a-n-a-n-1-a-1-c-Solve-S-lim-n-a-n-

Question Number 4597 by FilupSmith last updated on 10/Feb/16

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

a_{1} = c

Solve :

S = \lim_{n \to \infty} a_{n}

Commented by FilupSmith last updated on 10/Feb/16

Would a solution be :

for c > 0

a_{1} = c^{1 / 2}

a_{2} = {(c^{1 / 2})}^{1 / 2}

= (c^{1 / 2^{2}})

a_{3} = {(c^{1 / 2^{2}})}^{1 / 2}

= c^{1 / 2^{3}}

a_{n} = c^{\frac{1}{2^{n}}}

∴ a s n \to \infty

a_{\infty} = c^{1 / \infty} = c^{0}

a_{\infty} = 1

∴ \lim_{n \to \infty} a_{n} = 1

for c = 0

\lim_{n \to \infty} a_{n} = 0

Commented by Yozzii last updated on 10/Feb/16

T h i s a p p e a r s t o b e l e g i t i m a t e f o r i f

c = 0, a_{n} = 0^{1 / 2^{n - 1}} = 0 \forall n \in N \Rightarrow \lim_{n \to + \infty} a_{n} = \underset{n \to + \infty}{\lim 0} = 0 .

T h e t e r m a_{\infty} d o e s n o t e x i s t b y t h e w a y .

{I t}^{'} s b e t t e r t o s a y^{'} a s n \to \infty, 2^{n - 1} \to \infty

\Rightarrow \frac{1}{2^{n - 1}} \to 0 \Rightarrow c^{1 / 2^{n - 1}} \to c^{0} = 1 \Rightarrow a_{n} \to 1 .^{'}

T o f u r t h e r c o n v i n c e y o u r a u d i e n c e

o f t h e o b v i o u s p a t t e r n t h a t a_{n} = c^{1 / 2^{n - 1}},

g i v e n t h a t a_{1} = c > 0 a n d a_{n + 1} = \sqrt{a_{n}},

u s e i n d u c t i o n o n n . (I s a w t h a t y o u

w r o t e a_{1} = c^{1 / 2} w h e n t h e i n f o r m a t i o n

g a v e a_{1} = c . {I t}^{'} s r e a l l y a_{2} = \sqrt{a_{1}} = c^{1 / 2} \Rightarrow

a_{3} = \sqrt{a_{2}} = c^{1 / 2^{2}} \Rightarrow a_{4} = \sqrt{a_{3}} = c^{1 / 2^{3}} \Rightarrow \dots \Rightarrow a_{n} = c^{1 / 2^{n - 1}} (n \in N) .)

Commented by FilupSmith last updated on 11/Feb/16

Thanks . I do not know induction . I have

been meaning to learn but i {can}^{'} t

find an easy to understand explanation

Commented by Yozzii last updated on 11/Feb/16

T h e u n d e r l y i n g p r i n c i p l e o f

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

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

k + 1 i f y o u k n o w w h a t k i s, b e i n g a n

i n t e g e r . S o, g i v e n a n i n t e g e r k

y o u c a n g e t t h e n e x t i n t e g e r b y a d d i n g

1 t o k \Rightarrow k + 1 . S o, s i n c e y o u k n o w

w h a t y o u r i n i t i a l i n t e g e r i s, e . g 1,

y o u c a n f i n d a n y i n t e g e r g r e a t e r t h a n

(o r l e s s t h a n) 1 i f y o u c o n t i n u e o n

f r o m 1 t o 2, t h e n 2 t o 3, t h e n 3 t o 4,

\dots ., t h e n k t o k + 1, t h e n \dots . t o i n f i n i t y .

Y o u c a n a s s i g n a m a t h e m a t i c a l

i d e a P t h a t d e p e n d s o n n t o u s e

i n d u c t i o n t o s h o w t h a t P (n) i s t r u e

f o r a l l i n t e g e r s n s i n c e f r o m, s a y P (1),

y o u c a n d e d u c e P (k), P (k + 1), P (k + 2) \dots .

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