prove-the-convergence-of-n-n-such-that-tan-1-n-n-2-n-1-0-

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