This question is really important Prove or disprove that lim_(n→∞) ((3^n m+3^(n−1) )/2^(⌈(n/2)⌉) ) + (3^(n−1) /2^n ) the limit exists for m ∈ N \B where B = {n ∣ log


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Question Number 218208 by Marzuk last updated on 01/Apr/25
$This question is really important Prove or disprove that lim_(n→∞) ((3^n m+3^(n−1) )/2^(⌈(n/2)⌉) ) + (3^(n−1) /2^n ) the limit exists for m ∈ N \B where B = {n ∣ log_2 (n) ∈ N }$
$T h i s q u e s t i o n i s r e a l l y i m p o r t a n t$ $P r o v e o r d i s p r o v e t h a t$ $\lim_{n \to \infty} \frac{3^{n} m + 3^{n - 1}}{2^{⌈ \frac{n}{2} ⌉}} + \frac{3^{n - 1}}{2^{n}}$ $t h e l i m i t e x i s t s f o r m \in N ∖ B$ $w h e r e B = {n ∣ {l o g}_{2} (n) \in N}$
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
	$=lim_(n→∞) ((3^(n−1) (3m+1))/2^([n/2]) )+(3^(n−1) /2^n ) =lim_(n→∞) 3^(n−1) (((3m+1)/2^([n/2]) )+(1/2^n )) ∀n∈N,[(n/2)]is defined as ⌈(n/2)⌉= { (((n/2) if n∈even)),((((n+1)/2) if n is odd)) :} Case when n is even: 2^(⌈(n/2)⌉) =2^(n/2) ⇒((3^(n−1) (3m+1))/2^(n/2) )=((3^n (3m+1))/(3∙2^(n/2) ))=(((3m+1))/3)∙((9/2))^(n/2) n∈odd: 2^(⌈(n/2)⌉) =2^((n+1)/2) ⇒((3^(n−1) (3m+1))/2^((n+1)/2) )=((3^n (3m+1))/(3∙2^((n+1)/3) ))=(((3m+1))/(3(√2)))∙((9/2))^(n/2) Second term: (3^(n−1) /2^n )=(1/3)∙((3/4))^n Main term analysis (order comparison): ((9/2))^(n/2) ≫((3/4))^n ∵((9/2))^(1/2) =(3/( (√2)))>(3/4) ⇒lim_(n→∞) (((3m+1)/3)∙((9/2))^(n/2) +(1/3)∙((3/4))^n )=+∞ ⇒∄lim$
	$= \lim_{n \to \infty} \frac{3^{n - 1} (3 m + 1)}{2^{[n / 2]}} + \frac{3^{n - 1}}{2^{n}}$ $= {\underset{n \to \infty}{\lim 3}}^{n - 1} (\frac{3 m + 1}{2^{[n / 2]}} + \frac{1}{2^{n}}) \forall n \in N, [\frac{n}{2}] is defined as$ $⌈ \frac{n}{2} ⌉ = {\begin{cases} \frac{n}{2} if n \in even \\ \frac{n + 1}{2} if n is odd \end{cases}$ $Case when n is even :$ $2^{⌈ \frac{n}{2} ⌉} = 2^{n / 2}$ $\Rightarrow \frac{3^{n - 1} (3 m + 1)}{2^{n / 2}} = \frac{3^{n} (3 m + 1)}{3 \cdot 2^{n / 2}} = \frac{(3 m + 1)}{3} \cdot {(\frac{9}{2})}^{n / 2}$ $n \in odd :$ $2^{⌈ \frac{n}{2} ⌉} = 2^{(n + 1) / 2}$ $\Rightarrow \frac{3^{n - 1} (3 m + 1)}{2^{(n + 1) / 2}} = \frac{3^{n} (3 m + 1)}{3 \cdot 2^{(n + 1) / 3}} = \frac{(3 m + 1)}{3 \sqrt{2}} \cdot {(\frac{9}{2})}^{n / 2}$ $Second term :$ $\frac{3^{n - 1}}{2^{n}} = \frac{1}{3} \cdot {(\frac{3}{4})}^{n}$ $Main term analysis (order comparison) :$ ${(\frac{9}{2})}^{n / 2} ≫ {(\frac{3}{4})}^{n} ∵ {(\frac{9}{2})}^{1 / 2} = \frac{3}{\sqrt{2}} > \frac{3}{4}$ $\Rightarrow \lim_{n \to \infty} (\frac{3 m + 1}{3} \cdot {(\frac{9}{2})}^{n / 2} + \frac{1}{3} \cdot {(\frac{3}{4})}^{n}) = + \infty$ $\Rightarrow ∄ \lim_{―}$
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