prove it : times_n ; (√(4+(√(4+(√(4+...+(√4))))) ))


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Question Number 192062 by mehdee42 last updated on 07/May/23

$p r o v e i t :$ $t i m e s_n; \sqrt{4 + \sqrt{4 + \sqrt{4 + . . . + \sqrt{4}}}} < 3$
Commented by ajfour last updated on 08/May/23

$b u t 4 > 3$ $h o w c a n t h i s b e . . .$ $r e a l l y n o u s e a r g u i n g!$
Commented by ajfour last updated on 08/May/23
sometimes it just really dont make no sense et al.
Commented by mehdee42 last updated on 08/May/23

$p a y a t t e n t i o n t o q u e s t i o n$ $t h a t i s (\sqrt{4} < 3) r i g h t, n o t w h a t y o u w r o t e (4 > 3)$
	Answered by Frix last updated on 07/May/23

	$x = \sqrt{4 + \sqrt{4 + \sqrt{4 + . . .}}} > 0$ $x = \sqrt{4 + x}$ $x^{2} - x - 4 = 0$ $x = \frac{1 + \sqrt{17}}{2}$ $\frac{1 + \sqrt{17}}{2} < 3$ $1 + \sqrt{17} < 6$ $\sqrt{17} < 5$ $17 < 25 true$
	Commented by mehdee42 last updated on 07/May/23

	$p a y a t t e n t i o n :$ $\sqrt{4 + \sqrt{4 + \sqrt{4 + . . . .}}} \neq \sqrt{4 + \sqrt{4 + \sqrt{4 + . . . \sqrt{4}}}}$ $\Rightarrow i f x = \sqrt{4 + \sqrt{4 + \sqrt{4 + . . . \sqrt{4}}}} ⇏ x^{2} = 4 + x$ $t h e e x p r e s s i o n o n t h e l e f t c o n t a i n s t h e i n f i n i t i v e o f t h e$ $s e n t e n c e . w h i l e t h e n u m b e r o f s e n t e n e s$ $i n t h r r i g h t e x p e r e s s i o n i s f i n i t .$
	Commented by Frix last updated on 07/May/23

	$x = \frac{1 + \sqrt{17}}{2}$ $\sqrt{4} < \sqrt{4 + \sqrt{4}} < \sqrt{4 + \sqrt{4 + \sqrt{4}}} < . . . < x < 3$
	Commented by mehdee42 last updated on 07/May/23

	$w h y ‘ ‘ . . . < x < 3^{″} ?!$
	Commented by deleteduser1 last updated on 07/May/23

	$L e t 4_{n} = \sqrt{4 + \sqrt{4 + \sqrt{4 + . . . + \sqrt{4}}}} (w h e r e 4 a p p e a r s n$ $t i m e s)$ $4_{n} < 4_{p} w h e n n < p . . . (i)$ $T h i s i s t r u e s i n c e 4_{n} = \sqrt{4 + z} a n d z i n c r e a s e s a s$ $n i n c r e a s e s .$ $s i n c e 4_{\infty} < 3, (i) \Rightarrow 4_{n} < 4_{\infty} < 3$ $H e n c e, w e h a v e s h o w n t h a t f o r a l l n, 4_{n} < 3 .$
	Commented by mehdee42 last updated on 07/May/23

	$s i r$ $w h y 4_{n} < 4_{\infty} < 3 ? ?$
	Commented by mehdee42 last updated on 07/May/23

	$i t c a n b e p r o v e n b y a v e r y s i m p l e$ $m a h e m a t i c a l i n d u c t i o n m e t o d$ $g o o d l u c k$
	Answered by mehdee42 last updated on 08/May/23

	$a n s w e r t o q u e s t i o n n u m b e r$ $l e t : p_{n} = n_t e r m e s \sqrt{4 + \sqrt{4 + \sqrt{4 + . . . + \sqrt{4}}}} < 3$ $p_{1} = \sqrt{4} = 2 < 3 ✓ i . s$ $p_{k} = (k_t e r m e s) \sqrt{4 + \sqrt{4 + \sqrt{4 + . . . . + \sqrt{4}}}} < 3 i . h$ $p_{k + 1} = (k + 1_t e r m e s) \sqrt{4 + \sqrt{4 + \sqrt{4 + . . . + \sqrt{4}}}} < 3 ? i . r$ $p_{k + 1}^{2} = 4 + (k_t e r m e s) \sqrt{4 + \sqrt{4 + \sqrt{4 + . . . + \sqrt{4}}}} < 4 + 3 = 7$ $\Rightarrow p_{k + 1} < \sqrt{7} < 3 ✓$
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