n-0-arg-n-2-n-1-i-pi-2-

Question Number 196325 by sniper237 last updated on 22/Aug/23

\underset{n = 0}{\sum^{\infty}} a r g (n^{2} + n + 1 + i) = π / 2

Answered by witcher3 last updated on 22/Aug/23

= \sum_{n ⩾ 0} \tan^{- 1} (\frac{1}{1 + n + n^{2}}) = \sum_{n ⩾ 0} \tan^{- 1} (\frac{n + 1 - n}{1 + (n + 1) n})

= \sum_{n ⩾ 0} \tan^{- 1} (n + 1) - \tan^{- 1} (n)

= \lim_{x \to \infty} \underset{n = 0}{\sum^{x}} \tan^{- 1} (n + 1) - \tan^{- 1} (n) = {\underset{x \to \infty}{\lim t a n}}^{- 1} (1 + x) = \frac{π}{2}

Facebook Tweet Pin