Prove-that-lim-n-n-n-0-Where-1-5-2-x-is-the-floor-function-

Question Number 4911 by FilupSmith last updated on 21/Mar/16

Prove that :

\lim_{n \to \infty} (\emptyset^{n} - ⌊ ϕ^{n} ⌋) = 0

Where : ϕ = \frac{1 + \sqrt{5}}{2}

⌊ x ⌋ is the floor function

Commented by Yozzii last updated on 21/Mar/16

L e t f (x) = a^{x} w h e r e a > 1, x \in R^{+} .

\Rightarrow f^{^{'}} (x) = a^{x} l n a .

a > 1 \Rightarrow l n a > 0 a n d a > 1 \Rightarrow a^{x} > 1 > 0 f o r x \in R^{+} .

S o, f^{^{'}} (x) > 0 f o r a l l x \in R^{+} . H e n c e,

f i s p e r p e t u a l l y i n c r e a s i n g a s x i n c r e a s e s .

T r e a t i n g x a s a n n a t u r a l n u m b e r n,

f (n) = a^{n} i s i n c r e a s i n g f o r a l l n \in N .

N o w, ϕ \approx 1.618033989 > 1 .

\Rightarrow ϕ^{n} > 1 \Rightarrow f (n) > 1 . \Rightarrow ⌊ f (n) ⌋ ⩾ 1 > 0 f o r

a l l n \in N . S o, \lim_{n \to \infty} ⌊ f (n) ⌋ \neq 0 .

Commented by FilupSmith last updated on 21/Mar/16

s o r r y i m a d e a h o r r i b l e t y p o o n m y q u e s t i o n

Answered by Algebro last updated on 21/Mar/16

\emptyset \approx 1.618

\lim_{n \to \infty} (⌊ 1 . 6^{n} ⌋) = \infty

because

\lim_{n \to \infty} (1 . 6^{n}) = \infty

I am srry bro : /

peace Algebro

Facebook Tweet Pin