Problem 28: Sequence

by henderson, Jun 23, 2016, 2:07 PM

$\bf\color{red}Problem \ 28$ For $n=1,2,...,$ let ${\{u_n\} }$ be a sequence defined by
$u_1=1, u_2=2, u_{n+2}=3u_{n+1}-u_n.$ Prove that
$u_{n+2}+u_n \geq 2+\frac{u^2_{n+1}}{u_n}.$ $\bf\color{red}My \ solution$ At first, let's check the cases $n=1$ and $n=2.$ Both are trivial. Then,
$u_{n+2}+u_n \geq 2+\frac{u^2_{n+1}}{u_n}$ $\iff$
$3u_{n+1} \geq 2+\frac{u^2_{n+1}}{u_n}$ $\iff$
$2u_{n+1} \geq 2+\frac{u_{n+1}(u_{n+1}-u_n)}{u_n}.$ Now, recall the fact

$u_{n+1}=2u_n+u_{n-1}+...+u_1$ can be easily obtained using induction.
For $n=1,$ we have $u_2=2u_1.$ $\checkmark$
Now, let's assume that
$u_n=2u_{n-1}+u_{n-2}+...+u_1$ is true and try to prove of it for $n+1.$
According to $u_{n+2}=3u_{n+1}-u_n,$ we have $u_{n+1}=3u_n-u_{n-1}.$ So, we have both
$u_n=2u_{n-1}+u_{n-2}+...+u_1$ and
$u_{n+1}=3u_n-u_{n-1}.$ Summing up these two equations, the induction is completed. $\square$

$u_{n+1}=2u_n+u_{n-1}+...+u_1,$ which can be proved easily by induction.
So, $2u_{n+1} \geq 2+\frac{u_{n+1}(u_{n+1}-u_n)}{u_n}$ $\iff$
$2u_{n+1} \geq 2+\frac{u_{n+1}(u_n+u_{n-1}+...+u_1)}{u_n}$ $\iff$
$u_{n+1} \geq 2+\frac{u_{n+1}(u_{n-1}+...+u_1)}{u_n}$ $\iff$
$\frac{u_{n+1}}{u_n} \left(u_n-(u_{n-1}+...+u_1)\right) \geq 2$ $\iff$
$u_{n+1}\cdot u_{n-1} \geq 2u_n.$ The last inequality is obvious, because we have $u_{n+1}> u_n$ $($ because $u_{n+1}=2u_n+u_{n-1}+...+u_1$ $)$ and $u_{n-1}\geq 2$ for $n>2.$ $($ remember we've checked the cases $n=1$ and $n=2$ $)$

This post has been edited 2 times. Last edited by henderson, Sep 11, 2016, 5:54 PM

Sequence

No Comments
(Post Your Comment)

Comment

0 Comments

"Do not worry too much about your difficulties in mathematics, I can assure you that mine are still greater." - Albert Einstein

henderson

Math Path

Problem 28: Sequence

by henderson, Jun 23, 2016, 2:07 PM

0 Comments