The sequence $\{x_n\}$ of real numbers is defined by \[x_1=1 \quad\text{and}\quad x_{n+1}=x_n+\sqrt[3]{x_n} \quad\text{for}\quad n\geq 1.\] Show that there exist real numbers $a, b$ such that $\lim_{n \rightarrow \infty}\frac{x_n}{an^b} = 1$.
Problem
Source: Turkey IMO TST 1995 #6
Tags: limit, algebra unsolved, algebra
AYMANE
14.07.2011 19:26
a=2sqrt(2)/3 b=3/2
UK2019Project
24.11.2019 03:25
Bumpinggg
ftheftics
24.11.2019 14:11
Consider, $f(x_n )= x_{n+1}$.
so see $f(x_n)-f(x_{n-1}) >0$. So it is increasing
.now think that $ \lim_{n\to \infty} \frac{x_{n+1}}{x_n} =1 $.
So it can diverge .Now see that.
$\lim_{n\to \infty}\frac{x_n}{n}=\lim_{n\to \infty} \frac{x_{n+1}-x_n}{1}=\lim _{n\to \infty}x_n^{1/3}$
stolze Cesaro says these.....,... ....... Now look
$\frac {x_n}{n^b} = \frac {x_n^{b/3}}{x_n ^{b-1}}$ ...by stolze Cesaro again ....
So,we can say $\frac{b}{3}=b-1$.
I.e $ b=\frac{3}{2} $. and hence $a= 1$.
So,$\boxed{b=\frac{3}{2},a=1}$
QED $\square$.
Moubinool
24.11.2019 14:32
$y'.y^{-1/3}=1$