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Question Number 218208 by Marzuk last updated on 01/Apr/25

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

P r o v e o r d i s p r o v e t h a t

\lim_{n \to \infty} \frac{3^{n} m + 3^{n - 1}}{2^{⌈ \frac{n}{2} ⌉}} + \frac{3^{n - 1}}{2^{n}}

t h e l i m i t e x i s t s f o r m \in N ∖ B

w h e r e B = {n ∣ {l o g}_{2} (n) \in N}

Commented by Marzuk last updated on 01/Apr/25

T h i s q u e s t i o n i s h i g h l y r e l a t e d w i t h

C o l l a t z C o n j e c t u r e

Answered by MrGaster last updated on 05/Apr/25

= \lim_{n \to \infty} \frac{3^{n - 1} (3 m + 1)}{2^{[n / 2]}} + \frac{3^{n - 1}}{2^{n}}

= {\underset{n \to \infty}{\lim 3}}^{n - 1} (\frac{3 m + 1}{2^{[n / 2]}} + \frac{1}{2^{n}}) \forall n \in N, [\frac{n}{2}] is defined as

⌈ \frac{n}{2} ⌉ = {\begin{cases} \frac{n}{2} if n \in even \\ \frac{n + 1}{2} if n is odd \end{cases}

Case when n is even :

2^{⌈ \frac{n}{2} ⌉} = 2^{n / 2}

\Rightarrow \frac{3^{n - 1} (3 m + 1)}{2^{n / 2}} = \frac{3^{n} (3 m + 1)}{3 \cdot 2^{n / 2}} = \frac{(3 m + 1)}{3} \cdot {(\frac{9}{2})}^{n / 2}

n \in odd :

2^{⌈ \frac{n}{2} ⌉} = 2^{(n + 1) / 2}

\Rightarrow \frac{3^{n - 1} (3 m + 1)}{2^{(n + 1) / 2}} = \frac{3^{n} (3 m + 1)}{3 \cdot 2^{(n + 1) / 3}} = \frac{(3 m + 1)}{3 \sqrt{2}} \cdot {(\frac{9}{2})}^{n / 2}

Second term :

\frac{3^{n - 1}}{2^{n}} = \frac{1}{3} \cdot {(\frac{3}{4})}^{n}

Main term analysis (order comparison) :

{(\frac{9}{2})}^{n / 2} ≫ {(\frac{3}{4})}^{n} ∵ {(\frac{9}{2})}^{1 / 2} = \frac{3}{\sqrt{2}} > \frac{3}{4}

\Rightarrow \lim_{n \to \infty} (\frac{3 m + 1}{3} \cdot {(\frac{9}{2})}^{n / 2} + \frac{1}{3} \cdot {(\frac{3}{4})}^{n}) = + \infty

\Rightarrow ∄ \lim_{―}

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