0922数分作业

P26

2(2)

$$|\frac{3n^2+n}{2n^2-1}-\frac{3}{2}|= |\frac{6n^2+2n-6n^2+3}{4n^2-2}|= |\frac{2n+3}{4n^2-2}|\\ \because n\ge1\therefore |\frac{2n+3}{4n^2-2}|\le |\frac{2n+3n}{4n^2-2n^2}|=\frac{5}{2n}\\ \therefore对\forall\epsilon>0,令N=\lceil\frac{2}{5\epsilon}\rceil,当n>N,|\frac{3n^2+n}{2n^2-1}-\frac{3}{2}|\le\frac{5}{2n}<\frac{5}{2N}\le\epsilon.\\ \therefore\lim_{n\rightarrow\infty}\frac{3n^2+n}{2n^2-1}=\frac{3}{2}$$

2(5)

$$令h=a-1>0,\\ n\ge 2时,a^n=(1+h)^n=1+nh+\frac{n(n-1)}{2}h^2+\dots+h^n>\frac{n(n-1)}{2}h^2\\ \therefore 0<\frac{n}{a^n}<\frac{n}{\frac{n(n-1)}{2}h^2}=\frac{2}{(n-1)h^2}\\ \therefore对\forall\epsilon>0,令N=\lceil1+\frac{2}{\epsilon h^2}\rceil,当n>N,\\ |\frac{n}{a^n}-0|=\frac{n}{a^n}<\frac{2}{(n-1)h^2}<\frac{2}{(N-1)h^2}<\epsilon\\ \therefore \lim_{n\rightarrow\infty}\frac{n}{a^n}=0$$

9(3)

$$当n为偶数时:|a_n-1|=\frac{1}{n}\\ 当n为奇数时:|a_n-1|=\sqrt{1+\frac{1}{n}}-1\\ =\frac{(\sqrt{1+\frac{1}{n}}-1)(\sqrt{1+\frac{1}{n}}+1)}{\sqrt{1+\frac{1}{n}}+1}<(1+\frac{1}{n})-1=\frac{1}{n}\\ \therefore|a_n-1|\le\frac{1}{n}\\ \therefore对\forall\epsilon>0,令N=\lceil\frac{1}{\epsilon}\rceil,当n>N,|a_n-1|\le\frac{1}{n}<\frac{1}{N}\le\epsilon\\ \therefore\lim_{n\rightarrow\infty}a_n=1$$

10

$$必要性：\\ \because\lim_{n\rightarrow\infty}a_n=0\\ \therefore对\forall M>0,令\epsilon=\frac{1}{M},\exists N\ 使\ \forall n>N 有a_n<\epsilon即\frac{1}{a_n}>M\\ \therefore\lim_{n\rightarrow\infty}\frac{1}{a_n}=\infty\\ 充分性：\\ \because\lim_{n\rightarrow\infty}\frac{1}{a_n}=\infty\\ \therefore对\forall \epsilon>0,令M=\frac{1}{\epsilon},\exists N\ 使\ \forall n>N 有\frac{1}{a_n}>M即a_n<\epsilon\\ \therefore\lim_{n\rightarrow\infty}a_n=0\\ \therefore\lim_{n\rightarrow\infty}\frac{1}{a_n}=\infty是\lim_{n\rightarrow\infty}a_n=0充要条件.$$

P33

1(6)

$$原式=\lim_{n\rightarrow\infty}\frac{1-\frac{1}{2^n}}{\frac{1}{2}(1-\frac{1}{3^n})}\\ \because n\rightarrow\infty时,\frac{1}{2^n}\rightarrow0,\frac{1}{3^n}\rightarrow0\\ \therefore原式=\lim_{n\rightarrow\infty}\frac{1-\frac{1}{2^n}}{\frac{1}{2}(1-\frac{1}{3^n})}=\frac{1-0}{\frac{1}{2}(1-0)}=2$$

