Math Analysis Homework - Week 14

Class 1

T1

讨论下列无穷乘积的敛散性: (2) $\prod_{n=1}^{\infty} \sqrt[n]{1 + \frac{1}{n}};$

敛散性等价于

\begin{gathered} \sum_{n=1}^\infty \dfrac{\ln(1+\dfrac1n)}n \\ \because \lim_{n \to \infty} \dfrac{\dfrac{\ln(1+\dfrac{1}{n} )}{n} }{\dfrac{1}{n^2} } =1,\sum _{n = 1} ^{\infty} \dfrac{1}{n^2} <\infty \\ \xRightarrow{\text{ Comparison Test }} \text{convergent} \end{gathered}

T2

讨论下列无穷乘积的敛散性: (4) $\prod_{n=2}^{\infty} \left(\frac{n^2 - 1}{n^2 + 1}\right)^p \ (p \in \mathbb{R});$

\begin{gathered} \ln(\prod_{n=2}^\infty (\dfrac{n^2-1}{n^2+1} )^p) \\ =p\sum _{n = 2} ^{\infty} \ln(1+\dfrac{2}{n^2+1} ) \\ \because \lim_{n \to \infty} \dfrac{\ln(1+\dfrac{2}{n^2+1} )}{\dfrac{1}{n^2} } =1,\sum _{n = 1} ^{\infty} \dfrac{1}{n^2} <\infty \\ \xRightarrow{\text{ Comparison Test }} \text{convergent} \end{gathered}

T3

设数列 $\{a_n\}$ , 其中 $a_n = \begin{cases} -\frac{1}{\sqrt{k}}, & n = 2k - 1, \\ \frac{1}{\sqrt{k}} + \frac{1}{k} + \frac{1}{k\sqrt{k}}, & n = 2k. \end{cases}$ 证明: $\sum_{n=1}^{\infty} a_n$ 与 $\sum_{n=1}^{\infty} a_n^2$ 都发散, 但是 $\prod_{n=1}^{\infty} (1+a_n)$ 收敛.

\begin{gathered} \sum _{n = 1} ^{\infty} a_n^2 \\ =\sum _{n = 1} ^{\infty} (\dfrac{1}{n} +(\dfrac{1}{\sqrt n} +\dfrac{1}{n} +\dfrac{1}{n^{\frac23}} ^2)) >\sum _{n = 1} ^{\infty} \dfrac{1}{n} \\ =\infty \end{gathered}

\begin{gathered} \sum _{n = 1} ^{\infty} a_n=\sum _{n = 1} ^{\infty} -\dfrac{1}{\sqrt{n}} +\dfrac{1}{\sqrt n} +\dfrac{1}{n} +\dfrac{1}{n\sqrt n} \\ >\sum _{n = 1} ^{\infty} \dfrac{1}{n} =\infty \\ \end{gathered}

\begin{gathered} \ln(\prod _{n = 1} ^{\infty} (1+a_n)) \\ \ln(\prod _{n = 1} ^{\infty} (1-\dfrac{1}{\sqrt n} )(1+\dfrac{1}{\sqrt{n}} +\dfrac{1}{n} +\dfrac{1}{n\sqrt n} )) \\ =\ln(\prod _{n=1}^{\infty} (1-\dfrac{1}{n^2} )) \\ =\sum _{n = 1} ^{\infty} \ln(1-\dfrac{1}{n^2} ) \\ \sim -\sum _{n = 1} ^{\infty} \dfrac{1}{n^2} >-\infty \end{gathered}

Class 2

T1

1. 求下列数列的上、下极限: (1) $\frac{n+1}{n}(1 + (-1)^{n+1})$ ;

偶数列到 $0$ ,奇数列到 $2$ ,且覆盖了所有元素.

\begin{gathered} \limsup a_n=2,\liminf a_n=0 \end{gathered}

T2

1. 求下列数列的上、下极限: (2) $\sin \frac{n\pi}{2} + n \cos \frac{n\pi}{2}$ .

取 $n=4k$ 得到 $\limsup a_n=+\infty$ .

取 $n=-4k$ 得 $\liminf a_n=-\infty$ .

T3

3. 设 $x_n > 0, y_n > 0$ , 证明: $\displaystyle \liminf_{n\to\infty} x_n \cdot \liminf_{n\to\infty} y_n \leqslant \liminf_{n\to\infty} x_ny_n \leqslant \limsup_{n\to\infty} x_n \cdot \liminf_{n\to\infty} y_n$ .

$a_n=\inf_{k>n} x_k,b_n=\inf_{k>n} y_k$ .

则 $\liminf x_n \liminf y_n=\lim a_n \lim b_n=\lim a_nb_n$ .

