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sequence has a limit

Vali

Member
Dec 29, 2018
48
I have the sequence from the picture and I have to demonstrate that this sequence has a limit.
I always get stuck at this kind of exercises.How to approach an exercise like this?
 

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Wilmer

In Memoriam
Mar 19, 2012
376

Vali

Member
Dec 29, 2018
48
Sorry for my bad English, if that's what you mean..
 

Wilmer

In Memoriam
Mar 19, 2012
376
Sorry for my bad English, if that's what you mean..
No. Just a joke...
"walking on tip-toe" means "very carefully"!
 

Olinguito

Well-known member
Apr 22, 2018
251
I have the sequence from the picture and I have to demonstrate that this sequence has a limit.
I always get stuck at this kind of exercises.How to approach an exercise like this?
Hi Vali .

Unless I’ve missed something, I make it that the sequence does not converge.

Observe that
$$x_n^2=\left(x_{n-1}+\frac1{x_{n-1}}\right)^2=x_{n-1}^2+2+\frac1{x_{n-1}^2}>x_{n-1}^2+2.$$

Thus:
$$x_n^2>x_{n-1}^2+2>x_{n-2}^2+4>\cdots>x_0^2+2n>2n.$$

Thus $x_n>\sqrt{2n}\to\infty$ as $n\to\infty$ $\implies$ $x_n$ also $\to\infty$ as $n\to\infty$.
 
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Vali

Member
Dec 29, 2018
48
Thank you for the answer.
I tried to calculate x_(n+1) - x_(n) which is 1/x_(n) which is positive so x_(n) increases.Then, I assumed that x_(n)>0 and demonstrate that x_(n+1) > 0 but x_(n+1)=x_(n)+1/x_(n) so x_(n+1)>0 and now I don't know how to continue.
 

DavidCampen

Member
Apr 4, 2014
64
Olinguito gave you the complete answer.

As an aside, not for pre-calculus, I notice that the difference between any 2 consecutive terms is a convergent of [a; a, a. ...] where a = x_0 . Is this correct?
 
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Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
371
I would start by assuming that a limit exists and determining what that limit must be. Calling the limit "X" and taking the limit on both sides of \(\displaystyle x_{n+1}= x+\frac{1}{x_n} \) we get \(\displaystyle A= A+ \frac{1}{A}\). That reduces to \(\displaystyle \frac{1}{A}= 0\) which is not true for any A! As Olinguito said, this sequence does not converge.
 
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