3(3)

$$令s_n=\sum_{i=1}^n\frac{2i-1}{2^i},\\ 有s_n-\frac{1}{2}s_n=\frac{1}{2}+(\frac{3}{2^2}-\frac{1}{2^2})+\dots+(\frac{2n-1}{2^n}-\frac{2n-3}{2^n})-\frac{2n-1}{2^{n+1}}\\ =\frac{1}{2}+\sum_{i=1}^{n-1}\frac{1}{2^i}-\frac{2n-1}{2^{n+1}} =\frac{3}{2}-\frac{2n+3}{2^{n+1}}\\ 即s_n=3-\frac{2n+3}{2^n}.\\ 原式=\lim_{n\rightarrow\infty}s_n=3-\lim_{n\rightarrow\infty}\frac{2n+3}{2^n}=3-0=3$$

6(1)

$$不成立.\\ 令a_{2k-1}=0,a_{2k}=1,易知数列a不收敛.$$

6(2)

$$成立.\\ 令公共极限L,由定义知,对\forall\epsilon>0,有\\ \begin{cases} \exists N_1,\forall n>N_1,|a_{3n-2}-L|<\epsilon\\ \exists N_2,\forall n>N_2,|a_{3n-1}-L|<\epsilon\\ \exists N_3,\forall n>N_3,|a_{3n}-L|<\epsilon\\ \end{cases}\\ 取 N=3\times\max\{N_1,N_2,N_3\},有\forall n>N,|a_n-L|<\epsilon.\\ \therefore a_n收敛于L$$

7(2)

$$n!<\sum_{p=1}^n p!<n!+(n-1)!+(n-2)(n-2)!<n!+2(n-1)!\\ \therefore 1<\frac{\sum_{p=1}^n p!}{n!}<1+\frac{2}{n}\\ \because\lim_{n\rightarrow\infty}1+\frac{2}{n}=1+0=1\\ \therefore\lim_{n\rightarrow\infty}\frac{\sum_{p=1}^n p!}{n!}=1$$

7(3)

$$令f(x)=x^\alpha,f'(x)=\alpha x^{\alpha-1}>0,f''(x)=\alpha(\alpha-1)x^{\alpha-2}<0\\即f(x)单调增,f'(x)单调减.\\ \therefore0<(n+1)^\alpha-n^\alpha<f'(n)\times(n+1-n)=\alpha n^{\alpha-1}\\ \because 1-\alpha>0, RHS=\frac{\alpha}{n^{1-\alpha}}\\ \therefore\lim_{n\rightarrow\infty}\alpha n^{\alpha-1}=\lim_{n\rightarrow\infty}\frac{\alpha}{n^{1-\alpha}}=0\\ \therefore \lim_{n\rightarrow\infty}((n+1)^\alpha-n^\alpha)=0$$

8

$$令M=\max\{a_1,a_2,\dots,a_m\}\\ \therefore M^n\le \sum_{i=1}^m a_i^n\le mM^n\\ \therefore M\le \sqrt[n]{\sum_{i=1}^m a_i^n}\le\sqrt[n]{m}M\\ 当n\rightarrow\infty有:\\ \lim_{n\rightarrow\infty}\sqrt[n]mM=M\lim_{n\rightarrow\infty}m^{\frac{1}{n}}=M\times1=M\\ \therefore由夹逼定理，有\lim_{n\rightarrow\infty}\sqrt[n]{\sum_{i=1}^m a_i^n}=M$$

9(1)

$$\because na_n-1<\lfloor na_n\rfloor\le na_n\\ \therefore a_n-\frac{1}{n}<\frac{\lfloor na_n\rfloor}{n}\le a_n\\ 当n\rightarrow\infty有:\\ \lim_{n\rightarrow\infty}(a_n-\frac{1}{n})=\lim_{n\rightarrow\infty}a_n-\lim_{n\rightarrow\infty}\frac{1}{n}=a-0=a\\ \lim_{n\rightarrow\infty}a_n=a\\ \therefore由夹逼定理，有\lim_{n\rightarrow\infty}\frac{\lfloor na_n\rfloor}{n}=a$$

