谢惠民,恽自求,易法槐,钱定边编数学分析习题课讲义16.2.3练习题参考解答[来自陶哲轩小弟]…

1.设已知 $ \sum\limits_{n = 1}^\infty {
{
{\left( { – 1} \right)}^{n – 1}}{a_n}} = A,\sum\limits_{n = 1}^\infty {
{a_{2n – 1}}} = B $ ,证明: $ \sum\limits_{n = 1}^\infty {
{a_n}} $ 收敛并求其和.
解:显然有
\[\sum\limits_{n = 1}^\infty {
{a_n}} = 2\sum\limits_{n = 1}^\infty {
{a_{2n – 1}}} – \sum\limits_{n = 1}^\infty {
{
{\left( { – 1} \right)}^{n – 1}}{a_n}} = 2B – A.\]

2.设 $ P(x)=a_0+a_1x+\cdots+a_mx^m $ 为 $ m $ 次多项式,求级数 $ \sum\limits_{n = 0}^\infty {\frac{
{P\left( n \right)}}{
{n!}}} $ 的和.
解:事实上,
$$\begin{align*}{b_k} &= \sum\limits_{n = 0}^\infty {\frac{
{
{n^k}}}{
{n!}}} = \sum\limits_{n = 1}^\infty {\frac{
{
{n^{k – 1}}}}{
{\left( {n – 1} \right)!}}} = \sum\limits_{n = 0}^\infty {\frac{
{
{
{\left( {n + 1} \right)}^{k – 1}}}}{
{n!}}} \\&= {b_{k – 1}} + C_{k – 1}^1{b_{k – 2}} + \cdots + C_{k – 1}^{k – 2}{b_1} + {b_0},\end{align*}$$
其中 $ b_0=e $ .
由此得到的数叫Bell数,记为 $ B_n $ ,并且
\[B\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{
{B\left( n \right)}}{
{n!}}{x^n}} = {e^{
{e^x} – 1}}.\]

回到原题,我们有\[\sum\limits_{n = 0}^\infty {\frac{
{P\left( n \right)}}{
{n!}}} = e\sum\limits_{k = 0}^m {
{a_k}{B_k}} .\]

3.求 $ 1 – \frac{
{
{2^3}}}{
{1!}} + \frac{
{
{3^3}}}{
{2!}} – \frac{
{
{4^3}}}{
{3!}} + \cdots $ 的和.
解:事实上,
$$\begin{align*}{b_k} &= \sum\limits_{n = 0}^\infty {
{
{\left( { – 1} \right)}^n}\frac{
{
{n^k}}}{
{n!}}} = \sum\limits_{n = 1}^\infty {
{
{\left( { – 1} \right)}^n}\frac{
{
{n^{k – 1}}}}{
{\left( {n – 1} \right)!}}} = \sum\limits_{n = 0}^\infty {
{
{\left( { – 1} \right)}^{n+1}}\frac{
{
{
{\left( {n + 1} \right)}^{k – 1}}}}{
{n!}}} \\& =- {b_{k – 1}} – C_{k – 1}^1{b_{k – 2}} – \cdots – C_{k – 1}^{k – 2}{b_1} – {b_0},\end{align*}$$
其中 $ b_0=1/e $ .因此 $ b_1=-1/e,b_2=0,b_3=1/e $ .

因此
$$\begin{align*}& 1 – \frac{
{
{2^3}}}{
{1!}} + \frac{
{
{3^3}}}{
{2!}} – \frac{
{
{4^3}}}{
{3!}} + \cdots = \sum\limits_{n = 0}^\infty {
{
{\left( { – 1} \right)}^n}\frac{
{
{
{\left( {n + 1} \right)}^3}}}{
{n!}}} \\=& {b_3} + 3{b_2} + 3{b_1} + {b_0} = – \frac{1}{e}.\end{align*}$$

