calculate Σ_(n=0) ^∞ arctan((1/(n^2 +n+1)))


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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
$=lim_(n→∞) {arctsn(n+1)−arctan(0)}$
$= \lim_{n \to \infty} {a r c t s n (n + 1) - a r c t a n (0)}$
Commented by mathmax by abdo last updated on 19/Mar/20

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