# Thread: to prove that the sequence is unbounded.

1. ## to prove that the sequence is unbounded.

QUESTION: show that the sequence $(x_n)$ defined recursively as $x_1=1,\,x_{n+1}=x_n+\frac{1}{x_n} \, is \, unbounded.$

MY SOLUTION:
Let the sequence be bounded above. It then has a least upper bound say $L$.

now,
$L \geq x_{n+1}=x_n+\frac{1}{x_n} \Rightarrow {x_n}^2-Lx_n+1 \leq 0$

hence:
$(x_n-(L/2- \sqrt{L^2-4})/2)(x- (L/2 + \sqrt{L^2-4})/2) \leq 0$

It can be computed that $x_{10}>4 \, so\, L>4$

so we have:

$2x_n \leq L/2+\sqrt{L^2-4}$
This is true for all $n \in \mathbb{Z}^{+}$ so we have:

$2L \leq L/2 + \sqrt{L^2-4} \Rightarrow 5L^2 \leq -16$ Contradiction.

Is this correct?

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

2. Originally Posted by abhishekkgp
also if you have a good method to do it then please post it.

Prove by induction (very easy):

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

The sequence is increasing:

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

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

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

3. Originally Posted by FernandoRevilla
Prove by induction (very easy):

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

The sequence is increasing:

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

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

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

in my book the concept of limit has not so far been introduced. so i am not sure how you got $l=l+1/l$. i believe the way i did it was correct too. thanks anyway.

4. Originally Posted by abhishekkgp
now,
$L \geq x_{n+1}=x_n+\frac{1}{x_n} \Rightarrow {x_n}^2-Lx_n+1 \leq 0$

hence:
$(x_n-(L/2- \sqrt{L^2-4})/2)(x- (L/2 + \sqrt{L^2-4})/2) \leq 0$

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

so we have: $2x_n \leq L/2+\sqrt{L^2-4}$

It should be:

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

This is true for all $n \in \mathbb{Z}^{+}$ so we have:

$2L \leq L/2 + \sqrt{L^2-4} \Rightarrow 5L^2 \leq -16$ Contradiction.

That is true if you choose L as the least upper bound of the sequence.

5. 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:

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

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.

6. Originally Posted by abhishekkgp
yes i had mentioned that L is the least upper bound of the sequence.

Right.

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