How find this limits $\lim_{n\to\infty}\frac{a^2_{n}-n}{\ln{n}}$ 
let sequence $\{a_{n}\}$ such $a_{1}=1$,and such
  $$a_{n+1}=a_{n}+\dfrac{1}{2a_{n}}$$
find this limit
  $$\lim_{n\to\infty}\dfrac{a^2_{n}-n}{\ln{n}}$$

I think we must use Stolz lemma,But I can't it
I have prove
$$a_{n}=\sqrt{n}+o(\sqrt{(n)})$$
or mean
$$\lim_{n\to\infty}\dfrac{a_{n}}{\sqrt{n}}=1$$
 A: $${a_{n + 1}} = {a_n} + {1 \over {2{a_n}}} \to a_{n + 1}^2 = a_n^2 + {1 \over {4a_n^2}} + 1$$
Since you have proved $a_{n}=\sqrt{n}+o(\sqrt{(n)})$,
$$\mathop {\lim }\limits_{n \to \infty } {{a_n^2 - n} \over {\ln n}} = \mathop {\lim }\limits_{n \to \infty } {{a_{n + 1}^2 - a_n^2 - 1} \over {\ln {{n + 1} \over n}}} = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {4a_n^2}}} \over {{1 \over n}}} = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {4n}}} \over {{1 \over n}}} = {1 \over 4}$$
A: Ok, first: 
($a_n$) increases, since $(a_n)$ is >0. 
If ($a_n$) is bounded then it converges, which is impossible given the relation between the terms: $L = L+ \frac{1}{2L} $ => absurd
So $(a_n)$ -> +inf when n->+inf
We have : $ a_{n+1}^2 = a_n^2*( 1+\frac{1}{2a_n^2})^2 = a_n^2*(1+\frac{1}{a_n^2}+\frac{1}{4a_n^4})=a_n^2+1+\frac{1}{4a_n^2} $
First we get, by summing diverging series : $a_n^2$ = n+ o(n)
Now replacing it in the relation above:
Let $V_n = a_n^2-n$ -> $ V_{n+1} = a_{n+1}^2 -(n+1) = V_n +\frac{1}{4n(1+o(1))} $
Hence : $ V_{n+1} -V_n = \frac{1}{4n} + o(\frac{1}{n}) $ 
Summing diverging series , you get : $ V_{n+1} $ ~ $H_n$ , $H_n$ being the partial harmonic sum, and it is well known that $H_n$ ~ln(n) ~ln(n+1) ~$H_{n+1}$
So you get your limit : $\frac{V_n}{ln(n)} = \frac{a_n^2-n}{ln(n)} $ ~$\frac{1}{4}$