P39

3(2)

$$x=\sqrt{c+x}\ 有根\ x=\frac{1+\sqrt{4c+1}}{2}\\ 已知a_{n+1}=\sqrt{a_n+c}>0\\ 下证\{a_n\}有上界M=\frac{1+\sqrt{4c+1}}{2}:\\ 1.a_1=\sqrt{c}<\frac{1}{2}+\sqrt{c+\frac{1}{4}}=M\\ 2.当a_k<M,a_{k+1}=\sqrt{a_k+c}<\sqrt{M+c}=M即a_{k+1}<M也成立.\\ 综上,\{a_n\}有上界M.\\ 下证\{a_n\}递增.\\ \because a_n\in(\frac{1-\sqrt{4c+1}}{2},M) \therefore a_n^2-a_n-c<0\\ \therefore a_{n+1}^2=a_n+c>a_n^2,即a_{n+1}>a_n.\\ \therefore \{a_n\}单调递增且有上界,根据单调有界定理必有极限.\\ 令极限为 A,有 A=\sqrt{c+A},A=\frac{1+\sqrt{4c+1}}{2}\\ 即\lim_{n\rightarrow\infty}a_n=\frac{1+\sqrt{4c+1}}{2}$$

6

$$不妨\{a_n\}单调增,记其收敛子列a_{n_k},\exists M,对\forall k,a_{n_k}<M.\\ 对\forall n,\exists n_k>n使a_n<a_{n_k}<M\\ \therefore \{a_n\}有上界M.\\ \because \{a_n\}单调增，有上界\therefore \{a_n\}收敛.$$

7

$$取常数\ q\ 使1<q<l.\\ \because \lim_{n\rightarrow\infty}\frac{a_n}{a_{n+1}}=l \therefore\exists N,\forall n>N,\frac{a_n}{a_{n+1}}>q.\\ 令 r=\frac{1}{q}\in(0,1),有 a_n<a_{N+1}\times r^{n-N-1}=\frac{a_{N+1}}{r^{N+1}}\times r^n\\ 令常数C=\frac{a_{N+1}}{r^{N+1}}, \lim_{n\rightarrow\infty}C\times r^n=0\\ \therefore根据夹逼定理，由0<a_n<C\times r^n可得 \lim_{n\rightarrow\infty}a_n=0.$$

10

$$\because当b>a>0时,\\ b^{n+1}-a^{n+1}=(b-a)(b^n+ab^{n-1}+\dots+a^n)>(n+1)a^{n}(b-a)\\ \therefore 取b=\frac{n+1}{n},a=\frac{n+2}{n+1},有\\ (\frac{n+1}{n})^{n+1}-(\frac{n+2}{n+1})^{n+1}>(n+1)(\frac{n+2}{n+1})^n(\frac{1}{n}-\frac{1}{n+1}),\\ 即 (\frac{n+1}{n})^{n+1}>(\frac{n+2}{n+1})^n(\frac{n+2}{n+1}+\frac{1}{n})\\ \because (n+1)^2=n^2+2n+1>n(n+2)\\ \therefore \frac{1}{n}>\frac{n+2}{(n+1)^2}\\ \therefore \frac{n+2}{n+1}+\frac{1}{n}>\frac{n+2}{n+1}+\frac{n+2}{(n+1)^2}=(\frac{n+2}{n+1})^2\\ \therefore (\frac{n+1}{n})^{n+1}>(\frac{n+2}{n+1})^n(\frac{n+2}{n+1}+\frac{1}{n})>(\frac{n+2}{n+1})^{n+2}\\ \therefore (1+\frac{1}{n})^{n+1}>(1+\frac{1}{n+1})^{n+2},即(1+\frac{1}{n})^{n+1}单调减且有下界1，收敛.\\ \because\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n+1}=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n\times\lim_{n\rightarrow\infty}(1+\frac{1}{n})=e\times1=e\\ \therefore e<(1+\frac{1}{n})^{n+1}.\\ 已知(1+\frac{1}{n})^n递增,有上界e,即有上界3.\\ \therefore (1+\frac{1}{n})^{n+1}=(1+\frac{1}{n})^n+\frac{1}{n}(1+\frac{1}{n})^{n+1}<(1+\frac{1}{n})^n+\frac{3}{n}\\ \therefore e<(1+\frac{1}{n})^n+\frac{3}{n}\\ \because (1+\frac{1}{n})^n<e\\ \therefore |e-(1+\frac{1}{n})^n|=e-(1+\frac{1}{n})^n<(1+\frac{1}{n})^n+\frac{3}{n}-(1+\frac{1}{n})^n=\frac{3}{n}$$