已知数列 $\{a_n\}$ 满足$$a_1=\frac{1}{2}, a_{n+1}=\frac{a_n}{(1-\sqrt{2})^{n+1}a_n+\sqrt{2}+1} (n=1,2,\ldots).$$试求 $\lim_{n\to +\infty}\sqrt[n]{a_n}$．
【难度】
【出处】
全国高中数学联赛模拟试题(21)
【标注】
【答案】
略
【解析】
考虑更一般的数列 $\{a_n\}$：$$a_1=\frac{1}{p+q}, a_{n+1}=\frac{a_n}{p^{n+1}a_n+q} (n=1,2,\ldots).$$则$$\frac{1}{a_{n+1}}=\frac{p^{n+1}a_n+q}{a_n}=p^{n+1}+\frac{q}{a_n},$$所以$$\frac{1}{a_{n+1}}-\frac{q}{a_n}=p^{n+1}.$$因此，$\{\frac{1}{a_{n+1}}-\frac{q}{a_n}\}$ 是公比为 $p$ 的等比数列．进而$$\begin{aligned}&\frac{1}{a_{n+1}}-\frac{q}{a_n}=p\left(\frac{1}{a_n}-\frac{q}{a_{n-1}}\right)\\
&\Rightarrow \frac{1}{a_{n+1}}-\frac{p}{a_n}=\frac{q}{a_n}-\frac{pq}{a_{n-1}}=q\left(\frac{1}{a_n}-\frac{p}{a_{n-1}}\right).\\
\end{aligned}$$从而，$\left\{\frac{1}{a_{n+1}}-\frac{p}{a_n}\right\}$ 是公比为 $q$ 的等比数列．故$$\frac{1}{a_{n+1}}-\frac{p}{a_n}=\left(\frac{1}{a_2}-\frac{p}{a_1}\right)q^{n-1}=\left(p^2+\frac{q}{a_1}-\frac{p}{a_1}\right)q^{n-1}
=(p^2+(q-p)(p+q))q^{n-1}=q^{n+1}.$$由式 ①，② 消去 $\frac{1}{a_{n+1}}$，得$$\frac{1}{a_n}=\frac{p^{n+1}-q^{n+1}}{p-q}\Rightarrow a_n=\frac{p-q}{p^{n+1}-q^{n+1}}.$$令 $p=1-\sqrt{2}, q=1+\sqrt{2}$，得$$a_n=\frac{2\sqrt{2}}{(1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}}.$$因为$$\begin{aligned}\frac{1}{2}(1+\sqrt{2})^{n+1}&=(1+\sqrt{2})^{n+1}-\frac{1}{2}(1+\sqrt{2})^{n+1}\\
&<(1+\sqrt{2})^{n+1}-\frac{1}{2}<(1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}\\
&<(1+\sqrt{2})^{n+1}+\frac{1}{2}<(1+\sqrt{2})^{n+1}+\frac{1}{2}(1+\sqrt{2})^{n+1}\\
&=\frac{3}{2}(1+\sqrt{2})^{n+1},\\
\end{aligned}$$所以$$\begin{aligned}
&\frac{1}{4\sqrt{2}}(1+\sqrt{2})^{n+1}<\frac{1}{a_n}<\frac{3}{4\sqrt{2}}(1+\sqrt{2})^{n+1}\\
&\Rightarrow (1+\sqrt{2})\left(\frac{1+\sqrt{2}}{4\sqrt{2}}\right)^{\frac{1}{n}}<\frac{1}{\sqrt[n]{a_n}}
<(1+\sqrt{2})\left(\frac{3(1+\sqrt{2})}{4\sqrt{2}}\right)^{\frac{1}{n}}\\
\end{aligned}$$故$$\lim_{n\to \infty}\frac{1}{\sqrt{a_n}}=1+\sqrt{2}\Rightarrow \lim_{n\to \infty} \sqrt[n]{a_n}=\sqrt{2}-1.$$
&\Rightarrow \frac{1}{a_{n+1}}-\frac{p}{a_n}=\frac{q}{a_n}-\frac{pq}{a_{n-1}}=q\left(\frac{1}{a_n}-\frac{p}{a_{n-1}}\right).\\
\end{aligned}$$从而，$\left\{\frac{1}{a_{n+1}}-\frac{p}{a_n}\right\}$ 是公比为 $q$ 的等比数列．故$$\frac{1}{a_{n+1}}-\frac{p}{a_n}=\left(\frac{1}{a_2}-\frac{p}{a_1}\right)q^{n-1}=\left(p^2+\frac{q}{a_1}-\frac{p}{a_1}\right)q^{n-1}
=(p^2+(q-p)(p+q))q^{n-1}=q^{n+1}.$$由式 ①，② 消去 $\frac{1}{a_{n+1}}$，得$$\frac{1}{a_n}=\frac{p^{n+1}-q^{n+1}}{p-q}\Rightarrow a_n=\frac{p-q}{p^{n+1}-q^{n+1}}.$$令 $p=1-\sqrt{2}, q=1+\sqrt{2}$，得$$a_n=\frac{2\sqrt{2}}{(1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}}.$$因为$$\begin{aligned}\frac{1}{2}(1+\sqrt{2})^{n+1}&=(1+\sqrt{2})^{n+1}-\frac{1}{2}(1+\sqrt{2})^{n+1}\\
&<(1+\sqrt{2})^{n+1}-\frac{1}{2}<(1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}\\
&<(1+\sqrt{2})^{n+1}+\frac{1}{2}<(1+\sqrt{2})^{n+1}+\frac{1}{2}(1+\sqrt{2})^{n+1}\\
&=\frac{3}{2}(1+\sqrt{2})^{n+1},\\
\end{aligned}$$所以$$\begin{aligned}
&\frac{1}{4\sqrt{2}}(1+\sqrt{2})^{n+1}<\frac{1}{a_n}<\frac{3}{4\sqrt{2}}(1+\sqrt{2})^{n+1}\\
&\Rightarrow (1+\sqrt{2})\left(\frac{1+\sqrt{2}}{4\sqrt{2}}\right)^{\frac{1}{n}}<\frac{1}{\sqrt[n]{a_n}}
<(1+\sqrt{2})\left(\frac{3(1+\sqrt{2})}{4\sqrt{2}}\right)^{\frac{1}{n}}\\
\end{aligned}$$故$$\lim_{n\to \infty}\frac{1}{\sqrt{a_n}}=1+\sqrt{2}\Rightarrow \lim_{n\to \infty} \sqrt[n]{a_n}=\sqrt{2}-1.$$
答案
解析
备注
