2021秋 数分B1例题

求极限 \(\lim\limits_{n \to \infty} \sqrt[n]{a} (a > 0)\)

① 当 \(a = 1\) 时，显然 \(\lim\limits_{n \to \infty}\sqrt[n]{a} = 1\)

② 当 \(a > 1\) 时，设 \(\lambda_n = \sqrt[n]{a} - 1 > 0\Rightarrow a = (1 + \lambda_n)^n\)

由 \(\text{Bernoulii}\) 不等式：\(a = (1 + \lambda_n)^n \geqslant 1 + n\lambda_n \Rightarrow \lambda_n \leqslant \frac{a - 1}{n}\)

\(0 < \lambda_n \leqslant \frac{a - 1}{n}\)，根据夹逼定理可得，\(\lambda_n \to 0 \Rightarrow \sqrt[n]{a} \to 1\)

即 \(\lim\limits_{n \to \infty} \sqrt[n]{a} = 1\)

求极限 \(\lim\limits_{n \to \infty} \sqrt[n]{n}\)

根据均值不等式：

\(\begin{aligned}\sqrt[n]{n} & = \sqrt[n]{\begin{matrix}\sqrt n \times \sqrt n \times & \underbrace{1 \times \cdots \times 1}\\ & n-2个1\end{matrix}} \\ & < \frac{2\sqrt n + n - 2}{n} \\ & = \frac{2}{\sqrt n} - \frac{2}{n} + 1 \to 1\end{aligned}\)

又有 \(\sqrt[n]{n} > 1\)，根据夹逼定理，\(\lim\limits_{n \to \infty} \sqrt[n]{n} = 1\)

证明：数列 \(\{a_n\}\) 发散 \((a_n = \sin n)\)

考虑反证，假设 \(\sin n\) 收敛于 \(a\)

因为 \(\sin(n + 1) - \sin(n - 1) = 2\sin1\cos n\)，根据假设，两边同时取极限得到 \(\cos n \to 0(n \to \infty)\)

又 \(\sin n = 2 \sin \frac{n}{2} \cos \frac{n}{2}\)，同样地，两边同时取极限得到 \(\sin n \to 0(n \to \infty)\)

上述结果与 \(\sin^2n + \cos^2n = 1\) 矛盾，故 \(\sin n\) 发散。证毕。

求 \(\lim\limits_{n \to \infty}\sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots + \sqrt 2}}}\)

设 \(a_n\) 是 \(n\) 重根号的结果，则 \(a_1 = \sqrt{2},a_{n + 1}^2 = a_n + 2\)

解法一：

设 \(a_n = 2 \cos \theta_n\)，则有 \(4\cos^2 \theta_{n+1}=2\cos \theta_n + 2\)

整理可得：\(\cos2\theta_{n + 1} = \cos\theta_n\)

不妨直接令 \(\theta_{n+1} = \frac{1}{2}\theta_n\)，又有 \(\theta_1 = \frac{\pi}{4}\)

求得 \(\theta_n = \frac{\pi}{2^{n + 1}} \Rightarrow a_n = 2\cos \frac{\pi}{2^{n+1}}\)，不难求得 \(a_n \to 2\)

解法二：

\(a_1^2 = 2,a_2^2 = 2 + \sqrt 2 > a_1^2 \Rightarrow a_1 < a_2\)

一般的，对于 \(n > 1\)

\(\begin{aligned}a_{n + 1} - a_n & = \sqrt{a_n + \sqrt 2} - \sqrt{a_{n - 1} + \sqrt 2} \\ & =\frac{a_n - a_{n - 1}}{\sqrt{a_n + \sqrt 2} + \sqrt{a_{n - 1} + \sqrt 2}} \end{aligned}\)

由归纳公理可得 \(\{a_n\} \uparrow\) 又发现 \(a_1 = \sqrt 2 < 2,a_{n + 1} = \sqrt{a_n + 2} < \sqrt{2 + 2} = 2\)，归纳可得 \(\{a_n\}\) 有上界

所以 \(\{a_n\}\) 收敛，设该极限为 \(a\)，\(a_{n + 1}^2 = a_n + 2 \Rightarrow a^2 = a + 2\)，解得 \(a = 2\)（负根舍去）

\(a_n \to 2\) 即为所求

\(a_1 = 3, a_{n + 1} = \frac{1}{1 + a_n}\)，求 \(\lim\limits_{n \to \infty}a_n\)

（考虑分别分析奇数项和偶数项，发现两个子列均单调有界，进一步证明两个子列收敛于同一个值）

​ \(a_1 = 3, a_2 = \frac 1 4,a_3 = \frac 4 5,a_4 = \frac 5 9\)

待填坑

设 \(\{a_n\}\) 使得 \(\{2a_{n + 1} + a_n\}\) 收敛. 试证明 \(\{a_n\}\) 收敛

设 \(\lim\limits_{n \to \infty}(2a_{n + 1} + a_n) = a \Rightarrow \lim\limits_{n \to \infty}(2(a_{n + 1} - \frac a 3) + (a_n - \frac a 3)) = 0\)

记 \(b_n = a_n - \frac a 3 \Rightarrow \lim\limits_{n \to \infty}(2b_{n + 1} + b_n) = 0\)

要证 \(\{a_n\}\) 收敛 \(\iff\) 证 \(\{b_n\}\) 收敛 \(\iff\) 证 \(\{(-1)^nb_n\}\) 收敛

\((-1)^nb_n = {(-2)^nb_n \over 2^n}\)，且

\[{(-2)^{n + 1}b_{n + 1} - (-2)^nb_n \over 2^{n + 1} - 2^n} = (-1)^{n + 1}{2^n(2b_{n + 1} + b_n) \over 2^n} = (-1)^{n + 1}(2b_{n + 1} + b_n) \to 0 \]

根据 \(\text{Stolz}\) 定理，\(\lim\limits_{n \to \infty}(-1)^nb_n = 0 \Rightarrow \lim\limits_{n \to \infty}b_n = 0 \Rightarrow \lim\limits_{n \to \infty}a_n = \frac a 3\)

\(\{a_n\}\) 收敛，证毕

设 \(0 < x_1 < 1,x_{n + 1} = x_n(1 - x_n)(n \geqslant 2)\). 求证 \(\lim\limits_{n \to \infty}nx_n=1\)

\(x_{n + 1} = x_n(1 - x_n) < x_n \Rightarrow\) \(\{x_n\}\) 单调递减，又 \(0\) 是 \(\{x_n\}\) 的下界 \(\Rightarrow \{x_n\}\) 收敛

设 \(\lim\limits_{n \to \infty}x_n = x \Rightarrow x = x(1 - x) \Rightarrow x = 0\)

\[\lim\limits_{n \to \infty}{x_{n + 1}^{-1} - x_n^{-1} \over n + 1 - n} = \lim\limits_{n \to \infty}{x_n - x_{n + 1} \over x_nx_{n + 1}} = \lim\limits_{n \to \infty}{1 \over 1 - x_n} = 1 \]

根据 \(\text{Stolz}\) 定理，\(\lim\limits_{n \to \infty}nx_n = \lim\limits_{n \to \infty}{x_n^{-1} \over n} = 1\)，证毕