Math Analysis Homework - Week 1

Class 1 Homework

T1

$a_n\le b_n\le c_n, \lim_{n \to \infty} (c_n-a_n)=0 \implies a_n \text{收敛}$

Obviously wrong.

\begin{gathered} a_n=b_n=c_n=n \end{gathered}

T2

\begin{gathered} a_n\le b_n\le c_n,b_n \text{收敛} , \lim_{n \to \infty} (c_n-a_n)=0 \implies a_n \text{收敛} \end{gathered}

\begin{gathered} \forall \epsilon_1 > 0, \exists N_1 \ s.t.\ n>N_1 \implies c_n-a_n< \epsilon_1 \\ \therefore b_n-a_n\le c_n-a_n<\epsilon_1 \\ \text{又}\because \forall \epsilon_2 > 0\exists N_2 \ s.t.\ n>N_2 \implies b_n-B<\epsilon_2 \\ \therefore \vert a_n-B \vert \le \vert b_n-a_n \vert+\vert b_n-B \vert \le \epsilon+\epsilon_2 \\ \therefore \epsilon_1,\epsilon_2:=\frac{\epsilon}{2} \text{有} \\ \forall \epsilon,N:=\max(N_1,N_2), n>N \implies \vert a_n-B\vert < \epsilon \\ \text{Q.E.D} \end{gathered}

T3

\begin{gathered} \lim a_n = A, a_n\ne 0 \implies \lim \frac{a_{n+1}}{a_n} = 1 \end{gathered}

Wrong

$a_n=2^{-n}$

T4

\begin{gathered} \lim_{n \to \infty} a_nb_n = 0 \implies (\lim_{n \to \infty} a_n)(\lim_{n \to \infty} b_n) =0 \end{gathered}

Wrong

\begin{gathered} a_n=(n \bmod 2) \\ b_n= ((n+1) \bmod 2) \end{gathered}

T5

\begin{gathered} \lim_{n \to \infty} \dfrac{b_n}{a_n} =1, \lim_{n \to \infty} a_n=A \implies \lim_{n \to \infty} b_n = A \end{gathered}

不妨设 $A>0$ ,又因为取 $\epsilon<A$ 可以让 $n>N$ 时 $a_n>0$ ,故不妨设 $a_n>0$

又 $,\epsilon_1<1,\epsilon_2<A$

\begin{gathered} \forall \epsilon_1 \exists N_1 \ s.t.\ \vert \dfrac{a_n}{b_n} - 1 \vert < \epsilon_1 \\ \forall \epsilon_2 \exists N_2 \ s.t.\ \vert a_n - A \vert < \epsilon_2 \\ \therefore \dfrac{a_n}{b_n} \in (1-\epsilon_1,1+\epsilon_1) \\ a_n\in (A-\epsilon_2,A+\epsilon_2) \\ \therefore b_n= \frac{a_n}{\frac{a_n}{b_n}} \in (\dfrac{A-\epsilon_2}{1+\epsilon_1},\dfrac{A+\epsilon_2}{1-\epsilon_1})\\ \therefore b_n-A \in (\dfrac{-A\epsilon_1-\epsilon_2}{1+\epsilon_1},\dfrac{A\epsilon_1+\epsilon_2}{1-\epsilon_1} ), \\ \vert b_n-A \vert \le \dfrac{A\epsilon_1+\epsilon_2}{1-\epsilon_1}<\epsilon \\ \epsilon_1:= \dfrac{\epsilon}{100A} ,\epsilon_2:=\dfrac{\epsilon}{100} \\ \implies \vert b_n-A \vert \le \dfrac{A\epsilon_1+\epsilon_2}{1-\epsilon_1}=\dfrac{\dfrac{\epsilon}{50} }{1-\dfrac{\epsilon}{100A} } \\ \text{let } \epsilon<A \implies 1-\dfrac{\epsilon}{100A} >\dfrac{1}{50} \\ \therefore N=\max(N_1,N_2) \implies n>A \implies \vert b_n-A \vert < \epsilon \\ \text{Q.E.D} \end{gathered}

T6

\begin{gathered} \lim_{n \to \infty} \dfrac{3n^2+n}{2n^2-1} = \dfrac{3}{2} \end{gathered}

\begin{gathered} \dfrac{3n^2+n}{2n^2-1}-\dfrac{3}{2}>\dfrac{3n^2}{2n^2}-\dfrac{3}{2}=0 \\ \dfrac{3n^2+n}{2n^2-1}-\dfrac{3}{2}=\dfrac{3n^2+n-\frac{3}{2}(2n^2-1)}{2n^2-1} = \dfrac{n+\dfrac{3}{2} }{2n^2-1}\stackrel{n>1}{<} \dfrac{2n}{n^2} =\dfrac{2}{n} <\epsilon \\ \therefore N:=\dfrac{2}{\epsilon} +114514 \\ \text{Q.E.D} \end{gathered}

