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For sequence $u_n$ satisfing : $$\begin{cases} u_1=\sqrt{2015}\\ u_{n+1}=u_n^2-2\end{cases}$$ How to prove : $$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$$

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  • $\begingroup$ The Maple command $$rsolve(\{u(1) = 2015^{1/2}, u(n+1) = u(n)^2-2\}, u) $$ produces $$2\,\cosh \left( 1/2\,{\it arccosh} \left( 1/2\,\sqrt {2015} \right) {2 }^{n} \right). $$ $\endgroup$
    – user64494
    Sep 2, 2013 at 7:43
  • $\begingroup$ After simplification with Maple $$ \left( e^{\ln \left( 1/2\,\sqrt {2015}+1/4\,\sqrt {-4+2\, \sqrt {2015}}\sqrt {4+2\,\sqrt {2015}} \right) 2^n}+1 \right)* $$ $$ e^{-1/2\,\ln \left( 1/2\,\sqrt {2015}+1/4\,\sqrt {-4+2\,\sqrt {2015}}\sqrt {4+2\,\sqrt {2015}} \right) 2^n} $$ $\endgroup$
    – user64494
    Sep 2, 2013 at 8:13
  • $\begingroup$ I'm surprised 'Maple' didn't do the further simplification that $e^{-1/2\ln(??)}=1/\sqrt{??}$. And I wouldn't say that an expression of the form $e^{\ln(\dots)}$ is completely simplified.... $\endgroup$ Sep 2, 2013 at 14:10
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    $\begingroup$ Simplifying Maple's simplification in a few naive ways gives $u_n=\sqrt{2^n\left(\frac12\sqrt{2015}+\frac14\sqrt{2011}\right)}+\dfrac1 {\sqrt{2^n\left(\frac12\sqrt{2015}+ \frac14 \sqrt{2011}\right)}}$ (I wouldn't be surprised if you could simplify this further). $\endgroup$ Sep 2, 2013 at 14:26

3 Answers 3

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The recurrence relation may be expressed as $$ {u_{n}}^2 = \frac{{u_{n+1}}^2 - 4}{{u_{n}}^2 - 4}. $$

Which allows telescopic cancellation,

$$ \begin{align*} \left ({u_{1}}{u_{2}}\cdots{u_{n}}\right )^2 &= \left (\frac{\color{#036}{{u_{2}}^{2} - 4}}{{u_{1}}^2 - 4} \right ) \left ( \frac{\color{#063}{{u_{3}}^{2} - 4}}{\color{#036}{{u_{2}}^2 - 4}} \right ) \cdots \left ( \frac{{u_{n+1}}^{2} - 4}{\color{#630}{{u_{n}}^2 - 4}} \right )\\[10pt] &= \frac{{u_{n+1}}^2 - 4}{{u_{1}}^2 - 4}. \end{align*} $$

Hence,

$$ \begin{align*} \lim_{n\to\infty} \left ( \frac{u_{n+1}}{{u_{1}}{u_{2}}\cdots{u_{n}}} \right )^2 &= \lim_{n\to\infty} {u_{n+1}}^2 \cdot \frac{{u_{1}}^2 - 4}{{u_{n+1}}^2 - 4}\\ &= \lim_{n\to\infty} \left ( {u_{1}}^2 - 4\right ) \cdot \frac{1}{1 - 4/{u_{n+1}}^2}\\ &= {u_{1}}^2 - 4 &\text{since $\lim_{n\to\infty} 1/u_{n+1} = 0$}. \end{align*} $$

Substituting $u_1 = \sqrt{2015}$ yields the final result

$$\lim_{n\to\infty} \left ( \frac{u_{n+1}}{{u_{1}}{u_{2}}\cdots{u_{n}}} \right )^2 = 2011.$$

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Here I use its properties as an additive telescoping series: $$\left(\frac{u_n^2-2}{u_1\cdots u_n}\right)^2=\frac{u_n^4}{(u_1\cdots u_n)^2}-\frac{4u_n^2}{(u_1\cdots u_n)^2}+\frac{4}{(u_1\cdots u_n)^2}$$ $$=\frac{u_n^2}{(u_1\cdots u_{n-1})^2}-\frac{4}{(u_1\cdots u_{n-1})^2}+\frac{4}{(u_1\cdots u_n)^2}$$ Substitute again: $$=\left(\frac{u_{n-1}^2}{(u_1\cdots u_{n-2})^2}-\frac{4}{(u_1\cdots u_{n-2})^2}+\frac{4}{(u_1\cdots u_{n-1})^2} \right)-\frac{4}{(u_1\cdots u_{n-1})^2}+\frac{4}{(u_1\cdots u_n)^2}$$ $$=\frac{u_{n-1}^2}{(u_1\cdots u_{n-2})^2}-\frac{4}{(u_1\cdots u_{n-2})^2}+\frac{4}{(u_1\cdots u_n)^2}$$ And again: $$=\left(\frac{u_{n-2}^2}{(u_1\cdots u_{n-3})^2}-\frac{4}{(u_1\cdots u_{n-3})^2}+\frac{4}{(u_1\cdots u_{n-2})^2} \right)-\frac{4}{(u_1\cdots u_{n-2})^2}+\frac{4}{(u_1\cdots u_n)^2}$$

$$=\frac{u_{n-2}^2}{(u_1\cdots u_{n-3})^2}-\frac{4}{(u_1\cdots u_{n-3})^2}+\frac{4}{(u_1\cdots u_n)^2}$$

Reapeat this process ad infinitum (noting that $\lim_{n \rightarrow \infty}u_n= \infty$):

$$=\frac{u_{2}^2}{(u_1)^2}-\frac{4}{(u_1)^2}+\frac{4}{(u_1\cdots u_n)^2}$$

And take the limit $n \rightarrow \infty$:

$$=\frac{(2015-2)^2}{2015}-\frac{4}{2015}+0$$ $$=\frac{2015^2-4\cdot 2015 +4}{2015}-\frac{4}{2015}$$ $$=2015-4=2011$$

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\begin{aligned} u_{n+1} = {u_{n}}^2 - 2 \\ {u_{n+1}}^2 = ({u_{n}}^2 - 2)^2 \\ {u_{n+1}}^2 = {u_{n}}^4 - 4{u_{n}}^2 + 4 \\ {u_{n+1}}^2 - 4 = {u_{n}}^2({u_{n}}^2 - 4) \\ {u_{n}}^2 = \frac{{u_{n+1}}^2 - 4}{{u_{n}}^2 - 4} \end{aligned}

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