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Let $y_0\geqslant 2$, $y_n=y_{n-1}^2-2$, $n\in\mathbb{N}_+$, set $\displaystyle S_n=\sum_{k=0}^{n}\frac{1}{y_0\cdots y_k}$, how to prove $$\lim_{n\to\infty}S_n=\frac{y_0-\sqrt{y_0^2-4}}{2}.$$ Do you have some idea?

I can only get that $$y_n=\begin{cases}y_0, & n=0 \\ \left(\frac{y_0^2-2+\sqrt{y_0^4-4y_0^2}}{2}\right)^{2^{n-1}}+\left(\frac{y_0^2-2+\sqrt{y_0^4-4y_0^2}}{2}\right)^{-2^{n-1}}, & n\ge1\end{cases}.$$

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Since, $y_n=y_{n-1}^2-2$, we have $y_n^2=(y_{n-1}^2-2)^2=y_{n-1}^2(y_{n-1}^2-4)+4$.

or, $y_n^2-4=y_{n-1}^2(y_{n-1}^2-4)=y_{n-1}^2y_{n-2}^2(y_{n-2}^2-4)=\ldots=(\prod\limits_{i=0}^{n-1}y_i^2)(y_0^2-4)$.

Thus, $\dfrac{y_n^2-4}{\prod\limits_{i=1}^{n-1}y_i^2}=y_0^2-4$ for $n\ge1$.

$\implies \dfrac{\sqrt{y_n^2-4}}{\prod\limits_{i=0}^{n-1}y_i}=\sqrt{y_0^2-4}$

Thus, $\lim\limits_{n \to \infty} \dfrac{\sqrt{y_n^2-4}}{\prod\limits_{i=0}^{n-1}y_i} = \lim\limits_{n \to \infty} \dfrac{y_n}{\prod\limits_{i=0}^{n-1}y_i} = \sqrt{y_0^2-4}$

Now, $\displaystyle \frac{1}{y_0\cdots y_k} = \frac{1}{2}\bigg(\frac{y_k}{y_0\cdots y_{k-1}}-\frac{y_{k+1}}{y_0\cdots y_k}\bigg)$, for each $k\ge 1$.

Thus, it undergoes telescopic cancellation to yield,

$\displaystyle \lim\limits_{n \to \infty} S_n = \frac{1}{y_0} + \frac{y_1}{2y_0} - \frac{\sqrt{y_0^2-4}}{2} = \frac{y_0^2-\sqrt{y_0^2-4}}{2}$.

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  • $\begingroup$ @DaoyiPeng Welcome ,, :) $\endgroup$ – r9m Apr 1 '14 at 12:26

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