<s>[INST] An inequality about a sequence Let $(a_n)$ be a sequence such that  $a_0=1 , a_1=2 , a_{n+1}=a_n+\dfrac {a_{n-1}}{1+ a_{n-1}^2} , \forall n \ge1 $ , then is it true that $52 < a_{1371} < 65$ ? 
$ EDIT:-$  I am posing another question , so I'm not able to accept Oleg567 ' s very correct answer  : 
Let $(a_n)$ be a sequence such that  $a_0=1 , a_1=2 , a_{n+1}=a_n+\dfrac {a_{n-1}}{1+ (a_n-1)^2} , \forall n \ge1 $ , then is it true that $52 < a_{1371} < 65$ ?
 [/INST] First, note that next recurrent relation is true for sequence $(a_n)$:
$$
a_0=1,\\
a_{n+1} = a_n+\frac{1}{a_n}, \qquad n\ge 0.\tag{1}
$$
Yes, $a_0=1$, $a_1 = 1+\frac{1}{1}=2$ for both definitions; if $a_{n+1}$ is defined by $(1)$, then
$$
a_{n+1} = a_n+\dfrac{1}{a_n} = a_n + \dfrac{1}{a_{n-1}+\frac{1}{a_{n+1}}} = 
a_n + \dfrac{a_{n-1}}{1+a_{n-1}^2}, \qquad n\ge 1.\tag{2}
$$

Hmm, nice recurrent formula as finite continued fraction:
$$
a_{n+1} = a_n+\cfrac{1}{a_{n-1}+\cfrac{1}{a_{n-2}+\cfrac{1}{\cdots + \cfrac{1}{a_1+\cfrac{1}{a_0+1}}}}}\tag{3}
$$

Looking at $(1)$, one can show that
$$
\sqrt{2n}<a_{n-1}<\sqrt{\left(2+\frac{1}{6}\right)n}, \qquad n\ge 3.\tag{4}
$$
Proof of $(4)$ is here.

For $a_{1371}$ we get:
$$
52<52.383\approx\sqrt{2744}<a_{1371}<\sqrt{2972\frac{2}{3}}\approx 54.522<65.$$
Note:
looking at bounds $52$ and $65$, I think that there is enough to show weaker (than $(4)$) inequality: just that
$$
\sqrt{2n}<a_{n-1}<\sqrt{3n},\qquad n\ge 3.\tag{5}
$$ 
</s>