MHB Find the Limit of a Sequence: Tips & Techniques

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The discussion focuses on determining the limit of a specific sequence, with participants sharing their approaches and findings. One contributor demonstrates that the sequence does not converge by showing that each term increases without bound, specifically noting that \(x_n > \sqrt{2n}\) leads to \(x_n \to \infty\) as \(n \to \infty\). Another participant attempts to analyze the sequence by examining the difference between consecutive terms but struggles to progress further. The consensus is that the sequence diverges, as shown through various mathematical arguments. Ultimately, the conclusion reached is that the sequence does not have a limit.
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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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Vali said:
How to approach an exercise like this?
...by walking on tip-toe :)
 
Sorry for my bad English, if that's what you mean..
 
Vali said:
Sorry for my bad English, if that's what you mean..
No. Just a joke...
"walking on tip-toe" means "very carefully"!
 
Vali said:
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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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.
 
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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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 $$x_{n+1}= x+\frac{1}{x_n} $$ we get [math]A= A+ \frac{1}{A}[/math]. That reduces to [math]\frac{1}{A}= 0[/math] which is not true for any A! As Olinguito said, this sequence does not converge.
 
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