T7

\begin{gathered} \lim_{n \to \infty} \sqrt{n^2+n}-n=\dfrac{1}{2} \end{gathered}

\begin{gathered} \sqrt{n^2+n}-n=\dfrac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n} \\ =\dfrac{n}{\sqrt{n^2+n}+n} \\ =\dfrac{1}{1+\sqrt{1+\frac{1}{n}}} \\ {\left \vert \dfrac{1}{1+\sqrt{1+\frac{1}{n}}}-\dfrac{1}{2} \right \vert} \\ = \dfrac{1}{2}-\dfrac{1}{1+\sqrt{1+\frac{1}{n}}} \\ =\dfrac{\sqrt{1+\frac{1}{n}}-1}{2(1+\sqrt{1+\frac{1}{n}})} \\ <\sqrt{1+\frac{1}{n}}-1 \\ \stackrel{\text{Bernoulli Inequality}}{<}1+\dfrac{1}{2n} -1 \\ =\dfrac{1}{2n} \\ \therefore N:=\frac{1}{2\epsilon}+100 \\ \text{Q.E.D} \end{gathered}

T8

\begin{gathered} \dfrac{n^2\arctan(n)}{1+n^2}=\dfrac{\pi}{2} \end{gathered}

\begin{gathered} \dfrac{n^2\arctan(n)}{1+n^2} \\ =\dfrac{n^2}{1+n^2}\arctan(n) \\ \end{gathered}

\begin{gathered} \lim_{n \to \infty} \dfrac{n^2}{1+n^2}=1 \end{gathered}

\begin{gathered} \lim {\left \vert \dfrac{n^2}{1+n^2}-1 \right \vert} =\dfrac{1}{1+n^2}<\dfrac{1}{n} \\ N:=\dfrac{1}{\epsilon}+100 \\ \text{Q.E.D} \end{gathered}

\begin{gathered} \lim_{n \to \infty} \arctan(n)=\dfrac{\pi}{2} \end{gathered}

\begin{gathered} N:=\tan(\dfrac{\pi}{2}-\dfrac{\epsilon}{2}) \implies \\ \forall \epsilon, \vert \arctan(n)-\dfrac{\pi}{2}\vert =\dfrac{\pi}{2}-\arctan(n)=\dfrac{\epsilon}{2}<\epsilon \\ \text{Q.E.D} \end{gathered}

\begin{gathered} (\lim a_n)(\lim_{n \to \infty} b_n) = X,a_n>0,b_n>0 \implies \lim_{n \to \infty} a_nb_n=X \end{gathered}

\begin{gathered} A:=\lim a_n,B:=\lim b_n \\ \forall \epsilon_1, \exists N_1 \ s.t.\ n>N_1 \implies {\left \vert a_n-A \right \vert} < \epsilon \\ \forall \epsilon_2, \exists N_2 \ s.t.\ n>N_2 \implies {\left \vert b_n-B \right \vert} < \epsilon \\ \therefore n>\max(N_1,N_2) \implies \\ a_n \in (A-\epsilon_1,A+\epsilon_1),b_n\in (B-\epsilon_2,B+\epsilon_2) \\ \implies \\ a_nb_n \in ((A-\epsilon_1)(B-\epsilon_2),(A+\epsilon_1)(B+\epsilon_2)) \\ \vert a_nb_n-AB\vert < A\epsilon_2+B\epsilon_1+\epsilon_1\epsilon_2\\ \forall \epsilon, \epsilon_2:=\dfrac{\epsilon}{4A},\epsilon_1:=\dfrac{\epsilon}{4B} \\ \therefore \vert a_nb_n-AB\vert=\dfrac{\epsilon}{2}+\dfrac{\epsilon^2}{16AB}\stackrel{\epsilon<AB}{<}\epsilon \\ \text{Q.E.D} \end{gathered}

\begin{gathered} \lim_{n \to \infty} \dfrac{n^2\arctan(n)}{1+n^2}=\lim_{n \to \infty} \dfrac{n^2}{1+n^2} \lim_{n \to \infty} \arctan(n)=1\times \dfrac{\pi}{2} = \dfrac{\pi}{2} \\ \text{Q.E.D} \end{gathered}

