2021秋 数分B1例题

求极限 (limlimits_{n o infty} sqrt[n]{a} (a > 0))

① 当 (a = 1) 时,显然 (limlimits_{n o infty}sqrt[n]{a} = 1)

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

( ext{Bernoulii}) 不等式:(a = (1 + lambda_n)^n geqslant 1 + nlambda_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 o 0 Rightarrow sqrt[n]{a} o 1)

(limlimits_{n o infty} sqrt[n]{a} = 1)

求极限 (limlimits_{n o infty} sqrt[n]{n})

根据均值不等式:

(egin{aligned}sqrt[n]{n} & = sqrt[n]{egin{matrix}sqrt n imes sqrt n imes & underbrace{1 imes cdots imes 1}\ & n-2个1end{matrix}} \ & < frac{2sqrt n + n - 2}{n} \ & = frac{2}{sqrt n} - frac{2}{n} + 1 o 1end{aligned})

又有 (sqrt[n]{n} > 1),根据夹逼定理,(limlimits_{n o 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) = 2sin1cos n),根据假设,两边同时取极限得到 (cos n o 0(n o infty))

(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 o 0(n o infty))

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

(limlimits_{n o infty}sqrt{2 + sqrt{2 + sqrt{2 + cdots + sqrt 2}}})

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

解法一:

(a_n = 2 cos heta_n),则有 (4cos^2 heta_{n+1}=2cos heta_n + 2)

整理可得:(cos2 heta_{n + 1} = cos heta_n)

不妨直接令 ( heta_{n+1} = frac{1}{2} heta_n),又有 ( heta_1 = frac{pi}{4})

求得 ( heta_n = frac{pi}{2^{n + 1}} Rightarrow a_n = 2cos frac{pi}{2^{n+1}}),不难求得 (a_n o 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)

(egin{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 o 2) 即为所求

(a_1 = 3, a_{n + 1} = frac{1}{1 + a_n}),求 (limlimits_{n o 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

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

(limlimits_{n o infty}(2a_{n + 1} + a_n) = a Rightarrow limlimits_{n o infty}(2(a_{n + 1} - frac a 3) + (a_n - frac a 3)) = 0)

(b_n = a_n - frac a 3 Rightarrow limlimits_{n o infty}(2b_{n + 1} + b_n) = 0)

要证 ({a_n}) 收敛 (iff)({b_n}) 收敛 (iff)({(-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) o 0 ]

根据 ( ext{Stolz}) 定理,(limlimits_{n o infty}(-1)^nb_n = 0 Rightarrow limlimits_{n o infty}b_n = 0 Rightarrow limlimits_{n o infty}a_n = frac a 3)

({a_n}) 收敛,证毕

(0 < x_1 < 1,x_{n + 1} = x_n(1 - x_n)(n geqslant 2)). 求证 (limlimits_{n o infty}nx_n=1)

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

(limlimits_{n o infty}x_n = x Rightarrow x = x(1 - x) Rightarrow x = 0)

[limlimits_{n o infty}{x_{n + 1}^{-1} - x_n^{-1} over n + 1 - n} = limlimits_{n o infty}{x_n - x_{n + 1} over x_nx_{n + 1}} = limlimits_{n o infty}{1 over 1 - x_n} = 1 ]

根据 ( ext{Stolz}) 定理,(limlimits_{n o infty}nx_n = limlimits_{n o infty}{x_n^{-1} over n} = 1),证毕

原文地址:https://www.cnblogs.com/zsbzsb/p/15310002.html