Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1977 Canadian MO Problems/Problem 6"

(Created page with "== Problem == Let <math>0<u<1</math> and define <cmath>u_1=1+u\quad ,\quad u_2=\frac{1}{u_1}+u\quad \ldots\quad u_{n+1}=\frac{1}{u_n}+u\quad ,\quad n\ge 1</cmath> Show that ...")
 
(→‎Solution)
 
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
 +
Prove by induction that <math>\forall i>1</math> <math>1<u_i<\frac{1}{1-u}</math>
 +
 +
<math>u_1=1+k</math> that <math>1<u_1</math> and <math>1+u<\frac{1}{1-u} \to 1-u^2<1 \to -u^2<0</math>.
 +
 +
By induction if <math>u_i<\frac{1}{1-u}</math> then <math>u_{i+1}=\frac{1}{u_i}+u>(1-u)+u=1 \to u_{i+1}>1</math> and if <math>u_i>1</math> then <math>u_i>1>1-\frac{u^2}{1-u+u^2}=\frac{1-u}{1-u+u^2}</math> (<math>\frac{u^2}{1-u+u^2}>0</math> because <math>u^2>0</math> and <math>1-u+u^2>1-u>0</math> because <math>1>u</math>).
 +
 +
So <math>u_{i+1}=\frac{1}{u_i}+u</math> and so <math>u_{i+1}<\frac{1-u+u^2}{1-u}+u=\frac{1}{1-u} \to u_{i+1}<\frac{1}{1-u}</math>
 +
 +
So <math>1<u_{i+1}<\frac{1}{1-u}</math>, and in particular <math>u_{i+1}>1 \forall</math>  <math>i\geq 1</math>

Latest revision as of 08:18, 13 April 2025

Problem

Let $0<u<1$ and define \[u_1=1+u\quad ,\quad u_2=\frac{1}{u_1}+u\quad \ldots\quad u_{n+1}=\frac{1}{u_n}+u\quad ,\quad n\ge 1\] Show that $u_n>1$ for all values of $n=1,2,3\ldots$.

Solution

Prove by induction that $\forall i>1$ $1<u_i<\frac{1}{1-u}$

$u_1=1+k$ that $1<u_1$ and $1+u<\frac{1}{1-u} \to 1-u^2<1 \to -u^2<0$.

By induction if $u_i<\frac{1}{1-u}$ then $u_{i+1}=\frac{1}{u_i}+u>(1-u)+u=1 \to u_{i+1}>1$ and if $u_i>1$ then $u_i>1>1-\frac{u^2}{1-u+u^2}=\frac{1-u}{1-u+u^2}$ ($\frac{u^2}{1-u+u^2}>0$ because $u^2>0$ and $1-u+u^2>1-u>0$ because $1>u$).

So $u_{i+1}=\frac{1}{u_i}+u$ and so $u_{i+1}<\frac{1-u+u^2}{1-u}+u=\frac{1}{1-u} \to u_{i+1}<\frac{1}{1-u}$

So $1<u_{i+1}<\frac{1}{1-u}$, and in particular $u_{i+1}>1 \forall$ $i\geq 1$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1977_Canadian_MO_Problems/Problem_6&oldid=246593"