4.求下列级数的和:(1) $ \sum\limits_{n = 1}^\infty {\arctan \frac{1}{
{2{n^2}}}} $ ; (2) $ \sum\limits_{n = 1}^\infty {\arctan \frac{2}{
{
{n^2}}}} $ .
解:事实上
\[\sum\limits_{n = 1}^\infty {\arctan \frac{1}{
{2{n^2}}}} = \sum\limits_{n = 1}^\infty {\left( {\arctan \frac{1}{
{2n – 1}} – \arctan \frac{1}{
{2n + 1}}} \right)} = \frac{\pi }{4}.\]
而
\[\sum\limits_{n = 1}^\infty {\arctan \frac{2}{
{
{n^2}}}} = \sum\limits_{n = 1}^\infty {\left( {\arctan \frac{1}{
{n – 1}} – \arctan \frac{1}{
{n + 1}}} \right)} = \frac{\pi }{2} + \frac{\pi }{4} = \frac{
{3\pi }}{4}.\]
5.设 $ a>1 $ ,求 $ \sum\limits_{n = 0}^\infty {\frac{
{
{2^n}}}{
{
{a^{
{2^n}}} + 1}}} $ 的和.
解:事实上
$$\begin{align*}\sum\limits_{n = 0}^\infty {\frac{
{
{2^n}}}{
{
{a^{
{2^n}}} + 1}}} &= \frac{1}{
{a + 1}} + \sum\limits_{n = 1}^\infty {\frac{
{
{2^n}}}{
{
{a^{
{2^n}}} + 1}}} = \frac{1}{
{a + 1}} – \frac{1}{
{a – 1}} + \frac{1}{
{a + 1}} + \sum\limits_{n = 1}^\infty {\frac{
{
{2^n}}}{
{
{a^{
{2^n}}} + 1}}} \\&= \frac{1}{
{a + 1}} – \frac{2}{
{
{a^2} – 1}} + \sum\limits_{n = 1}^\infty {\frac{
{
{2^n}}}{
{
{a^{
{2^n}}} + 1}}} = \frac{1}{
{a + 1}} – \frac{
{
{2^2}}}{
{
{a^{
{2^2}}} – 1}} + \sum\limits_{n = 2}^\infty {\frac{
{
{2^n}}}{
{
{a^{
{2^n}}} + 1}}} \\&= \frac{1}{
{a + 1}} – \mathop {\lim }\limits_{n \to \infty } \frac{
{
{2^{n + 1}}}}{
{
{a^{
{2^{n + 1}}}} – 1}} = \frac{1}{
{a + 1}}.\end{align*}$$

6.求 $ 1 + \frac{1}{3} – \frac{1}{5} – \frac{1}{7} + \frac{1}{9} + \frac{1}{
{11}} – \cdots $ 的和.
解:
$$\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{
{8n – 7}} + \frac{1}{
{8n – 5}} – \frac{1}{
{8n – 3}} – \frac{1}{
{8n – 1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {
{x^{8n – 8}} + {x^{8n – 6}} – {x^{8n – 4}} – {x^{8n – 2}}} \right)} } \\=& \int_0^1 {\sum\limits_{n = 1}^\infty {\left( {
{x^{8n – 8}} + {x^{8n – 6}} – {x^{8n – 4}} – {x^{8n – 2}}} \right)} dx} = \int_0^1 {\frac{
{1 + {x^2} – {x^4} – {x^6}}}{
{1 – {x^8}}}dx} \\= &\left. {\frac{
{\arctan \left( {1 + \sqrt 2 x} \right) – \arctan \left( {1 – \sqrt 2 x} \right)}}{
{\sqrt 2 }}} \right|_0^1 = \frac{\pi }{
{2\sqrt 2 }}.\end{align*}$$