Class 2

T1

$\lim_{n\to\infty} \frac{(-2)^n + 3^n}{(-2)^{n+1} + 3^{n+1}}$

\lim_{n \to \infty} \frac{(-2)^n + 3^n}{(-2)^{n+1} + 3^{n+1}} \\ \lim_{n \to \infty} =\frac{1}{3} \frac{(\frac{-2}{3})^n +1}{(\frac{-2}{3})^{n+1} +1} \\ \lim_{n \to \infty} =\frac{1}{3}

T2

$\lim_{n\to\infty} \left[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)} \right]$

=\lim_{n \to \infty} \sum_{i=1}^n \frac{1}{i(i+1)} \\ =\lim_{n \to \infty} \sum_{i=1}^n \frac{1}{i} -\frac{1}{i+1} \\ =\lim_{n \to \infty} 1-\frac{1}{n+1} =1

T3

$\lim_{n\to\infty} \left[ \frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + \dots + \frac{1}{\sqrt{n^2+n}} \right]$

\begin{gathered} \sum _{i = 1} ^{n} \frac{1}{\sqrt{n^2+1}} \le \sum _{i = 1} ^{n} \frac{1}{\sqrt{n^2+i}} \le \sum _{i = 1} ^{n} \frac{1}{\sqrt{n^2+n}} \\ \stackrel{\text{Squeeze Theorem}}{\Longrightarrow } \\ \lim_{n \to \infty} \sum _{i = 1} ^{n} \frac{1}{\sqrt{n^2+1}} \le L=\lim_{n \to \infty} \sum _{i = 1} ^{n} \frac{1}{\sqrt{n^2+i}} \le \lim_{n \to \infty} \sum _{i = 1} ^{n} \frac{1}{\sqrt{n^2+n}} \\ \implies \lim_{n \to \infty} \frac{n}{\sqrt{n^2+1}} \le L \le \lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}} \\ \implies \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{1}{n^2} }} \le L \le \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{1}{n} }} \\ \implies L=1 \end{gathered}

T4

$\lim_{n\to\infty} \sqrt[n]{n^2-n+2}$

Obviously: $n>2 \implies n^2-n+2>8>1$

\begin{gathered} \vert (n^2-n+2)^{\frac{1}{n} }-1 \vert =(n^2-n+2)^{\frac{1}{n} }-1<\epsilon \\ \impliedby n^2-n+2<(1+\epsilon)^n \\ \impliedby F(x)=n^2-n+2<1+n\epsilon+\frac{(n^2-n)\epsilon^2}{2}+\frac{n(n-1)(n-2)}{6}\epsilon^3=G(x) \\ \exists C_1(\epsilon),C_2(\epsilon) \\ s.t.\\ F(x)<C_1(\epsilon)n^2,G(x)>C_2(\epsilon)n^3 \\ \therefore N:=\frac{C_1(\epsilon)}{C_2(\epsilon)} +1 \implies (n>N \implies (n^2-n+2)^{\frac{1}{n} }-1<\epsilon) \\ \therefore \lim_{n\to\infty} \sqrt[n]{n^2-n+2}=1 \end{gathered}

T5

\lim_{n \to \infty} \sqrt[n]{\arctan(n)}

\begin{gathered} 1<\arctan(n)<\frac{\pi}{2} \\ \implies 1<\sqrt[n]{\arctan(n)}<\sqrt[n]{\frac{\pi}{2} } \\ \begin{cases} 1<\sqrt[n]{\arctan(n)}<\sqrt[n]{\frac{\pi}{2} } \\ \lim_{n \to \infty} 1=\lim_{n \to \infty} \sqrt[n]{\frac{\pi}{2} } \end{cases} \stackrel{\text{Squeeze Theorem}}{\Longrightarrow } \lim_{n \to \infty} \sqrt[n]{\arctan(n)}=1 \end{gathered}

T6

\lim_{n \to \infty} \sqrt[n]{2\sin^2(n)+\cos^2(n)}

同上一题,里面有界,是 $1$

T7

$\lim_{n\to\infty} \frac{[na_n]}{n}$ , 这里 $\lim_{n\to\infty} a_n = a$ .

