let a_(n+1) =(√(2+(√a_n ))) a_0 =(√2) find lim_(n→∞) a


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Question Number 93892 by frc2crc last updated on 15/May/20

$l e t$ $a_{n + 1} = \sqrt{2 + \sqrt{a_{n}}} a_{0} = \sqrt{2}$ $f i n d \lim_{n \to \infty} a_{n + 1}$
Commented by Tony Lin last updated on 16/May/20
$let f(x)=(√(2+(√x))) f ′(x)=(1/(4(√x)∙(√(2+(√x)))))>0 ∴f(x) is an increasing function similarily ⟨a_n ⟩ is an increasing series let a_k ≤1.84 a_(k+1) ≤(√(2+(√(1.84))))≤1.84 ⟨a_n ⟩ has upper bound ∴ lim_(n→∞) a_(n+1) exists and equals lim_(n→∞) a_n let lim_(n→∞) a_(n+1) =x x=(√(2+(√x))) x^2 =2+(√x) (x^2 −2)^2 =x x^4 −4x^2 −x+4=0 (x−1)(x^3 +x^2 −3x−4)=0 1<a_0 =(√2)(false) x^3 +x^2 −3x−4=0 let t=x+(1/3) ,x=t−(1/3) t^3 −((10)/3)t−((79)/(27))=0 let t=u+v (u+v)^3 −((10)/3)(u+v)−((79)/(27))=0 u^3 +v^3 −((79)/(27))=(((10)/3)−3uv)(u+v) suppose uv=((10)/9) →u^3 +v^3 =((79)/(27)) { ((u^3 +v^3 =((79)/(27))=−(b/a))),((u^3 v^3 =((1000)/(729))=(c/a))) :} ⇒729y^2 −2133y+1000=0 y=−((79)/(54))±((√((83)/3))/6) (one is u^3 ,another is v^3 ) t=_3 (√(−((79)/(54))+((√((83)/3))/6)))+_3 (√(−((79)/(54))−((√((83)/3))/6))) (real root one) x=t−(1/3) =_3 (√(−((79)/(54))+((√((83)/3))/6)))+_3 (√(−((79)/(54))−((√((83)/3))/6)))−(1/3) ≈1.8312 ⇒lim_(n→∞) a_(n+1) =x≈1.8312$
$l e t f (x) = \sqrt{2 + \sqrt{x}}$ $f^{'} (x) = \frac{1}{4 \sqrt{x} \cdot \sqrt{2 + \sqrt{x}}} > 0$ $∴ f (x) i s a n i n c r e a s i n g f u n c t i o n$ $s i m i l a r i l y$ $⟨ a_{n} ⟩ i s a n i n c r e a s i n g s e r i e s$ $l e t a_{k} ⩽ 1.84$ $a_{k + 1} ⩽ \sqrt{2 + \sqrt{1.84}} ⩽ 1.84$ $⟨ a_{n} ⟩ h a s u p p e r b o u n d$ $∴ \lim_{n \to \infty} a_{n + 1} e x i s t s a n d e q u a l s \lim_{n \to \infty} a_{n}$ $l e t \lim_{n \to \infty} a_{n + 1} = x$ $x = \sqrt{2 + \sqrt{x}}$ $x^{2} = 2 + \sqrt{x}$ ${(x^{2} - 2)}^{2} = x$ $x^{4} - 4 x^{2} - x + 4 = 0$ $(x - 1) (x^{3} + x^{2} - 3 x - 4) = 0$ $1 < a_{0} = \sqrt{2} (f a l s e)$ $x^{3} + x^{2} - 3 x - 4 = 0$ $l e t t = x + \frac{1}{3}, x = t - \frac{1}{3}$ $t^{3} - \frac{10}{3} t - \frac{79}{27} = 0$ $l e t t = u + v$ ${(u + v)}^{3} - \frac{10}{3} (u + v) - \frac{79}{27} = 0$ $u^{3} + v^{3} - \frac{79}{27} = (\frac{10}{3} - 3 u v) (u + v)$ $s u p p o s e u v = \frac{10}{9}$ $\to u^{3} + v^{3} = \frac{79}{27}$ ${\begin{cases} u^{3} + v^{3} = \frac{79}{27} = - \frac{b}{a} \\ u^{3} v^{3} = \frac{1000}{729} = \frac{c}{a} \end{cases}$ $\Rightarrow 729 y^{2} - 2133 y + 1000 = 0$ $y = - \frac{79}{54} \pm \frac{\sqrt{\frac{83}{3}}}{6} (o n e i s u^{3}, a n o t h e r i s v^{3})$ $t =_{3} \sqrt{- \frac{79}{54} + \frac{\sqrt{\frac{83}{3}}}{6}} +_{3} \sqrt{- \frac{79}{54} - \frac{\sqrt{\frac{83}{3}}}{6}}$ $(r e a l r o o t o n e)$ $x = t - \frac{1}{3}$ $=_{3} \sqrt{- \frac{79}{54} + \frac{\sqrt{\frac{83}{3}}}{6}} +_{3} \sqrt{- \frac{79}{54} - \frac{\sqrt{\frac{83}{3}}}{6}} - \frac{1}{3}$ $\approx 1.8312$ $\Rightarrow \lim_{n \to \infty} a_{n + 1} = x \approx 1.8312$
Commented by Tony Lin last updated on 16/May/20

$T h e o n l y u s e o f a_{0} = \sqrt{2}$ $i s t o v e r i f y t h a t \lim_{n \to \infty} a_{n + 1} s h o u l d b e$ $b i g g e r t h a n a_{0} = \sqrt{2} b e c a u s e$ $⟨ a_{n} ⟩ i s a n i n c r e a s i n g s e r i e s$ $a n d 1.8312 > \sqrt{2}$ $w h i c h i s t h e u p p e r b o u n d o f t h e s e r i e s$
Commented by frc2crc last updated on 16/May/20

$w h a t i s 1.8312 i t t e r m s o f \sqrt{2} ?$
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