7.求 $ 1 – \frac{1}{7} + \frac{1}{9} – \frac{1}{
{15}} + \cdots $ 的和.
解:
$$\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{
{8n – 7}} – \frac{1}{
{8n – 1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {
{x^{8n – 8}} – {x^{8n – 2}}} \right)} } \\=& \int_0^1 {\sum\limits_{n = 1}^\infty {\left( {
{x^{8n – 8}} – {x^{8n – 2}}} \right)} dx} = \int_0^1 {\frac{
{1 – {x^6}}}{
{1 – {x^8}}}dx} \\= &\left. {\frac{
{2\arctan x + \sqrt 2 \arctan \left( {1 + \sqrt 2 x} \right) – \arctan \left( {1 – \sqrt 2 x} \right)}}{4}} \right|_0^1 = \frac{
{\sqrt 2 + 1}}{8}\pi .\end{align*}$$

8.求 $ 1 – \frac{1}{4} + \frac{1}{7} – \frac{1}{
{10}} + \cdots $ 的和.
解:
$$\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{
{6n – 5}} – \frac{1}{
{6n – 2}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {
{x^{6n – 6}} – {x^{6n – 3}}} \right)} } \\= &\int_0^1 {\sum\limits_{n = 1}^\infty {\left( {
{x^{6n – 6}} – {x^{6n – 3}}} \right)} dx} = \int_0^1 {\frac{
{1 – {x^3}}}{
{1 – {x^6}}}dx} = \int_0^1 {\frac{1}{
{1 + {x^3}}}dx} \\=& \left. {\left( { – \frac{1}{6}\ln \left( {
{x^2} – x + 1} \right) + \frac{1}{3}\ln \left( {x + 1} \right) + \frac{
{\arctan \frac{
{2x – 1}}{
{\sqrt 3 }}}}{
{\sqrt 3 }}} \right)} \right|_0^1 = \frac{
{\sqrt 3 \pi + 3\ln 2}}{9}.\end{align*}$$

9.设 $ {a_n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n},n = 1,2, \cdots $ ,求 $ \sum\limits_{n = 1}^\infty {\frac{
{
{a_n}}}{
{n\left( {n + 1} \right)}}} $ 的和.
解:
$$\begin{align*}&\sum\limits_{n = 1}^\infty {\frac{
{
{a_n}}}{
{n\left( {n + 1} \right)}}} = \sum\limits_{n = 1}^\infty {\frac{
{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{
{n\left( {n + 1} \right)}}} \\=&\sum\limits_{n = 1}^\infty {\left( {\frac{
{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{n} – \frac{
{1 + \frac{1}{2} + \cdots + \frac{1}{
{n + 1}}}}{
{n + 1}}} \right)} + \sum\limits_{n = 1}^\infty {\frac{1}{
{
{
{\left( {n + 1} \right)}^2}}}} \\= & 1 – \mathop {\lim }\limits_{n \to \infty } \frac{
{1 + \frac{1}{2} + \cdots + \frac{1}{
{n + 1}}}}{
{n + 1}} + \left( {\frac{
{
{\pi ^2}}}{6} – 1} \right) = \frac{
{
{\pi ^2}}}{6} – \mathop {\lim }\limits_{n \to \infty } \frac{
{\frac{1}{
{n + 2}}}}{1} = \frac{
{
{\pi ^2}}}{6}.\end{align*}$$

10.求 $ \sum\limits_{n = 0}^\infty {\left( {\frac{1}{
{4n + 1}} + \frac{1}{
{4n + 3}} – \frac{1}{
{2n + 2}}} \right)} $ 的和.
解:
$$\begin{align*}&\sum\limits_{n = 0}^\infty {\left( {\frac{1}{
{4n + 1}} + \frac{1}{
{4n + 3}} – \frac{1}{
{2n + 2}}} \right)} = \sum\limits_{n = 0}^\infty {\int_0^1 {\left( {
{x^{4n}} + {x^{4n + 2}} – {x^{2n + 1}}} \right)} } \\= &\int_0^1 {\sum\limits_{n = 0}^\infty {\left( {
{x^{4n}} + {x^{4n + 2}} – {x^{2n + 1}}} \right)} dx} = \int_0^1 {\left( {\frac{
{1 + {x^2}}}{
{1 – {x^4}}} – \frac{x}{
{1 – {x^2}}}} \right)dx} \\=& \int_0^1 {\frac{1}{
{1 + x}}dx} = \ln 2.\end{align*}$$

