问题2012102801

证明:$\{a_n\}$收敛的充分必要条件是$\{a_{2k}\}$和$\{a_{2k-1}\}$收敛于同一极限.

来源

数学分析习题课讲义,作者:谢惠民,恽自求,易法槐,钱定边.

解答

(1)设$a_n\rightarrow a$,$\forall \epsilon>0$,$\exists N$,当$n>N$时
\[|a_n-a|<\epsilon\]
于是$k>N$时,$2k>N$,$|a_{2k}-a|<\epsilon$,$a_{2k} \rightarrow a$,$2k-1>N$,$|a_{2k-1}-a|<\epsilon$,$a_{2k-1} \rightarrow a$.

(2)反之,$a_{2k} \rightarrow a$,$a_{2k-1} \rightarrow a$,$\forall \epsilon>0$,$\exists N_1$,当$k>N_1$时,$|a_{2k}-a|<\epsilon$,$\exists N_2$,当$k>N_2$时,$|a_{2k-1}-a|<\epsilon$.于是令$N=\max\{2N_1,2N_2\}$,则$n>N$时有
\[|a_n-a|<\epsilon.\]
获证.

问题2012102802

以下是可以应用夹逼定理的几个题:
(1)给定$p$个正数$a_1,\cdots,a_p$,求$\lim\limits_{n \rightarrow \infty}{\sqrt[n]{a_1^n+\cdots+a_p^n}}$;
(2)设$x_n=\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + \cdots + \frac{1}{\sqrt{(n+1)^2}}$,$n \in N_+$,求$\lim\limits_{n \rightarrow \infty}{x_n}$;
(3)设$a_n=(1 + \frac{1}{2} + \cdots + \frac{1}{n})^{1/n}$,$n \in N_+$,求$\lim\limits_{n \rightarrow \infty}{a_n}$;
(4)设$\{a_n\}$为正数列,并且已知它收敛于$a>0$,证明$\lim\limits_{n \rightarrow \infty}{\sqrt[n]{a_n}} = 1$;

来源

数学分析习题课讲义,作者:谢惠民,恽自求,易法槐,钱定边.

解答

(1)记$a = \max\{a_1,\cdots,a_p\}$,那么有
\[\begin{aligned}\sqrt[n]{a_1^n + \cdots + a_p^n} &\ge \sqrt[n]{a^n}=a \\\sqrt[n]{a_1^n + \cdots + a_p^n} &\le \sqrt[n]{pa^n}=\sqrt[n]{p}a\end{aligned}\]
而$\sqrt[n]{p} \rightarrow 1$,于是$\sqrt[n]{p}a \rightarrow a$,由此
\[\lim_{n \rightarrow \infty}{\sqrt[n]{a_1^n+\cdots+a_p^n}} = \max\{a_1,\cdots,a_p\}.\]

(2)注意到如下夹逼方式即可
\[\begin{aligned}x_n &> \frac{1}{n+1} + \cdots + \frac{1}{n+1} \\&= \frac{2n+1}{n+1} \rightarrow 2 \\x_n &< \frac{1}{\sqrt{n^2+1}} + \cdots + \frac{1}{\sqrt{n^2+1}} \\&=\frac{2n+1}{\sqrt{n^2+1}} \rightarrow 2\end{aligned}\]
因此$\lim\limits_{n\rightarrow \infty}{x_n}=2$.

(3)首先有$a_n>1$,另一方面$a_n < n^{1/n}\rightarrow 1$,因此$\lim\limits_{n \rightarrow \infty}{a_n}=1$.

(4)对于$a/2 > 0$,$\exists N$,当$n>N$时,$|a_n-a|<a/2$,由此得到
\[\begin{aligned}&0 < \frac{a}{2} < a_n < \frac{3a}{2}\\\Rightarrow &\sqrt[n]{\frac{a}{2}} < \sqrt[n]{a_n} \sqrt[n]{\frac{3a}{2}}\end{aligned}\]
而$\lim\limits_{n \rightarrow \infty}{\sqrt[n]{\frac{a}{2}}}=1$,$\lim\limits_{n \rightarrow \infty}{\sqrt[n]{\frac{3a}{2}}}=1$,因此$\lim\limits_{n \rightarrow \infty}{a_n}=1$.

问题2012102803

求以下极限:
(1)$\lim\limits_{n \rightarrow \infty}{(1+x)(1+x^2)\cdots(1+x^{2^n})}$,其中$|x|<1$.
(2)$\lim\limits_{n \rightarrow \infty}{(1-\frac{1}{2^2})(1-\frac{1}{3^2})\cdots(1-\frac{1}{n^2})}$;
(3)$\lim\limits_{n \rightarrow \infty}{(1 – \frac{1}{1+2})(1 – \frac{1}{1+2+3})\cdots(1 – \frac{1}{1+2+\cdots+n})}$;
(4)$\lim\limits_{n \rightarrow \infty}{(\frac{1}{1\cdot2\cdot3} + \frac{1}{2\cdot3\cdot4} + \cdots + \frac{1}{n\cdot(n+1)\cdot(n+2)})}$;
(5)$\lim\limits_{n \rightarrow \infty}{\sum_{k=1}^{n}{\frac{1}{k(k+1)\cdots(k+v)}}}$.

来源

数学分析习题课讲义,作者:谢惠民,恽自求,易法槐,钱定边.

