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I am given the sequence $(x_n)_{n \ge 0}$ with the recurrence relation

$$x_{n+1}=x_n + \dfrac{2}{x_n}$$

and $x_0=1.$ I have to find the following limit:

$$\lim\limits_{n \to \infty} \dfrac{x_n}{\sqrt{n}}$$

In the first part of the problem, I had to find the limit of the sequence itself. This is what I did:

Let $$\lim\limits_{n \to \infty} x_n = a$$

My recurrence relation is:

$$x_{n+1}=x_n + \dfrac{2}{x_n}$$

If I take the limit of both sides I get:

$$\hspace{2cm} a=a+\dfrac{2}{a} \hspace{2cm} -|a$$

$$\hspace{2cm} \dfrac{2}{a}=0 \hspace{4cm}$$

Which means:

$$a=\pm \infty \hspace{1.5cm}$$

Now, since the terms of the sequence are clearly positive,

$$a= + \infty$$

Which means:

$$\hskip{6cm} \lim\limits_{n \to \infty}x_n = \infty \hskip{6cm} (1)$$

Great. I think I got this right. If not, please correct me. Now, the second part of the problem asks me to find:

$$\lim\limits_{n \to \infty} \dfrac{x_n}{\sqrt{n}}$$

And I don't know how to approach this. I can see that since we have $(1)$, this is a limit of the type $\dfrac{\infty}{\infty}$, so L'Hospital comes to mind. However I don't see any way of applying it.

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    $\begingroup$ Hint: apply Stolz-Cesaro to $\frac{x_n^2}{n}$ with the fact $x_{n+1}^2 = x_n^2 + 4 + \frac{4}{x_n^2}$. $\endgroup$ – achille hui Nov 10 '19 at 23:40
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Your first part is not rigorous. When you write let $\lim_{n\to\infty} x_n=a$ you have made an implicit assumption that the limit exists but unless this assumption is justified the approach can't be considered rigorous. Also one can't write $a=\pm\infty $.

First of all note that the sequence is consisting of positive terms and the sequence is increasing. Therefore it either tends to a limit or to $\infty $. If it tends to a limit $L$ then we must have $L\geq x_0=1$ and taking limit of the recurrence relation we get $L=L+(2/L)$ which can't hold. Thus $x_n\to\infty $.

For the next part use the hint given in comments. We have $$x_{n+1}^2=x_n^2+4+\frac{4}{x_n^2}$$ so that $$x_{n+1}^2-x_n^2\to 4$$ By Cesaro-Stolz we have $x_n^2/n\to 4$ and hence $x_n/\sqrt{n} \to 2$.

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  • $\begingroup$ You said that the method I used for finding the first limit is not rigorous and I agree. I am not that great in sequences/limits. Can you please show me a step by step method of finding that limit that is considered rigorous enough? $\endgroup$ – user592938 Nov 11 '19 at 22:57
  • $\begingroup$ @user1502: if your method is based on standard definitions and theorems given in your books and there is no logical flaw in your steps then the method is rigorous. Unfortunately the teaching of calculus in most schools is devoid of rigor. So that most students really don't know the definitions and theorems. $\endgroup$ – Paramanand Singh Nov 12 '19 at 1:46

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