11.求 $ 1 – \frac{1}{4} + \frac{1}{6} – \frac{1}{9} + \frac{1}{
{11}} – \frac{1}{
{14}} + \cdots $ 的和.
解:
$$\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{
{5n – 4}} – \frac{1}{
{5n – 1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {
{x^{5n – 5}} – {x^{5n – 2}}} \right)dx} } \\=& \int_0^1 {\sum\limits_{n = 1}^\infty {\left( {
{x^{5n – 5}} – {x^{5n – 2}}} \right)} dx} = \int_0^1 {\frac{
{1 – {x^3}}}{
{1 – {x^5}}}dx} \\= &\left. {\left( {\frac{
{\left( {5 – \sqrt 5 } \right)/10}}{
{
{x^2} + \frac{
{\sqrt 5 + 1}}{2}x + 1}} + \frac{
{\left( {5 + \sqrt 5 } \right)/10}}{
{
{x^2} + \frac{
{ – \sqrt 5 + 1}}{2}x + 1}}} \right)} \right|_0^1 = \frac{
{\sqrt {25 + 10\sqrt 5 }}}{
{25}}\pi .\end{align*}$$

12.求 $ \frac{
{
{x^3}}}{
{3!}} + \frac{
{
{x^9}}}{
{9!}} + \frac{
{
{x^{15}}}}{
{15!}} + \cdots $ 的和函数.
解:事实上,方程 $ \omega^3=1 $ 有三个根 $ 1,{ – \frac{1}{2} + \frac{
{\sqrt 3 i}}{2}},{ – \frac{1}{2} – \frac{
{\sqrt 3 i}}{2}} $ .利用 $ \sinh $ 便可得到所需函数
$$\begin{align*}&\frac{
{\sinh x + \sinh \left( { – \frac{1}{2} + \frac{
{\sqrt 3 i}}{2}} \right)x + \sinh \left( { – \frac{1}{2} – \frac{
{\sqrt 3 i}}{2}} \right)x}}{3}\\= & – \frac{2}{3}\sinh \frac{x}{2}\cos \frac{
{\sqrt 3 x}}{2} + \frac{
{\sinh x}}{3} = \frac{
{
{x^3}}}{
{3!}} + \frac{
{
{x^9}}}{
{9!}} + \frac{
{
{x^{15}}}}{
{15!}} + \cdots .\end{align*}$$

我们还有

$$\begin{align*}&{\frac{
{\sin x +\sin \left( { – \frac{1}{2} + \frac{
{\sqrt 3 i}}{2}} \right)x + \sin \left( { – \frac{1}{2} – \frac{
{\sqrt 3 i}}{2}} \right)x}}{
{ – 3}}}\\= &\frac{2}{3}\sin \frac{x}{2}\cosh \frac{
{\sqrt 3 x}}{2} – \frac{
{\sin x}}{3} = \frac{
{
{x^3}}}{
{3!}} – \frac{
{
{x^9}}}{
{9!}} + \frac{
{
{x^{15}}}}{
{15!}} – \frac{
{
{x^{21}}}}{
{21!}} + \cdots .\end{align*}$$

