Find-the-limit-of-this-sequence-2-2-2-2-2-2-2-2-2-2-Show-that-the-sum-of-the-terms-of-this-infinite-sequence-does-not-converge-

Question Number 1608 by 112358 last updated on 26/Aug/15

F i n d t h e l i m i t o f t h i s s e q u e n c e .

\sqrt{2}, \sqrt{2 \sqrt{2}}, \sqrt{2 \sqrt{2 \sqrt{2}}}, \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}, \dots

S h o w t h a t t h e s u m o f t h e

t e r m s o f t h i s i n f i n i t e s e q u e n c e

d o e s n o t c o n v e r g e .

Commented by Rasheed Ahmad last updated on 26/Aug/15

L e t \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \dots}}}} = x

S q u a r i n g b o t h s i d e s,

2 \sqrt{2 \sqrt{2 \sqrt{2 \dots}}} = x^{2}

2 . x = x^{2}

x = 2

T h e l i m i t o f t h e s e q u e n c e = 2

Commented by 112358 last updated on 26/Aug/15

I n d e e d i t i s! L = 2 .

Answered by 123456 last updated on 26/Aug/15

a_{n} = \sqrt{2 a_{n - 1}}

a_{0} = \sqrt{2}

a_{n} > 0 \Rightarrow 2 a_{n} > 0 \Rightarrow \sqrt{2 a_{n}} > 0 \Rightarrow a_{n + 1} > 0

0 < a_{n} < 2 \Rightarrow 0 < 2 a_{n} < 4 \Rightarrow 0 < \sqrt{2 a_{n}} < 2 \Rightarrow 0 < a_{n + 1} < 2

0 < a_{n} < a_{n + 1} < 2 \Rightarrow 0 < 2 a_{n} < 2 a_{n + 1} < 4 \Rightarrow 0 < \sqrt{2 a_{n}} < \sqrt{2 a_{n + 1}} < 2 \Rightarrow 0 < a_{n + 1} < a_{n + 2} < 2

0 < a_{n + 1} < a_{n} < 2 \Rightarrow 0 < 2 a_{n + 1} < 2 a_{n} < 4 \Rightarrow 0 < \sqrt{2 a_{n + 1}} < \sqrt{2 a_{n}} < 2 \Rightarrow 0 < a_{n + 2} < a_{n + 1} < 2

a_{0} = \sqrt{2}, a_{1} = \sqrt{2 \sqrt{2}} \Rightarrow a_{1} > a_{0}, the a is crescent

since a is bounced and crescent, \lim_{n \to + \infty} a_{n} exists

a_{n + 1} = a_{n} = a

a = \sqrt{2 a} \Rightarrow a^{2} = 2 a \Rightarrow a (a - 2) = 0 \Rightarrow a = 0 \lor a = 2

since it crescent them a = 2, \lim_{n \to + \infty} a_{n} = a = 2

S = \underset{n = 0}{\sum^{+ \infty}} a_{n}

since \lim_{n \to + \infty} a_{n} \neq 0, then the serie diverge

Commented by Rasheed Ahmad last updated on 27/Aug/15

W h a t i s m e a n t b y^{'} {c r e s c e n t}^{'} ?

I a m a s k i n g f o r t h e s a k e o f

k n o w l e d g e .

Commented by 123456 last updated on 27/Aug/15

growing or something like this (sorry about my bad english)

as shown

0 < a_{n} < a_{n + 1} < 2 \Rightarrow 0 < a_{n + 1} < a_{n + 2} < 2

since 0 < a_{0} < a_{1} < 2, them (0 < 2 < 2 \sqrt{2} < 4 \Rightarrow 0 < \sqrt{2} < \sqrt{2 \sqrt{2}} < 2)

0 < a_{0} < a_{1} < a_{2} < a_{3} < \cdot \cdot \cdot < 2

Commented by 112358 last updated on 27/Aug/15

H e / {s h e}^{'} s s a y i n g t h a t t h e t e r m s

o f t h e s e q u e n c e a r e p e r p e t u a l l y

i n c r e a s i n g t o a l i m i t i n g v a l u e o f

2 . H e n c e, t h e s e q u e n c e i s

m o n o t o n i c a n d u p p e r b o u n d e d .

i . e a_{n} ⩽ M f o r a l l n a n d

a_{n + 1} ⩾ a_{n} f o r a l l n .

G e n e r a l l y, a l l s e q u e n c e s w h i c h

a r e m o n o t o n i c (o n l y i n c r e a s i n g

o r d e c r e a s i n g b u t n o t b o t h) a n d b o u n d e d

(u p p e r o r l o w e r [a_{n} > M f o r a n y

f i n i t e n]) a r e c o n v e r g e n t . T h i s

i s c a l l e d t h e m o n o t o n i c

s e q u e n c e t h e o r e m .

Commented by Rasheed Ahmad last updated on 27/Aug/15

Thanks to both of you!