# Recursive Limit

1. Oct 8, 2008

### ILikePizza

1. The problem statement, all variables and given/known data
(x_n) is a sequence and x_1 > 2. From then on, x_{n+1} = x_n + 1/x_n

Prove that (x_n) is divergent.

2. Relevant equations
n/a

3. The attempt at a solution
I first tried assuming that a limit existed, but I didn't get a contradiction. (I had x = 2 + 1/x, x = (2 \pm \sqrt{5})/2, which could make sense.)

thankS!

2. Oct 8, 2008

You can immediately rule out the possibility

$$\frac{2 - \sqrt 5} 2$$

since that is a negative number and all terms in the sequence are positive.

to show it doesn't converge show

1) The sequence is monotone increasing (easy, you need to show $$a_{n+1} - a_n > 0$$ for every $$n$$
2) Show that for every $$n$$ it is true that $$x_n > n$$ - this will show that the sequence is not bounded above

these two points will show the sequence does not converge

3. Oct 8, 2008

### ILikePizza

I am aware that this will work, but I still have no idea where to start.

It is easy to show that this sequence is monotone increasing, but what about the second part? That's where I'm stuck.

Thanks

4. Oct 8, 2008

Think along these lines

You are told that $$x_1 > 2$$

Consider the function

$$f(x) = x + 1/x$$

Calculate $$f'(x)$$
* What sign does this have for $$x > 2$$?
* Is $$f$$ continuous for $$x > 2$$? Is it bounded?
* What does the sign of $$f'(x)$$ say about the behavior of $$f$$?
* Note that $$x_{n+1} = f(x_n)$$

5. Oct 8, 2008

### Dick

Ok, you know the sequence is monotone increasing and always bigger than 2. If the sequence were bounded above then you know it would have a limit. If it had a limit, the limit would have to satisfy L=L+1/L (not L=2+1/L). So?

6. Oct 8, 2008

### ILikePizza

L+1/L = x_n = L
x _n+1 > L,

a contradiction? Is that what you were fishing for?

7. Oct 8, 2008

### Dick

If x_n has a limit L, then the limit of x_(n+1) is L and the limit of x_n+1/x_n is L+1/L. Are there any solutions at all to L=L+1/L?? I don't think there are. So yes, that's a contradiction. What do you conclude from the presence of a contradiction. Which assumption must be wrong? That's what I'm fishing for. Bite.