13.求 $ \sum\limits_{n = 1}^\infty {\frac{
{
{
{\left[ {\left( {n – 1} \right)!} \right]}^2}}}{
{\left( {2n} \right)!}}{
{\left( {2x} \right)}^{2n}}} $ 的和函数.
解:在 $ |x|<1 $ 上对 $ S(x) $ 逐项求导,知 $ S’\left( x \right) = 2\sum\limits_{n = 1}^\infty {\frac{
{
{
{\left[ {\left( {n – 1} \right)!} \right]}^2}}}{
{\left( {2n – 1} \right)!}}{
{\left( {2x} \right)}^{2n – 1}}} $ ,且 $ S”\left( x \right) = 4\sum\limits_{n = 1}^\infty {\frac{
{
{
{\left[ {\left( {n – 1} \right)!} \right]}^2}}}{
{\left( {2n – 2} \right)!}}{
{\left( {2x} \right)}^{2n – 2}}} $ .由此可得 $ (1-x^2)S”(x)-xS'(x)=4 $ .在两端乘以 $ {(1-x^2)}^{-1/2} $ ,我们有
\[{\left( {\sqrt {1 – {x^2}} S’\left( x \right)} \right)^\prime } = \frac{4}{
{\sqrt {1 – {x^2}} }},\]故
\[S\left( x \right) = \frac{
{4\arcsin x}}{
{\sqrt {1 – {x^2}} }} + \frac{1}{
{\sqrt {1 – {x^2}} }},\quad \left| x \right| < 1.\]

14.求 $ \sum\limits_{n = 1}^\infty {\frac{
{
{x^{n + 1}}}}{
{\left( {1 – {x^n}} \right)\left( {1 – {x^{n + 1}}} \right)}}} $ 的和函数.
解:注意到
$$\begin{align*}&\left( {1 – \frac{1}{x}} \right)\sum\limits_{n = 1}^\infty {\frac{
{
{x^{n + 1}}}}{
{\left( {1 – {x^n}} \right)\left( {1 – {x^{n + 1}}} \right)}}} \\=& \sum\limits_{n = 1}^\infty {\frac{
{
{x^{n + 1}}}}{
{\left( {1 – {x^n}} \right)\left( {1 – {x^{n + 1}}} \right)}}} – \sum\limits_{n = 1}^\infty {\frac{
{
{x^n}}}{
{\left( {1 – {x^n}} \right)\left( {1 – {x^{n + 1}}} \right)}}} \\= &\sum\limits_{n = 1}^\infty {\frac{
{
{x^{n + 1}} – {x^n}}}{
{\left( {1 – {x^n}} \right)\left( {1 – {x^{n + 1}}} \right)}}} = \sum\limits_{n = 1}^\infty {\left( {\frac{1}{
{1 – {x^{n + 1}}}} – \frac{1}{
{1 – {x^n}}}} \right)} \\=& \mathop {\lim }\limits_{n \to \infty } \frac{1}{
{1 – {x^{n + 1}}}} – \frac{1}{
{1 – x}} = \begin{cases}\frac{1}{
{x – 1}},&\left| x \right| > 1\\\frac{x}{
{x – 1}},&\left| x \right| < 1\end{cases} .\end{align*}$$
因此
\[\sum\limits_{n = 1}^\infty {\frac{
{
{x^{n + 1}}}}{
{\left( {1 – {x^n}} \right)\left( {1 – {x^{n + 1}}} \right)}}} = \begin{cases}\frac{x}{
{
{
{\left( {x – 1} \right)}^2}}}, &\left| x \right| > 1\\\frac{
{
{x^2}}}{
{
{
{\left( {x – 1} \right)}^2}}}, &\left| x \right| < 1\end{cases} .\]

