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Question Number 165349 by mkam last updated on 31/Jan/22

	Answered by puissant last updated on 31/Jan/22
	$(1/(2n^2 )) = (((2n+1)−(2n−1))/(1+(2n+1)(2n−1))) ⇒Σ_(n=1) ^∞ arctan((1/(2n^2 )))=Σ_(n=1) ^∞ {arctan(2n+1)−arctan(2n−1)} ⇒ Σ_(n=1) ^∞ arctan((1/(2n^2 )))=lim_(n→∞) Σ_(k=1) ^n {arctan(2k+1)−arctan(2k−1)} = lim_(n→∞) (−arctan(1)+arctan(2n+1)) ( somme telescopique). = −(π/4)+(π/2) = (π/4). ...............Le puissant............$
	$\frac{1}{2 n^{2}} = \frac{(2 n + 1) - (2 n - 1)}{1 + (2 n + 1) (2 n - 1)}$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} a r c t a n (\frac{1}{2 n^{2}}) = \underset{n = 1}{\sum^{\infty}} {a r c t a n (2 n + 1) - a r c t a n (2 n - 1)}$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} a r c t a n (\frac{1}{2 n^{2}}) = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} {a r c t a n (2 k + 1) - a r c t a n (2 k - 1)}$ $= \lim_{n \to \infty} (- a r c t a n (1) + a r c t a n (2 n + 1))$ $(s o m m e t e l e s c o p i q u e) .$ $= - \frac{π}{4} + \frac{π}{2}$ $= \frac{π}{4} .$ $. . . . . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . .$
	Commented by puissant last updated on 31/Jan/22
	$=lim_(k→∞) Σ_(t=1) ^k arctan(2t+1)−arcan(2t−1) = lim_(t→∞) [arcan(3)−arctan(1)+arctan(5)−arctan(3)+ arctan(7)−arctan(5)....+arctan(2k+1)−arctan(2k−1)] = lim_(k→∞) {−arctan(1)+arctan(2k+1)} = −(π/4)+(π/2) = (π/4).$
	$= \lim_{k \to \infty} \underset{t = 1}{\sum^{k}} a r c t a n (2 t + 1) - a r c a n (2 t - 1)$ $= \lim_{t \to \infty} [a r c a n (3) - a r c t a n (1) + a r c t a n (5) - a r c t a n (3) +$ $a r c t a n (7) - a r c t a n (5) . . . . + a r c t a n (2 k + 1) - a r c t a n (2 k - 1)]$ $= \lim_{k \to \infty} {- a r c t a n (1) + a r c t a n (2 k + 1)}$ $= - \frac{π}{4} + \frac{π}{2} = \frac{π}{4} .$
	Commented by mkam last updated on 31/Jan/22
	$sir how Σ_(k=1) ^∞ {arctan(2k+1)−arctan(2k−1)}= −arctan(1)+arctan(2n+1)$
	$s i r h o w \underset{k = 1}{\sum^{\infty}} {a r c t a n (2 k + 1) - a r c t a n (2 k - 1)} = - a r c t a n (1) + a r c t a n (2 n + 1)$
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