\begin{cases} \frac{na_n-1}{n} <\frac{[na_n]}{n}<\frac{na_n+1}{n} \\ \lim_{n \to \infty} \frac{na_n+1}{n}=\lim_{n \to \infty} a_n+\lim_{n \to \infty} \frac{1}{n}=a \\ \lim_{n \to \infty} \frac{na_n-1}{n}=\lim_{n \to \infty} a_n-\lim_{n \to \infty} \frac{1}{n}=a \end{cases} \\ \stackrel{\text{Squeeze Theorem}}{\Longrightarrow } \lim_{n\to\infty} \frac{[na_n]}{n}=a_n

T8

证明 $a_n=\frac{2n+(-1)^n n}{3n+1}$ 发散

\begin{gathered} \begin{cases} \lim_{n \to \infty} a_{2n} = \lim_{n \to \infty} \frac{6n}{6n+1} =1 \\ \lim_{n \to \infty} a_{2n+1} = \lim_{n \to \infty} \frac{2n+1}{6n+4} =\frac{1}{3} \ne 1 \end{cases} \\ \implies \{ a_n \} \text{发散} \end{gathered}

T9

a_n\ne 0,\frac{a_{n+1}}{a_n} >0,\lim_{n \to \infty} \frac{a_{n+1}}{a_n} =0 \implies \exists N \ s.t.\ n>N \implies \{a_n\} \text{单调}

\begin{gathered} \text{use } \epsilon_1=1,\exists N_1 \\ s.t.\\ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} <1 \\ \text{exact } a_{n+1}<a_n \end{gathered}

T10

设 $\lim_{n\to\infty} (x_n - x_{n-1}) = d$ , 证明: $\lim_{n\to\infty} \frac{x_n}{n} = d$ .

\begin{gathered} \forall \epsilon_1, \exists N_1 \ s.t.\ n>N_1 \implies \vert x_n-x_{n-1}-d \vert <\epsilon_1 \\ \iff x_n-x_{n-1} \in [d-\epsilon_1,d+\epsilon_1] \\ \therefore x_n=x_{N_1}+\sum _{i = N_1+1} ^{n} (x_i-x_{i-1}) \\ \in [x_{N_1}+(n-N_1)(d-\epsilon_1),x_{N_1}+(n-N_1)(d+\epsilon_1)] \\ \therefore \frac{x_n}{n} \in [\frac{x_{N_1}-N_1(d-\epsilon_1)}{n}+d-\epsilon_1,\frac{x_{N_1}-N_1(d+\epsilon_1)}{n}+d+\epsilon_1 ] \\ \forall \epsilon,\frac{x_n}{n} -d\in[\frac{x_{N_1}-N_1(d-\epsilon_1)}{n}-\epsilon_1,\frac{x_{N_1}-N_1(d+\epsilon_1)}{n}+\epsilon_1] \\ \epsilon_1:=\frac{\epsilon}{2} ,n:=\frac{2x_{N_1}-N_1(d-\epsilon_1)}{\epsilon} \\ \implies \vert \frac{x_n}{n} -d \vert < \epsilon \end{gathered}

T11

设 $\lim_{n\to\infty} a_n = a (a>0, n \in \mathbb{N})$ , 证明: $\lim_{n\to\infty} \sqrt[n]{a_1 a_2 \dots a_n} = a$ , 并由此证明:

若 $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = a (a>0, n \in \mathbb{N})$ ,则 $\lim_{n\to\infty} \sqrt[n]{a_n} = a$ ;
$\lim_{n \to \infty} \frac{\sqrt[n]{ n! } }{n} =\frac{1}{e}$

(1)

Solution 1

\begin{gathered} {\left \vert \sqrt[n]{ \prod_i a_i } -a \right \vert} <\epsilon \\ \iff a-\epsilon<\sqrt[n]{ \prod_i a_i }<a+\epsilon \\ \iff (a-\epsilon)^n<\prod_i a_i<(a+\epsilon)^n \\ \forall \epsilon_1,\exists N_1 \ s.t.\ n>N_1 \implies a_n\in [a-\epsilon_1,a+\epsilon_1] \\ \prod_i a_i=(\prod_{i=1}^{N_1} a_i) (\prod_{i=N_1+1}^{n}a_i)\in [A(a-\epsilon_1)^{n-N_1},A(a+\epsilon_1)^{n-N_1}] \\ \text{let} \epsilon_1:=\frac{\epsilon}{2} ,\text{Consider } A(a-\epsilon_1)^{n-N_1}>(a-\epsilon)^n: \\ \iff \frac{A}{(a-\epsilon_1)^{N_1}} >(\frac{a-\epsilon}{a-\epsilon_1} )^n \\ \text{Since }\frac{a-\epsilon}{a-\epsilon_1}<1,\exists N_2 \ s.t.\ n>N_2 \implies \text{不等式成立} \\ \text{右侧同理有} N_3 \\ \therefore N:=N_1+N_2+N_3 \\ s.t.\\ n>N \\ \implies (a-\epsilon)^n<A(a-\epsilon_1)^{n-N_1}<\prod_i a_i<A(a+\epsilon_1)^{n-N_1}<(a+\epsilon)^n\\ \implies {\left \vert \sqrt[n]{ \prod_i a_i } -a \right \vert} <\epsilon \\ \text{Q.E.D} \end{gathered}