15.设 $ \sum\limits_{n = 1}^\infty {\frac{1}{
{
{a_n}}}} $ 为发散的正项级数, $ x>0 $ ,求 $ \sum\limits_{n = 1}^\infty {\frac{
{
{a_1}{a_2} \cdots {a_n}}}{
{\left( {
{a_2} + x} \right) \cdots \left( {
{a_{n + 1}} + x} \right)}}} $ 的和函数.
解:首先,
$$\begin{align*}&\sum\limits_{n = 1}^\infty {\frac{
{
{a_1}{a_2} \cdots {a_n}}}{
{\left( {
{a_2} + x} \right) \cdots \left( {
{a_{n + 1}} + x} \right)}}} \\=& \frac{
{
{a_1}}}{
{
{a_2} + x}} + \frac{1}{x}\sum\limits_{n = 2}^\infty {\left[ {\frac{
{
{a_1}{a_2} \cdots {a_n}}}{
{\left( {
{a_2} + x} \right) \cdots \left( {
{a_n} + x} \right)}} – \frac{
{
{a_1}{a_2} \cdots {a_{n + 1}}}}{
{\left( {
{a_2} + x} \right) \cdots \left( {
{a_{n + 1}} + x} \right)}}} \right]} \\=& \frac{
{
{a_1}}}{
{
{a_2} + x}} + \frac{1}{x}\left[ {\frac{
{
{a_1}{a_2}}}{
{
{a_2} + x}} – \mathop {\lim }\limits_{n \to \infty } \frac{
{
{a_1}{a_2} \cdots {a_{n + 1}}}}{
{\left( {
{a_2} + x} \right) \cdots \left( {
{a_{n + 1}} + x} \right)}}} \right].\end{align*}$$
当 $ n $ 足够大时,\[1 + \frac{x}{
{
{a_{n + 1}}}} \sim {e^{x/{a_{n + 1}}}}.\]
因此 $ {\left( {1 + \frac{x}{
{
{a_2}}}} \right) \cdots \left( {1 + \frac{x}{
{
{a_{n + 1}}}}} \right)} $ 与 $ \exp \left\{ {x\sum\limits_{n = 1}^\infty {\frac{1}{
{
{a_n}}}} } \right\} $ 具有相同的收敛性,均发散,故
\[\mathop {\lim }\limits_{n \to \infty } \frac{
{
{a_1}{a_2} \cdots {a_{n + 1}}}}{
{\left( {
{a_2} + x} \right) \cdots \left( {
{a_{n + 1}} + x} \right)}} = \mathop {\lim }\limits_{n \to \infty } \frac{
{
{a_1}}}{
{\left( {1 + \frac{x}{
{
{a_2}}}} \right) \cdots \left( {1 + \frac{x}{
{
{a_{n + 1}}}}} \right)}} = 0.\]
从而
\[\sum\limits_{n = 1}^\infty {\frac{
{
{a_1}{a_2} \cdots {a_n}}}{
{\left( {
{a_2} + x} \right) \cdots \left( {
{a_{n + 1}} + x} \right)}}} = \frac{
{
{a_1}}}{
{
{a_2} + x}} + \frac{
{
{a_1}{a_2}}}{
{x\left( {
{a_2} + x} \right)}} = \frac{
{
{a_1}}}{x}.\]

16.设 $ x>1 $ ,求 $ \frac{x}{
{x + 1}} + \frac{
{
{x^2}}}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right)}} + \frac{
{
{x^4}}}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right)\left( {
{x^4} + 1} \right)}} + \cdots $ 的和函数.
解:$$\begin{align*}I &= \left( {1 – \frac{1}{
{x + 1}}} \right) + \frac{
{
{x^2}}}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right)}} + \frac{
{
{x^4}}}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right)\left( {
{x^4} + 1} \right)}} + \cdots \\&= 1 + \left( { – \frac{1}{
{x + 1}} + \frac{
{
{x^2}}}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right)}}} \right) + \frac{
{
{x^4}}}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right)\left( {
{x^4} + 1} \right)}} + \cdots \\&= 1 – \frac{1}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right)}} + \frac{
{
{x^4}}}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right)\left( {
{x^4} + 1} \right)}} + \cdots \\&= 1 – \frac{1}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right)\left( {
{x^4} + 1} \right)}} + \cdots \\&= \cdots = 1 – \mathop {\lim }\limits_{n \to \infty } \frac{1}{
{\left( {x + 1} \right)\left( {
{x^2} + 1} \right) \cdots \left( {
{x^{
{2^{n – 1}}}} + 1} \right)}} = 1.\end{align*}$$