \sum_{k = 0}^{n} a r c t a n (\frac{1}{k^{2} + k + 1}) \Rightarrow S_{n} = \sum_{k = 0}^{n} a r c t a n (\frac{k + 1 - k}{1 + k (k + 1)})

l e t k = t a n (u_{k}) \Rightarrow \frac{k + 1 - k}{1 + k (k + 1)} = \frac{t a n (u_{k + 1}) - t a n (u_{k})}{1 + t a n (u_{k}) t a n (u_{k + 1})}

= t a n (u_{k + 1} - u_{k}) \Rightarrow S_{n} = \sum_{k = 0}^{n} (u_{k + 1} - u_{k})

= u_{n + 1} - u_{0} = a r c t a n (n + 1) \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{π}{2}