Solution 2

唐. 可以用调和均值/算数均值夹两边.

(2)

令 $b_n=\frac{a_{n+1}}{a_n}$ ,则问题转化为(1).

(3)

\begin{gathered} e=\lim_{n \to \infty} (1+\frac{1}{n})^n \\ \implies \frac{1}{e} =\lim_{n \to \infty} (\frac{n}{n+1} )^n \\ a_n:=(\frac{n}{n+1})^n \text{检验符合引理} \\ \text{Q.E.D} \end{gathered}

T12

设 $\lim_{n\to\infty} x_n = +\infty$ , 证明: $\lim_{n\to\infty} \frac{x_1+x_2+\dots+x_n}{n} = +\infty$ .

\begin{gathered} \forall X,X_1:=X+1 \\ \lim_{n \to \infty} x_n=\infty \implies \exists N_1 \ s.t.\ n>N_1 \implies x_n>X_1=X+1 \\ \text{for }n>N_1,\frac{\sum _{i = 1} ^{n} x_i}{n} =\frac{\sum _{i = 1} ^{N_1} x_i+\sum _{i = N_1+1} ^{n} X+1}{n} \\ >(1-\frac{N_1}{n})(X+1) \\ n:=N_1(X+1)+100 \\ \implies \frac{\sum _{i = 1} ^{n} x_i}{n} >X \end{gathered}

T13

a_n>0,\lim_{n \to \infty} \frac{a_n}{a_{n+1}+a_{n+2}}=0 \implies a_n \text{is unbounded}

反证,设 $\exists M$ 令 $0<a_n<M$ .

\begin{gathered} \epsilon=\frac{1}{5} ,\frac{a_n}{a_{n+1}+a_{n+2}} <\frac{1}{5} \implies a_{n+1}+a_{n+2}>5a_n \\ \therefore \max(\{ a_{n+1},a_{n+2} \} )>2a_n \\ \therefore b_1=N+1,b_i=a_{b_{i-1}+1},a_{b_{i-1}+2} \text{中较大的一个的下标} \implies a_{b_i}>2^ia_{b_1} \end{gathered}

故 $a$ 有发散子列, $a$ 发散.

T14

设数列 $\{x_n\}$ 单调增加, $\lim_{n\to\infty} \frac{x_1+x_2+\dots+x_n}{n} = a$ , 证明: $\lim_{n\to\infty} x_n = a$ .

若存在 $x_N>a$ ,则 $n>N$ 时 $x_n>a$ , $\lim_{n \to \infty} x_N>a$ ,则 $\exists A\in (x_N,A)$ . 并有 $\forall i\ge N, x_i>A>a$ ,则

\begin{gathered} \lim_{n \to \infty} \frac{\sum _{i = 1} ^{n} x_i}{n} \\ >\lim_{n \to \infty} \frac{\sum _{i = 1} ^{N} a_i}{n}+\frac{n-N}{n}(a+(A-a)) =A>a \end{gathered}

矛盾,故 $\forall i,x_i<a$

于是

\begin{cases} \frac{\sum _{i = 1} ^{n} x_i}{n} <x_n<a \\ \lim_{n \to \infty} a =a \\ \lim_{n \to \infty} \frac{\sum _{i = 1} ^{n} x_i}{n}=a \end{cases} \implies \lim_{n \to \infty} x_n=a

Class 3

T1

$\lim_{n \to \infty} \frac{1 + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}}{\ln \sqrt{n}}$

\begin{gathered} \lim_{n \to \infty} \frac{1 + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}}{\ln \sqrt{n}} \\ \stackrel{\text{Stolz Theorem}}{\impliedby }\lim_{n \to \infty} \dfrac{\dfrac{2}{\sqrt n} }{\ln\dfrac{n}{n-1} } \\ \stackrel{x>1 \implies \ln(x)>2\frac{x-1}{x+1} }{>}\lim_{n \to \infty} \dfrac{\dfrac{2}{\sqrt{ n } } }{\dfrac{2}{2n-1} }\\ =+\infty \end{gathered}

