浙江大学 2023年强基第13题

$\sqrt{2}-1$ ，解析：由递推公式取倒数知，$\displaystyle \frac{1}{a_{n+1}}=(1-\sqrt{2})^{n+1}+\frac{\sqrt{2}+1}{a_{n}}$ ，即 $\displaystyle \frac{1}{(\sqrt{2}+1)^{n+1} a_{n+1}}= (2 \sqrt{2}-3)^{n+1}+\frac{1}{(\sqrt{2}+1)^{n} a_{n}}$ ，故 $\displaystyle \frac{1}{(\sqrt{2}+1)^{n} a_{n}}=\frac{1}{(\sqrt{2}+1)^{1} a_{1}}+\sum_{i=2}^{n}(2 \sqrt{2}-3)^{i}=2(\sqrt{2}-1)+ \frac{(\sqrt{2}-1)^{3}}{2 \sqrt{2}}\left(1-(2 \sqrt{2}-3)^{n-1}\right)$ ，因此 $\displaystyle a_{n}=\frac{2 \sqrt{2}}{(1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}}$ ．计算 $$ \begin{gathered} \lim _{n \rightarrow+\infty} \sqrt[n]{a_{n}}=\lim _{n \rightarrow+\infty} \frac{(2 \sqrt{2})^{\frac{1}{n}}}{\left((1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}\right)^{\frac{1}{n}}}=\lim _{n \rightarrow+\infty} \frac{1}{(1+\sqrt{2})\left(1-(2 \sqrt{2}-3)^{n+1}\right)^{\frac{1}{n}}} \\ =\frac{1}{1+\sqrt{2}}=\sqrt{2}-1 \end{gathered} $$