prove the convergence of (α_n )_(n ) such that tan^(−1) ((α_n /n))−2α


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Question Number 127501 by slahadjb last updated on 30/Dec/20

$p r o v e t h e c o n v e r g e n c e o f {(α_{n})}_{n} s u c h t h a t$ $\tan^{- 1} (\frac{α_{n}}{n}) - 2 α_{n} + 1 = 0$
	Answered by mindispower last updated on 31/Dec/20

	$\Rightarrow 2 a_{n} = 1 + {t a n}^{-} (\frac{a_{n}}{n}) \Rightarrow a_{n} \in [- \frac{π}{4} + \frac{1}{2}, \frac{π}{4} + \frac{1}{2}]$ $b o u n d e d \Rightarrow \lim_{n \to \infty} \frac{a_{n}}{n} \to 0$ $\Rightarrow {t a n}^{-} (\frac{α_{n}}{n}) + 1 \Rightarrow 1 s i n c e g : x \to \tan^{- 1} (x) + 1 i s c o n t i n u s$ $2 a_{n} = g (\frac{a_{n}}{n}), 2 a_{n} \to g (0) = 1$ $a_{n} \to \frac{1}{2}$
	Commented by slahadjb last updated on 31/Dec/20

	$T h a n k y o u . C a n w e p r o v e t h a t (α_{n}) i s m o n o t o n e ?$
	Commented by mindispower last updated on 31/Dec/20

	$2 a_{n} = 1 + {t a n}^{-} (\frac{a_{n}}{n})$ $1 = 2 a_{n} - {t a n}^{-} (\frac{a n}{n})$ $f (x) = 2 x - {t a n}^{-} (\frac{x}{n}),$ $2 a_{n} - {t a n}^{-} (\frac{a_{n}}{n + 1}) = 1 + {t a n}^{-} (\frac{a_{n}}{n}) - {t a n}^{-} (\frac{a_{n}}{n + 1}) > 1$ $a_{n} > 0, \forall n ⩾ 4$ $⩾ 2 a_{n + 1} - {t a n}^{-} (\frac{a_{n + 1}}{n + 1}) = f (a_{n + 1})$ $f (a_{n}) ⩾ f (a_{n + 1}) \Rightarrow a_{n} > a_{n + 1} n > 4, c a u s e f i s i n c r e a s e$
	Commented by slahadjb last updated on 31/Dec/20

	$t h a n k y o u s o m u c h .$
	Commented by mindispower last updated on 31/Dec/20

	$p l e a s u r$
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