T2

$\lim_{n \to \infty} \frac{1 + \sqrt{2} + \sqrt[3]{3} + \cdots + \sqrt[n]{n}}{n}$

\begin{gathered} \lim_{n \to \infty} \frac{1 + \sqrt{2} + \sqrt[3]{3} + \cdots + \sqrt[n]{n}}{n} \\ \stackrel{\text{Stolz Theorem}}{\impliedby }\lim_{n \to \infty} \dfrac{\sqrt[n]{n}}{1} \\ =1 \end{gathered}

T3

$\lim_{n \to \infty} \frac{a_1 + 2a_2 + \cdots + na_n}{\sum_{i=1}^n i}$ (已知 $\lim_{n \to \infty} a_n = a$ )

\begin{gathered} \lim_{n \to \infty} \frac{a_1 + 2a_2 + \cdots + na_n}{\sum_{i=1}^n i} \\ \stackrel{\text{Stolz Theorem}}{\impliedby } \lim_{n \to \infty} \dfrac{na_n}{n} \\ =a \end{gathered}

T4

计算极限 $\lim_{n \to \infty} (n!)^{\frac{1}{n^2} }$ .

\begin{gathered} \begin{cases} 1<(n!)^{\frac{1}{n^2} }<(n^n)^{\frac{1}{n^2} }=n^{\frac{1}{n} } \\ \lim_{n \to \infty} 1=\lim_{n \to \infty} n^{\frac{1}{n} }=1 \end{cases} \\ \stackrel{\text{Squeeze Theorem}}{\Longrightarrow } \lim_{n \to \infty} (n!)^{\frac{1}{n^2} }=1 \end{gathered}

T5

设 $x_n = \frac{1}{n^2} \sum_{k=0}^n \ln \binom{n}{k}$ , $n=1,2,\cdots$ , 求极限 $\lim_{n \to \infty} x_n$ .

\begin{gathered} \frac{1}{n^2} \sum_{k=0}^n \ln \binom{n}{k} \\ \stackrel{\text{Stolz Theorem}}{\impliedby } \dfrac{\sum _{i = 0} ^{n} \ln\binom{n}{i}-\sum _{i = 0} ^{n-1} \ln \binom{n-1}{i}}{2n-1} \\ =\dfrac{\sum _{i = 0} ^{n-1} \ln(\dfrac{n}{n-i} )}{2n-1} \\ =\dfrac{\ln(\dfrac{n^n}{n!} )}{2n-1} \\ = \dfrac{n\ln(\dfrac{n}{\sqrt[n]{ n! } } )}{2n-1} \\ \text{According to homework class-2:} \\ \lim_{n \to \infty} \dfrac{\sqrt[n]{ n! } }{n} =\dfrac{1}{e} \\ \implies \lim_{n \to \infty} \dfrac{n\ln(\dfrac{n}{\sqrt[n]{ n! } } )}{2n-1} \\ =\lim_{n \to \infty} \dfrac{n}{2n-1} \lim_{n \to \infty} \ln(\dfrac{n}{\sqrt[n]{ n! } } ) \\ =\dfrac{1}{2} \end{gathered}

还是用了连续性/kk

T6

设 $\lim_{n \to \infty} n(A_n - A_{n-1}) = 0$ , 试证: 当极限 $\lim_{n \to \infty} \frac{A_1 + A_2 + \cdots + A_n}{n}$ 存在时, $\lim_{n \to \infty} A_n = \lim_{n \to \infty} \frac{A_1 + A_2 + \cdots + A_n}{n}$ .

\begin{gathered} x_n:=\sum _{i = n} ^{n} \Delta A_i i \\ =\sum _{i = 1} ^{n} i(A_i-A_{i-1}) \\ =nA_n-\sum_{i=1}^{n-1} A_i \\ \lim_{n \to \infty} \dfrac{x_n}{n} \\ \stackrel{\text{Stolz Theorem}}{\impliedby } \dfrac{x_n-x_{n-1}}{n-(n-1)} =n\Delta A_n =0 \\ \therefore \lim_{n \to \infty} A_n- \lim_{n \to \infty} \dfrac{\sum _{i = 1} ^{n-1} A_i}{n-1} = \lim_{n \to \infty} \dfrac{x_n}{n} =0 \\ \lim_{n \to \infty} A_n=a \end{gathered}