calculer Σ_(n=1) ^(+oo) artan((2/n^2 ))


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Question Number 181238 by SANOGO last updated on 23/Nov/22

$c a l c u l e r$ $\underset{n = 1}{\sum^{+ o o}} a r t a n (\frac{2}{n^{2}})$
	Answered by qaz last updated on 23/Nov/22
	$arctan (2/n^2 )=arctan (n+1)−arctan (n−1) Σarctan (2/n^2 )=Σa_n =Σa_(2n−1) +Σa_(2n) =Σ{[arctan (2n)−arctan (2n−2)]+[arctan (2n+1)−arctan (2n−1)]} =[arctan (2∙+∞)−arctan (2∙1−2)]+[arctan (2∙+∞+1)−arctan (2∙1−1)] =(3/4)π$
	$\arctan \frac{2}{n^{2}} = \arctan (n + 1) - \arctan (n - 1)$ $Σ \arctan \frac{2}{n^{2}} = Σ a_{n} = Σ a_{2 n - 1} + Σ a_{2 n}$ $= Σ {[\arctan (2 n) - \arctan (2 n - 2)] + [\arctan (2 n + 1) - \arctan (2 n - 1)]}$ $= [\arctan (2 \cdot + \infty) - \arctan (2 \cdot 1 - 2)] + [\arctan (2 \cdot + \infty + 1) - \arctan (2 \cdot 1 - 1)]$ $= \frac{3}{4} π$
	Commented by mr W last updated on 23/Nov/22

	$s e e m s t o b e w r o n g . p l e a s e r e c h e c k s i r .$
	Commented by qaz last updated on 23/Nov/22

	$0 < \underset{n = 1}{\sum^{\infty}} \arctan \frac{2}{n^{2}} < \underset{n = 1}{\sum^{\infty}} \frac{2}{n^{2}} \approx 3.29$ $\frac{3}{4} π \approx 2.35, \frac{5}{4} π \approx 3.92$ $\frac{3}{4} π i s c o r r e c t, i t h i n k$
	Commented by MJS_new last updated on 23/Nov/22

	$I also think {it}^{'} s correct$
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