而 $c_n=\inf_{k>n} x_ky_k\ge a_nb_n$ :假设 $c_n<a_nb_n$ ,取 $\epsilon<\dfrac{a_nb_n-c_n}2$ ,则能取到 $x_ky_k<c_n+\epsilon$ ,但显然 $a_n\le x_k,b_n\le y_k$ ,于是 $c_n+\epsilon>a_nb_n,c_n<a_nb_n$ ,与 $\epsilon<a_nb_n-c_n$ 矛盾.

于是 $c_n=\inf_{k>n} x_ky_k,\lim_{n \to \infty} c_n<\lim_{n \to \infty} a_nb_n$ ,左边得证.

右边,设 $d_n=\sup_{k>n} x_k$ .则 $\forall \epsilon,\exists k,b_n>y_k-\epsilon$ 同时 $c_n<x_ky_k$ , $d_n>x_k$ ,则 $c_n<x_ky_k<d_n(b_n+\epsilon)$ 对任意 $\epsilon$ ,于是 $c_n\le d_nb_n$ ,于是右边得证.

Class 3

T1

5. 设 $x_1 > 0, x_{n+1} = 1 + \frac{1}{x_n} (n=1, 2, \cdots)$ , 证明:

(1) $1 \leqslant \liminf_{n\to\infty} x_n \leqslant \limsup_{n\to\infty} x_n \leqslant 2$ ;

(2) $\lim_{n\to\infty} x_n$ 存在, 并求其极限值.

\begin{gathered} \text{let } S=\limsup_{n \to \infty} x_n,I=\liminf_{n \to \infty} x_n \\ \begin{cases} S=1+\dfrac{1}{I} \\ I=1+\dfrac{1}{S} \\ S\ge I>0 \end{cases} \implies S=I=\dfrac{1+\sqrt 5}{2} \in [1,2] \end{gathered}

得证.

T2

6. 设 $a_n > 0$ , 证明: $\limsup_{n\to\infty} n \left(\frac{1+a_{n+1}}{a_n} - 1\right) \geqslant 1$ .

反证,假设 $\limsup_{n \to \infty} n(\dfrac{1+a_{n+1}}{a_n} -1)<1$ ,则存在 $N$ 使得 $n>N$ 时:

\begin{gathered} n(\dfrac{1+a_{n+1}}{a_n} -1)<1 \\ \implies \dfrac{a_{n+1}}{n+1} <\dfrac{a_n}{n} -\dfrac{1}{n+1} \\ \implies \dfrac{a_{n}}{n}<\dfrac{a_{N+1}}{N+1}-\sum _{i = N+2} ^{n} \dfrac{1}{i} \\ \end{gathered}

调和级数发散,所以存在 $n$ 使得右边为负,左边 $\dfrac{a_n}n<0$ ,与 $a_n>0$ 矛盾

T3

7. 设数列 $\{x_n\}$ 满足: $x_n + x_m - 1 \leqslant x_{n+m} \leqslant x_n + x_m + 1$ , 证明: $\{\frac{x_n}{n}\}$ 收敛.

取 $n=pk+r$ :

\begin{gathered} x_n=x_{pk+r}\in [kx_p+x_r-k,kx_p+x_r+k] \\ \dfrac{x_n}{n} \in [\dfrac{k(x_p-1)}{pk+r}+\dfrac{x_r}{pk+r} ,\dfrac{k(x_p+1)}{pk+r}+\dfrac{x_r}{pk+r} ] \\ \limsup_{n \to \infty} \dfrac{x_n}{n} \in [\dfrac{x_p-1}{p} ,\dfrac{x_p+1}{p}] \\ \limsup_{n \to \infty} \dfrac{x_n}{n} \in [\liminf_{p \to \infty} \dfrac{x_p}{p} ,\liminf_{np \to \infty} \dfrac{x_p}{p} ] \end{gathered}

即上下极限相等,收敛.

T4

8. 设正数列 $\{a_n\}$ . 证明: $\limsup_{n\to\infty} \sqrt[n]{a_n} \leqslant 1$ 的充分必要条件是: 对任意的 $l > 1$ , 成立 $\lim_{n\to\infty} \frac{a_n}{l^n} = 0$ .

首先前推后:反证,假设 $\exists l>1,\lim_{n \to \infty} \dfrac{a_n}{l^n} =a\ne 0$ .则 $\exists N,\forall n>N,a_n>\epsilon l^n,\limsup_{n \to \infty} \sqrt[ n ]{ a_n } \ge l>1$ .

后推前,反证,假设 $\limsup_{n \to \infty} \sqrt[ n ]{ a_n }=A >1$ ,则存在子列 $a_{p_n}$ 使得 $\sqrt[n]{a_{p_n}}>B,B\in (1,A)$ ,于是

\begin{gathered} \text{let } l=B,\limsup_{n \to \infty} \dfrac{a_n}{l^n} \ge \limsup_{n \to \infty} \dfrac{a_{p_n}}{l^n} >\limsup_{n \to \infty} \dfrac{B^n}{l^n} =1 \end{gathered}

矛盾,得证.