Proving $(a_n)$ diverges For  a given sequence $(a_n)$, where $a_1>0$ and for each natural number $n\ge 1$ :
$$a_{n+1}= \dfrac{a_n^2 + 1}{a_n}$$
Prove that a sequence $(a_n)$ diverges.
Proof: 
(objectionable evidence)
Lower barrier 0, prove. 
$a_n>0$   $\forall n\in N$. $N$ is natural number and $N>1$.
$n=1$ 
$a_{1+1}=\frac{a_1\cdot a_1+1}{a_1}=a_1+\frac{1}{a_1} >0$
$a_n>0$ $\Rightarrow a_{n+1}>0$ : 
$a_{n+1}=a_n+\frac{1}{a_n}>0$, because $a_n > 0$
Monotonics:
$a_n\geq a_{n+1}$
$a_n-a_{n+1}=a_n-a_n-\frac{1}{a_n} < 0$, because $\frac{1}{a_n}>0$ 
Sequence is growing.
lim $a_{n+1} = A$
lim $a_{n+1} =$ lim $(a_n+\frac{1}{a_n}) =$ lim $a_n +$ lim $\frac{1}{a_n} = A + \frac{1}{A} = \frac{A\cdot A+1}{A}$
....
Is this proof correct? Where are mistakes? The correct end of proof? 

$a_{n+1} = a_n + \dfrac{1}{a_n}\implies a_{n+1}^2 = a_n^2 + 2+\dfrac{1}{a_n^2}\implies a_{n+1}^2 - a_n^2 > 2, \forall n \ge 1 \implies a_n^2 = (a_n^2- a_{n-1}^2)+(a_{n-1}^2-a_{n-2}^2)+(a_{n-2}^2-a_{n-3}^2)+\cdots+(a_2^2-a_1^2)+a_1^2> 2(n-1)+a_1^2 > 2(n-1)\implies a_n > \sqrt{2(n-1)} \implies a_n \to \infty$. 
