设 $p , q$ 是一元二次方程 ${x^2} + 2ax - 1 = 0\left( {a > 0} \right)$ 的两个根，其中 $p > 0$．令 ${y_1} = p - q$，${y_{n + 1}} = y_n^2 - 2$，$n=1,2,\cdots$．证明：$\lim\limits_{n \to + \infty } \left( {\dfrac{1}{{{y_1}}} + \dfrac{1}{{{y_1}{y_2}}} + \cdot \cdot \cdot + \dfrac{1}{{{y_1}{y_2} \cdot \cdot \cdot {y_n}}}} \right) = p$．
【难度】
【出处】
无
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【答案】
略
【解析】
根据题意，$q=-\dfrac 1p$ 且 $0<p<1$．由 $y_n=p^{2^n}+\dfrac{1}{p^{2^n}}$，于是\[\begin{split} \dfrac{1}{y_1y_2\cdots y_n}&=\dfrac{1}{\left(p+\dfrac 1p\right)\left(p^2+\dfrac{1}{p^2}\right)\cdots \left(p^{2^{n-1}}+\dfrac{1}{p^{2^{n-1}}}\right)}\\
&=\dfrac{p^{2^n-1}}{\left(p^2+1\right)\left(p^4+1\right)\cdots \left(p^{2^n}+1\right)}\\
&=\dfrac 1p\cdot\left[\dfrac{1}{\left(p^2+1\right)\left(p^4+1\right)\cdots \left(p^{2^{n-1}}+1\right)}-\dfrac{1}{\left(p^2+1\right)\left(p^4+1\right)\cdots \left(p^{2^n}+1\right)}\right],\end{split} \]从而$$\dfrac{1}{y_1}+\dfrac{1}{y_1y_2}+\cdots +\dfrac{1}{y_1y_2\cdots y_n}=\dfrac 1p\left[1-\dfrac{1}{\left(p^2+1\right)\left(p^4+1\right)\cdots \left(p^{2^n}+1\right)}\right]=\dfrac 1p\left(1-\dfrac{p^2-1}{p^{2^{n+1}}-1}\right),$$从而原式的值为 $p$．
&=\dfrac{p^{2^n-1}}{\left(p^2+1\right)\left(p^4+1\right)\cdots \left(p^{2^n}+1\right)}\\
&=\dfrac 1p\cdot\left[\dfrac{1}{\left(p^2+1\right)\left(p^4+1\right)\cdots \left(p^{2^{n-1}}+1\right)}-\dfrac{1}{\left(p^2+1\right)\left(p^4+1\right)\cdots \left(p^{2^n}+1\right)}\right],\end{split} \]从而$$\dfrac{1}{y_1}+\dfrac{1}{y_1y_2}+\cdots +\dfrac{1}{y_1y_2\cdots y_n}=\dfrac 1p\left[1-\dfrac{1}{\left(p^2+1\right)\left(p^4+1\right)\cdots \left(p^{2^n}+1\right)}\right]=\dfrac 1p\left(1-\dfrac{p^2-1}{p^{2^{n+1}}-1}\right),$$从而原式的值为 $p$．
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