Proving Divergence of (x_n): A Homework Challenge

In summary, the sequence (x_n) is divergent because it is both monotone increasing and unbounded, and a contradiction arises when assuming it has a limit. This means that one of the initial assumptions, such as the presence of a limit, must be incorrect.
  • #1
ILikePizza
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Homework Statement


(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.


Homework Equations


n/a


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!
 
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  • #2
You can immediately rule out the possibility

[tex]
\frac{2 - \sqrt 5} 2
[/tex]

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 [tex] a_{n+1} - a_n > 0 [/tex] for every [tex] n [/tex]
2) Show that for every [tex] n [/tex] it is true that [tex] x_n > n [/tex] - this will show that the sequence is not bounded above

these two points will show the sequence does not converge
 
  • #3
Thanks for the reply.

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
Think along these lines

You are told that [tex] x_1 > 2 [/tex]

Consider the function

[tex]
f(x) = x + 1/x
[/tex]

Calculate [tex] f'(x) [/tex]
* What sign does this have for [tex] x > 2 [/tex]?
* Is [tex] f [/tex] continuous for [tex] x > 2 [/tex]? Is it bounded?
* What does the sign of [tex] f'(x) [/tex] say about the behavior of [tex] f [/tex]?
* Note that [tex] x_{n+1} = f(x_n) [/tex]
 
  • #5
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
L+1/L = x_n = L
x _n+1 > L,

a contradiction? Is that what you were fishing for?
 
  • #7
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.
 

FAQ: Proving Divergence of (x_n): A Homework Challenge

What is "Proving Divergence of (x_n)"?

"Proving Divergence of (x_n)" is a mathematical concept that involves determining whether a sequence of numbers, denoted as (x_n), diverges or approaches infinity as n (the number of terms in the sequence) increases.

How can I prove the divergence of (x_n)?

There are several methods for proving the divergence of (x_n), such as the comparison test, the ratio test, and the root test. Each method involves evaluating the behavior of the sequence and determining whether it approaches infinity or not.

What is the purpose of proving the divergence of (x_n)?

Proving the divergence of (x_n) is important in understanding the behavior of a sequence and its limit. It can also be used to determine the convergence or divergence of more complex series and to solve various mathematical problems.

Can the divergence of (x_n) be proven for all sequences?

No, the divergence of (x_n) can only be proven for certain types of sequences. For example, it cannot be proven for sequences that approach a finite limit or oscillate between two or more values.

Are there any real-world applications of proving the divergence of (x_n)?

Yes, proving the divergence of (x_n) has applications in fields such as physics, engineering, and economics. It can be used to model the behavior of various systems and predict their future outcomes.

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