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Let $a_{n}>1$ ,and such $$\begin{cases} a_{1}=2\\ a^2_{n+1}-a_{n+1}-a^2_{n}+1=0 \end{cases}$$ show that $$\lim_{n\to+\infty}\dfrac{a_{n}}{n}=1$$

Try $$(a_{n+1}-a_{n})(a_{n+1}+a_{n})=a_{n+1}-1>0$$ and $$a_{n+1}-a_{n}=\dfrac{a_{n+1}-1}{a_{n+1}+a_{n}}<1$$ thus we have $$a_{n}<a_{1}+(n-1)=n+1$$

$a_{n}>?$

Jack D'Aurizio
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math110
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  • you solved the question $a_n<n+1$ thus $a_n/n<1+1/n$ and $a_n>1$ given thus by sandwich theorem you have your answer – avz2611 May 19 '16 at 10:00
  • $a_{n}>1$ $\dfrac{a_{n}}{n}>\dfrac{1}{n}$ then？ – math110 May 19 '16 at 10:01
  • you have this $1<a_n<1+1/n$ now apply sandwich theorem – avz2611 May 19 '16 at 10:03
  • That is not enough as you want $ 1 < \dfrac{a_n}{n} < 1+\dfrac{1}{n}$ – DeepSea May 19 '16 at 10:04
  • @avz2611,you must be carefull – math110 May 19 '16 at 10:05
  • ah i messed up i see – avz2611 May 19 '16 at 10:06
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    Note that $(a_{n+1}-\frac12)^2=a_{n+1}^2-a_{n+1}+\frac14=a_n^2-\frac34=\frac14(4a_n^2-3)$ hence $$a_{n+1}=\frac12+\frac12\sqrt{4a_n^2-3}=a_n+\frac12-\frac32\frac1{b_n}$$ with $$b_n=2a_n+\sqrt{4a_n^2-3}.$$ If $a_n\geqslant2$, $b_n\geqslant4+\sqrt{13}\geqslant7$ hence $$a_{n+1}>a_n+\frac12-\frac32\frac17=a_n+\frac27,$$ in particular, for every $a_1\geqslant2$, we know that $a_n\geqslant2$ for every $n\geqslant1$ and that $a_n\to\infty$. Now, $b_n\to\infty$ hence $$a_{n+1}=a_n+\frac12+o(1),$$ which in turn implies that $$\lim_{n\to\infty}\frac{a_n}n=\frac12.$$ – Did May 19 '16 at 10:59

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