prove-that-2-3-3-1993-1993-lt-2-

Question Number 10539 by malwaan last updated on 17/Feb/17

p r o v e t h a t

\sqrt{2 + \overset{3}{} \sqrt{3 + \dots + \overset{1993}{} \sqrt{1993}}} < 2

Answered by mrW1 last updated on 17/Feb/17

f u n c t i o n y = {(x + a)}^{\frac{1}{x}} w i t h a > 1 i s

f o r x > 0 s t r i c t l y d e c r e a s i n g, t h a t

m e a n s

^{3} \sqrt{3 + a} < \sqrt{2 + a}

^{4} \sqrt{4 + a} < \sqrt{2 + a}

\dots \dots

^{1993} \sqrt{1993 + a} < \sqrt{2 + a}

\dots \dots

^{n} \sqrt{n + a} < \sqrt{2 + a}

\dots \dots

A = \sqrt{2 + \overset{3}{} \sqrt{3 + \dots + \overset{1993}{} \sqrt{1993}}} (f i n i t e)

< \sqrt{2 + \overset{3}{} \sqrt{3 + \dots + \overset{1993}{} \sqrt{1993 +^{1994} \sqrt{1994 + \dots .}}}} (i n f i n i t e)

< \sqrt{2 + \sqrt{2 + \dots + \sqrt{2 + \sqrt{2 + \dots}}}} = x

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

2 + x = x^{2}

x^{2} - x - 2 = 0

(x - 2) (x + 1) = 0

s i n c e x > 0,

\Rightarrow x = 2

∴ A = \sqrt{2 + \overset{3}{} \sqrt{3 + \dots + \overset{1993}{} \sqrt{1993}}} < x = 2

Commented by FilupS last updated on 18/Feb/17

Another way of writing this is (exact same proof) :

^{n} \sqrt{a} <^{m} \sqrt{a} for n > m

let a = 3 +^{4} \sqrt{4 +^{5} \sqrt{5 + \dots +^{1993} \sqrt{1993}}}

A = \sqrt{2 + \overset{3}{} \sqrt{3 + \dots + \overset{1993}{} \sqrt{1993}}}

A = \sqrt{2 + \overset{3}{} \sqrt{a}}

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

x^{2} = 2 + x

x^{2} - x - 2 = 0

x = 2 (from previois post)

let b = 2 +^{2} \sqrt{2 + \dots}

x =^{2} \sqrt{2 +^{2} \sqrt{b}}

b > a

\sqrt{2 + \sqrt{b}} > \sqrt{2 +^{3} \sqrt{a}}

b^{3} > a^{2}

∴ x > A

∴ A < 2

∴ \sqrt{2 + \overset{3}{} \sqrt{3 + \dots + \overset{1993}{} \sqrt{1993}}} < 2

Commented by malwaan last updated on 20/Feb/17

t h a n k y o u s o m u c h

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