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:$$\begin{align*}I
&= \left( {1 - \frac{1}{{x + 1}}} \right) + \frac{{{x^2}}}{{\left( {x + 1}
\right)\left( {{x^2} + 1} \right)}} + \frac{{{x^4}}}{{\left( {x + 1}
\right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)}} + \cdots \\&=
1 + \left( { - \frac{1}{{x + 1}} + \frac{{{x^2}}}{{\left( {x + 1} \right)\left(
{{x^2} + 1} \right)}}} \right) + \frac{{{x^4}}}{{\left( {x + 1} \right)\left(
{{x^2} + 1} \right)\left( {{x^4} + 1} \right)}} + \cdots \\&= 1 -
\frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} + \frac{{{x^4}}}{{\left(
{x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)}} + \cdots
\\&= 1 - \frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)\left(
{{x^4} + 1} \right)}} + \cdots \\&= \cdots = 1 - \mathop {\lim }\limits_{n
\to \infty } \frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right) \cdots
\left( {{x^{{2^{n - 1}}}} + 1} \right)}} = 1.\end{align*}$$

源自: http://www.math.org.cn/forum.php?mod=viewthread&tid=35174 [未验证其正确性, 仅供参考]