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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 $$\lim_{n\to\infty}\dfrac{a_{n}}{\sqrt{n}}=1$$

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$${a_{n + 1}} = {a_n} + {1 \over {2{a_n}}} \to a_{n + 1}^2 = a_n^2 + {1 \over {4a_n^2}} + 1$$

Since you have proved $a_{n}=\sqrt{n}+o(\sqrt{(n)})$, $$\mathop {\lim }\limits_{n \to \infty } {{a_n^2 - n} \over {\ln n}} = \mathop {\lim }\limits_{n \to \infty } {{a_{n + 1}^2 - a_n^2 - 1} \over {\ln {{n + 1} \over n}}} = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {4a_n^2}}} \over {{1 \over n}}} = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {4n}}} \over {{1 \over n}}} = {1 \over 4}$$

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  • $\begingroup$ I don't get how you can pass from your first step to your second, using an= $\sqrt{n} + o(\sqrt{n})$ $\endgroup$ – mvggz Oct 21 '14 at 13:11
  • $\begingroup$ 1st equation:By applying Stolz lemma as you said;$\\$2nd equation:$a_{n}\sim \sqrt n$ $\endgroup$ – Shine Mic Oct 21 '14 at 13:12
  • $\begingroup$ Ah ok, but I don't know this lemma sorry. You've mistaken me for the OP :) $\endgroup$ – mvggz Oct 21 '14 at 13:14
  • $\begingroup$ @mvggz Refer to imomath.com/index.php?options=686, You'll find both L’Hopital’s theorem and Cesaro-Stolz’s theorem $\endgroup$ – Shine Mic Oct 21 '14 at 13:22
  • $\begingroup$ Yeah I found it thanks, both lemmas are completely unused in Classe Prepa in France (that's where I learned), although I have the feeling that everybody on this site know them very well :) $\endgroup$ – mvggz Oct 21 '14 at 13:25
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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)} $ ~$\frac{1}{4}$

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