to prove that the sequence is unbounded.

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May 6th 2011, 12:37 AM
abhishekkgp

to prove that the sequence is unbounded.

QUESTION: show that the sequence http://latex.codecogs.com/png.latex?(x_n) defined recursively as http://latex.codecogs.com/png.latex?... \, unbounded.

MY SOLUTION:
Let the sequence be bounded above. It then has a least upper bound say http://latex.codecogs.com/png.latex?L.

now,
http://latex.codecogs.com/png.latex?...-Lx_n+1 \leq 0

hence:
http://latex.codecogs.com/png.latex?...-4})/2) \leq 0

It can be computed that http://latex.codecogs.com/png.latex?...>4 \, so\, L>4

so we have:

http://latex.codecogs.com/png.latex?...2+\sqrt{L^2-4}
This is true for all http://latex.codecogs.com/png.latex?...\mathbb{Z}^{+} so we have:

http://latex.codecogs.com/png.latex?... 5L^2 \leq -16 Contradiction.

Is this correct?

also if you have a good method to do it then please post it.
May 6th 2011, 01:09 AM
FernandoRevilla

Quote:

Originally Posted by abhishekkgp

also if you have a good method to do it then please post it.

Prove by induction (very easy):

$x_n>0\;\forall n=1,2,\ldots$

The sequence is increasing:

$x_{n+1}-x_n=\dfrac{1}{x_n}>0$

If the sequence is upper bounded, denote l > 0 its limit. Then,

$x_{n+1}=x_n+\dfrac{1}{x_n}\Rightarrow l=l+\dfrac{1}{l}\Rightarrow 0=\dfrac{1}{l}\neq 0$

Contradiction.
May 6th 2011, 01:43 AM
abhishekkgp

Quote:

Originally Posted by FernandoRevilla

Prove by induction (very easy):

$x_n>0\;\forall n=1,2,\ldots$

The sequence is increasing:

$x_{n+1}-x_n=\dfrac{1}{x_n}>0$

If the sequence is upper bounded, denote l > 0 its limit. Then,

$x_{n+1}=x_n+\dfrac{1}{x_n}\Rightarrow l=l+\dfrac{1}{l}\Rightarrow 0=\dfrac{1}{l}\neq 0$

Contradiction.

in my book the concept of limit has not so far been introduced. so i am not sure how you got http://latex.codecogs.com/png.latex?l=l+1/l. i believe the way i did it was correct too. thanks anyway.
May 6th 2011, 03:18 AM
FernandoRevilla

Quote:

Originally Posted by abhishekkgp

now,
http://latex.codecogs.com/png.latex?...-Lx_n+1 \leq 0

hence:
http://latex.codecogs.com/png.latex?...-4})/2) \leq 0

You need to prove that the terms of the sequence are positive in order to obtain an equivalent inequality.

Quote:

so we have: http://latex.codecogs.com/png.latex?...2+\sqrt{L^2-4}

It should be:

$2x_n \leq L+\sqrt{L^2-4}$

Quote:

This is true for all http://latex.codecogs.com/png.latex?...\mathbb{Z}^{+} so we have:

http://latex.codecogs.com/png.latex?... 5L^2 \leq -16 Contradiction.

That is true if you choose L as the least upper bound of the sequence.
May 6th 2011, 04:02 AM
abhishekkgp

Quote:

Originally Posted by FernandoRevilla

You need to prove that the terms of the sequence are positive in order to obtain an equivalent inequality.

It should be:

$2x_n \leq L+\sqrt{L^2-4}$

oh yes right!

That is true if you choose L as the least upper bound of the sequence.
yes i had mentioned that L is the least upper bound of the sequence.

thanks for that.
May 6th 2011, 04:15 AM
FernandoRevilla

Quote:

Originally Posted by abhishekkgp

yes i had mentioned that L is the least upper bound of the sequence.

Right. :)