calculate-n-0-arctan-2n-1-n-4-2n-3-n-2-1-

Question Number 145165 by mathmax by abdo last updated on 02/Jul/21

calculate \sum_{n = 0}^{\infty} \arctan (\frac{2 n + 1}{n^{4} + {2 n}^{3} + n^{2} + 1})

Answered by mnjuly1970 last updated on 02/Jul/21

Ω_{n} = \underset{k = 0}{\sum^{n}} a r c t a n (\frac{{(k + 1)}^{2} - k^{^{2}}}{k^{2} {(k + 1)}^{2} + 1})

= \underset{k = 0}{\sum^{n}} {a r c t a n {(k + 1)}^{2} - a r c t a n (k^{2})}

\overset{Telescopic series}{=} a r c t a n {(n + 1)}^{2} - a r c t a n (0)

Ans : = \lim_{n \to \infty} Ω_{n} = a r c t a n (\infty) - a r c t a n (0) = \frac{π}{2}