TITLE: Limit of $\frac{x_{n}}{n}$ if $ x_{n} = \sqrt{{n \over 2} + x_{n - 1}x_{n - 2}\,}$
QUESTION [13 upvotes]: Let real sequence  $\left\{x_{n}\right\}$ such $x_{0} = x_{1} =1$, and
  $$
x_{n} = \sqrt{{n \over 2} + x_{n - 1}x_{n - 2}\,}\,,\qquad n \geq 2\,;\qquad
\mbox{Find}\quad\lim_{n\to\infty}{x_{n} \over n} =\ {\large ?}$$

My try:
$\displaystyle{\mbox{Since}\quad
\left(~x^{2}_{n} = {n \over 2} + x_{n - 1}x_{n - 2}
\quad\Longrightarrow\quad 2x^{2}_{n} = n + 2x_{n - 1}x_{n - 2}~\right)}$
and I know this $x_{n}$ don't have simple form. I guess 
$\displaystyle{\lim_{n\to \infty}{x_{n} \over n} = {\sqrt{6} \over 6}}$. So I ask: How find this limit is $\displaystyle{\sqrt{6} \over6}$ ?. Then I can't. Thank you.  very much!

REPLY [8 votes]: Show that
$$
\frac{n}{\sqrt{6}} \leq x_n \leq \frac{n}{\sqrt{6}} + 1
$$
by induction.

To guess this inequality I performed an analysis similar to Did's.  I substituted
$$
x_n = an + b + o(1)
$$
into the recurrence and found it to be consistent when
$$
a = \frac{1}{\sqrt{6}},
$$
and
$$
b = \frac{1}{3} \sqrt{\frac{2}{3}} \approx 0.27,
$$
so it would be reasonable to expect that
$$
x_n = \frac{n}{\sqrt{6}} + \frac{1}{3} \sqrt{\frac{2}{3}} + o(1).
$$
Based on this, I guessed that the inequalities
$$
x_n \geq \frac{n}{\sqrt{6}} \qquad \text{and} \qquad x_n \leq \frac{n}{\sqrt{6}} + x_0 = \frac{n}{\sqrt{6}} + 1
$$
might hold, which I then verified by induction.
In fact, this analysis suggests the tighter inequality
$$
\frac{n}{\sqrt{6}} + \frac{1}{3} \sqrt{\frac{2}{3}} \leq x_n \leq \frac{n}{\sqrt{6}} + 1,
$$
which can also be shown by induction.