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.


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

$F i n d t h e l i m i t o f t h i s s e q u e n c e .$ $\sqrt{2}, \sqrt{2 \sqrt{2}}, \sqrt{2 \sqrt{2 \sqrt{2}}}, \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}, . . .$ $S h o w t h a t t h e s u m o f t h e$ $t e r m s o f t h i s i n f i n i t e s e q u e n c e$ $d o e s n o t c o n v e r g e .$
Commented by Rasheed Ahmad last updated on 26/Aug/15

$L e t \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 . . .}}}} = x$ $S q u a r i n g b o t h s i d e s,$ $2 \sqrt{2 \sqrt{2 \sqrt{2 . . .}}} = x^{2}$ $2 . x = x^{2}$ $x = 2$ $T h e l i m i t o f t h e s e q u e n c e = 2$
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

	$Thanks to both of you!$
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