if-a-n-1-a-1-a-n-find-lim-x-a-n-

Question Number 147529 by mathdanisur last updated on 21/Jul/21

i f a_{n + 1} = \sqrt{a_{1} + a_{n}}

Double subscripts: use braces to clarify

Commented by mathdanisur last updated on 21/Jul/21

S o r r y a_{1} = 12

Answered by gsk2684 last updated on 21/Jul/21

L = \lim_{n \to \infty} a_{n + 1} = \lim_{n \to \infty} \sqrt{a_{1} + a_{n}} > 0

L = \sqrt{a_{1} + L}

L^{2} - L - a_{1} = 0

L = \frac{1 \pm \sqrt{1 + 4 a_{1}}}{2} > 0

L = \frac{1 + \sqrt{1 + 4 a_{1}}}{2} = \frac{1 + \sqrt{1 + 48}}{2} = 4

Commented by Snail last updated on 21/Jul/21

Y o u a r e t e c h n i c a l l y w r o n g i n t h e s e c o n d l i n e

w h i l e w r i t i n g L u h a v d t o s h o w a t f i r s t t h e s e q

i s e i t h e r m o n o t o n i c i n c r e a s i n g o r d e c r e a s i n g

Answered by Snail last updated on 21/Jul/21

C l a i m a_{n + 1} ⩾ a_{n}

G o t h r o u g h b a s i c s t e p s o f i n d u c t i o n

a n d t h e n

F o r (m + 1) c a s e

w e h a v e a_{m + 1} ⩾ a_{m}

\sqrt{(} a_{1} + a_{m + 1}) ⩾ \sqrt{(} a_{1} + a_{m})

a_{m + 2} ⩾ a_{m} h e n c e t h e s e q i s m o n o t o n i c i n c r e a s

N o w u c a n d o a s l i k e i t i s d o n e

Commented by gsk2684 last updated on 21/Jul/21

y e s, t h a n k y o u

Answered by mathmax by abdo last updated on 21/Jul/21

a_{n + 1} = f (a_{n}) with f (x) = \sqrt{x + 12} (a_{1} = 12)

f is defined continue on [- 12, + \infty [

f^{^{'}} (x) = \frac{1}{2 \sqrt{x + 12}} > 0 \Rightarrow f is increazing on] - 12, + \infty [

so \lim_{n \to + \infty} a_{n} = x_{0} / f (x_{0}) = x_{0} (fix point of f)

\Rightarrow x_{0} = \sqrt{x_{0} + 12} \Rightarrow x_{0}^{2} - x_{0} - 12 = 0

Δ = 1 + 4 (12) = 49 \Rightarrow x_{1} = \frac{1 + 7}{2} = 4 and x_{2} = \frac{1 - 7}{2} = - 3 < 0

let prove [that a_{n} > 0 \forall n

a_{2} = \sqrt{a_{1} + 12} > 0 (true) let suppose a_{n} > 0 \Rightarrow

a_{n + 1} = \sqrt{a_{n} + 12} > 0 \Rightarrow \lim_{n \to + \infty} a_{n} = 4

Commented by mathdanisur last updated on 21/Jul/21

t h a n k y o u S i r