两道有意思的数列极限

1

设 $a_1=\sqrt{\dfrac{1}{2}},\; a_n=\sqrt{\dfrac{1+a_{n-1}}{2}}$，求 $\displaystyle\lim_{n\to\infty}a_1a_2\cdots a_n$.

【Sol】: $a_!=\cos\dfrac{\pi}{4},\;a_2=\sqrt{\dfrac{1+\cos\frac{\pi}{4}}{2}}=\sqrt{\cos^2\dfrac{\pi}{4\cdot2}}=\cos\dfrac{\pi}{4\cdot2},\;a_3\sqrt{\dfrac{1+a_3}{2}}=\cos\dfrac{\pi}{4\cdot2^2},\;\cdots,\;a_n=\cos\dfrac{\pi}{4\cdot2^{n-1}}$

因此
$\begin{aligned} &\lim_{n\to\infty}a_1a_2\cdots a_n\\
=&\lim_{n\to\infty}\cos\dfrac{\pi}{4}\cdot\cos\dfrac{\pi}{4\cdot2}\cdot\cdots\cos\dfrac{\pi}{4\cdot2^{n-1}}\\
=&\lim_{n\to\infty}\dfrac{\cos\dfrac{\pi}{4}\cdot\cos\dfrac{\pi}{4\cdot2}\cdot\cdots\cos\dfrac{\pi}{4\cdot2^{n-1}}(2\sin\dfrac{\pi}{4\cdot2^{n-1}})}{2\sin\dfrac{\pi}{4\cdot2^{n-1}}}\\
=&\lim_{n\to\infty}\dfrac{\cos\dfrac{\pi}{4}\cdot\cos\dfrac{\pi}{4\cdot2}\cdot\cdots\cos\dfrac{\pi}{4\cdot2^{n-2}}\cdot\sin\dfrac{\pi}{4\cdot2^{n-2}}}{2\sin\dfrac{\pi}{4\cdot2^{n-1}}}\\
=&\lim_{n\to\infty}\dfrac{\sin\dfrac{\pi}{2}}{2^{n}\sin\dfrac{\pi}{4\cdot2^{n-1}}}\\
=&\lim_{n\to\infty}\dfrac{1}{2^n\cdot\dfrac{\pi}{2^{n+1}}}\\
=&\dfrac{2}{\pi} \end{aligned}$

这里将 $\sqrt{\dfrac{1}{2}}=\cos\dfrac{\pi}{4}$ 如果是第一次见的话可能比较难以想到.

2

设 $a_1=1,\;a_n=a_{n-1}+\dfrac{1}{a_{n-1}}(n\geqslant2)$，求

$$ \lim_{n\to\infty}\dfrac{\sqrt{2n}(a_n-\sqrt{2n})}{\ln n} $$

【Sol】:
(1) 由于 $a_1>0$ 因此 ${a_n}$ 为单调递增序列，且趋向于无穷大.
$a_{n+1}^2=a_{n}^2+\dfrac{1}{a_{n}^2}+2$

(2) $\displaystyle\lim_{n\to\infty}\dfrac{a_{n}^2}{2n}\xlongequal{\text{Stolz}}\lim_{n\to\infty}\dfrac{a_{n+1}^2-a_{n}^2}{2}=\lim_{n\to\infty}\dfrac{\dfrac{1}{a_n}^2+2}{2}=1\Rightarrow \lim_{n\to\infty}\dfrac{a_n}{\sqrt{2n}}=1$.

(3)
$\begin{aligned} &\lim_{n\to\infty}\dfrac{\sqrt{2n}(a_n-\sqrt{2n})}{\ln n}\\
=&\lim_{n\to\infty}\dfrac{\sqrt{2n}(a_{n}^2-2n)}{\ln n(a_n+\sqrt{2n})}\\
=&\lim_{n\to\infty}\dfrac{\sqrt{2n}}{a_n+\sqrt{2n}}\cdot\lim_{n\to\infty}\dfrac{a_n^2-2n}{\ln n}\\
\xlongequal{\text{Stolz}}&\dfrac{1}{2}\lim_{n\to\infty}\dfrac{a_{n+1}^2-a_{n}^2-2}{\ln(n+1)-\ln n}\\
=&\dfrac{1}{2}\lim_{n\to\infty}\dfrac{\dfrac{1}{a_n^2}}{\dfrac{1}{n}}\\
=&\dfrac{1}{2}\lim_{n\to\infty}\dfrac{n}{a_n^2}=\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{4} \end{aligned}$

这题的解答来自一位群友，MSE上有老哥写的更长的渐进展开.