# Given the sequence $x_{n+1}=x_n + \frac{2}{x_n}$ and $x_0 = 1$, find $\lim\limits_{n \to \infty} \frac{x_n}{\sqrt{n}}$

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.

• Hint: apply Stolz-Cesaro to $\frac{x_n^2}{n}$ with the fact $x_{n+1}^2 = x_n^2 + 4 + \frac{4}{x_n^2}$. – achille hui Nov 10 '19 at 23:40

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 $$x_n/\sqrt{n} \to 2$$.