解答

(1)由于$|x|<1$,于是$1-x \neq 0$,由此
\[\frac{(1-x)(1+x)(1+x^2)\cdots(1+x^{2^n})}{1-x} = \frac{1-x^{2^{n+1}}}{1-x} \rightarrow \frac{1}{1-x},\]
也就是所求的极限为$\frac{1}{1-x}$.

(2)进行变形:
\[\begin{aligned}&\lim_{n \rightarrow \infty}{(1-\frac{1}{2^2})(1-\frac{1}{3^2})\cdots(1-\frac{1}{n^2})} \\= &\lim_{n \rightarrow \infty}{\prod_{2}^{n}{(1 – \frac{1}{k^2})}} = \lim_{n \rightarrow \infty}{\prod_{2}^{n}{\frac{(k-1)(k+2)}{k^2}}} \\= &\lim_{n \rightarrow \infty}{\frac{1\cdot3}{2^2} \cdot \frac{2 \cdot 4}{3^2} \cdots \frac{(n-2)n}{(n-1)^2}} \cdot \frac{(n-1)(n+1)}{n^2} \\= &\lim_{n \rightarrow \infty}{\frac{1}{2}\cdot\frac{n+1}{n}} = \frac{1}{2}.\end{aligned}\]

(3)进行变形
\[\begin{aligned}&\lim_{n \rightarrow \infty}{(1 – \frac{1}{1+2})(1 – \frac{1}{1+2+3})\cdots(1 – \frac{1}{1+2+\cdots+n})}\\=&\lim_{n \rightarrow \infty}{\prod_{2}^{n}{1 – \frac{2}{k(k+1)}}} = \lim_{n \rightarrow \infty}{\prod_{2}^{n}{\frac{(k-1)(k+2)}{k(k+1)}}}\\=&\lim_{n \rightarrow \infty}{\frac{1 \cdot 4}{2 \cdot 3} \cdot \frac{2 \cdot 5}{3 \cdot 4} \cdot \frac{3 \cdot 6}{4 \cdot 5} \cdots \frac{(n-2)(n+1)}{(n-1)n} \cdot \frac{(n-1)(n+2)}{n(n+1)}} \\=&\lim_{n \rightarrow \infty}{\frac{(n-1)! \cdot \frac{1}{6}(n+2!)}{n! \cdot \frac{1}{2}(n+1)!}} = \lim_{n \rightarrow \infty}{\frac{n+2}{3n}} = \frac{1}{3}.\end{aligned}\]

(4)进行变形
\[\begin{aligned}&\lim_{n \rightarrow \infty}{(\frac{1}{1\cdot2\cdot3} + \frac{1}{2\cdot3\cdot4} + \cdots + \frac{1}{n\cdot(n+1)\cdot(n+2)})} \\=&\lim_{n \rightarrow \infty}{\sum_{1}^{n}{\frac{1}{k(k+1)(k+2)}}} = \lim_{n \rightarrow \infty}{\frac{1}{2}(\frac{1}{k(k+1)} – \frac{1}{(k+1)(k+2)})} \\=&\frac{1}{2}\lim_{n \rightarrow \infty}{(\frac{1}{1 \cdot 2} – \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3} – \frac{1}{3 \cdot 4} + \cdots + \frac{1}{n(n+1)} – \frac{1}{(n+1)(n+2)})} \\=&\frac{1}{2}\lim_{n \rightarrow \infty}{(\frac{1}{2} – \frac{1}{(n+1)(n+2)})} = \frac{1}{4}.\end{aligned}\]

(5)和上一小题类似:
\[\begin{aligned}&\lim\limits_{n \rightarrow \infty}{\sum_{k=1}^{n}{\frac{1}{k(k+1)\cdots(k+v)}}} \\=&\lim_{n \rightarrow \infty}{\sum_{k=1}^{n}{\frac{1}{v}(\frac{1}{k(k+1)\cdots(k+v-1)} – \frac{1}{(k+1)(k+2)\cdots(k+v)})}} \\=&\frac{1}{v}\lim_{n \rightarrow \infty}{[\frac{1}{1\cdot 2 \cdots v} – \frac{1}{2 \cdot 3 \cdots v(v+1)} + cdots + \frac{1}{n(n+1)\cdots(n+v-1)} – \frac{1}{(n+1)\cdots(n+v)}]}\\=&\frac{1}{v} \cdot \frac{1}{1 \cdot 2 \cdots v} = \frac{1}{vv!}.\end{aligned}\]

问题2012102804

设$S_n = a + 3a^2 + \cdots + (2n-1)a^n$,$|a|<1$,求$\{S_n\}$的极限.

来源

数学分析习题课讲义,作者:谢惠民,恽自求,易法槐,钱定边.

解答

令$a$乘以$S_n$,得到
\[aS_n = a^2 + 3a^3 + \cdots + (2n-3)a^n + (2n-1)a^{n+1},\]
$S_n$减去该式子得到
\[\begin{aligned}(1-a)S_n &= a + 2(a^2 + a^3 + \cdots + a^n) – (2n-1)a^{n+1}\\&=\frac{2a(1-a^n)}{1-a}-a-(2n-1)a^{n+1} \\S_n = \frac{2a(1-a^n)}{(1-a)^2} – \frac{a+(2n-1)a^{n+1}}{1-a},\end{aligned}\]
当$n \rightarrow \infty$,$|a|<1$时,$(2n-1)a^{n+1} \rightarrow 0$,$a^n \rightarrow 0$,因此
\[\lim_{n \rightarrow \infty}{S_n} = \frac{2a}{(1-a)^2} – \frac{1}{1-a} = \frac{a+a^2}{(1-a)^2}.\]