许以超第二版书上P73页留了个行列式求解的思考题,还是比较棘手的,下面给出自己的解答.

求

\[{\Delta _n} = \det \left( {\begin{array}{*{20}{c}}{1 + {x_1}{y_1}}&{1 + {x_1}{y_2}}& \cdots &{1 + {x_1}{y_n}}\\{1 + {x_1}y_1^2}&{1 + {x_1}y_2^2}& \cdots &{1 + {x_1}y_n^2}\\\vdots & \vdots &{}& \vdots \\{1 + {x_1}y_1^n}&{1 + {x_1}y_2^n}& \cdots &{1 + {x_1}y_n^n}\end{array}} \right).\]

解:先来个引理(许以超自己给出的).

事实上,我们有

\begin{align*}&\det A + x\sum\limits_{j,k = 1}^n {{A_{jk}}} = \det \left( {\begin{array}{*{20}{c}}{{a_{11}} + x}& \cdots &{{a_{1n}} + x}\\\vdots &{}& \vdots \\{{a_{n1}} + x}& \cdots &{{a_{nn}} + x}\end{array}} \right)\\=& \det A + x\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1&1\\{{a_{21}} - {a_{11}}}&{{a_{22}} - {a_{12}}}& \cdots &{{a_{2,n - 1}} - {a_{1,n - 1}}}&{{a_{2n}} - {a_{1n}}}\\{{a_{31}} - {a_{21}}}&{{a_{32}} - {a_{22}}}& \cdots &{{a_{3,n - 1}} - {a_{2,n - 1}}}&{{a_{3n}} - {a_{2n}}}\\\vdots & \vdots &{}& \vdots & \vdots \\{{a_{n1}} - {a_{n - 1,1}}}&{{a_{n2}} - {a_{n - 1,2}}}& \cdots &{{a_{n,n - 1}} - {a_{n - 1,n - 1}}}&{{a_{nn}} - {a_{n - 1,n}}}\end{array}} \right).\end{align*}

其中$A_{kj}$是方阵$A=(a_{jk})$的第$k$行,第$j$列位置的元素的代数余子式.

因此

\begin{align*}&{\Delta _n} = \det \left( {\begin{array}{*{20}{c}}{1 + {x_1}{y_1}}&{1 + {x_1}{y_2}}& \cdots &{1 + {x_1}{y_n}}\\{1 + {x_1}y_1^2}&{1 + {x_1}y_2^2}& \cdots &{1 + {x_1}y_n^2}\\\vdots & \vdots &{}& \vdots \\{1 + {x_1}y_1^n}&{1 + {x_1}y_2^n}& \cdots &{1 + {x_1}y_n^n}\end{array}} \right)\\= &\det A + \det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\{{x_1}\left( {y_1^2 - {y_1}} \right)}&{{x_1}\left( {y_2^2 - {y_2}} \right)}& \cdots &{x_1\left( {y_n^2 - {y_n}} \right)}\\\vdots & \vdots &{}& \vdots \\{{x_1}\left( {y_1^n - y_1^{n - 1}} \right)}&{{x_1}\left( {y_2^n - y_2^{n - 1}} \right)}& \cdots &{{x_1}\left( {y_n^n - y_n^{n - 1}} \right)}\end{array}} \right)\\=& \det A + x_1^{n - 1}\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\{y_1^2 - {y_1}}&{y_2^2 - {y_2}}& \cdots &{y_n^2 - {y_n}}\\\vdots & \vdots &{}& \vdots \\{y_1^n - y_1^{n - 1}}&{y_2^n - y_2^{n - 1}}& \cdots &{y_n^n - y_n^{n - 1}}\end{array}} \right).\end{align*}

对上面的行列式进行升阶:

\[\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\{y_1^2 - {y_1}}&{y_2^2 - {y_2}}& \cdots &{y_n^2 - {y_n}}\\\vdots & \vdots &{}& \vdots \\{y_1^n - y_1^{n - 1}}&{y_2^n - y_2^{n - 1}}& \cdots &{y_n^n - y_n^{n - 1}}\end{array}} \right) = \det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\0&1&1& \cdots &1\\0&{y_1^2 - {y_1}}&{y_2^2 - {y_2}}& \cdots &{y_n^2 - {y_n}}\\\vdots & \vdots & \vdots &{}& \vdots \\0&{y_1^n - y_1^{n - 1}}&{y_2^n - y_2^{n - 1}}& \cdots &{y_n^n - y_n^{n - 1}}\end{array}} \right).\]

将第一行加到第三行,第三行加到第四行,$\cdots$,最后将第$n$行加到第$n+1$行:

\begin{align*}\det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\0&1&1& \cdots &1\\1&{y_1^2}&{y_2^2}& \cdots &{y_n^2}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{y_1^n}&{y_2^n}& \cdots &{y_n^n}\end{array}} \right) &= \det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\1&1&1& \cdots &1\\1&{y_1^2}&{y_2^2}& \cdots &{y_n^2}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{y_1^n}&{y_2^n}& \cdots &{y_n^n}\end{array}} \right) + \det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\{ - 1}&0&0& \cdots &0\\1&{y_1^2}&{y_2^2}& \cdots &{y_n^2}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{y_1^n}&{y_2^n}& \cdots &{y_n^n}\end{array}} \right)\\& = \left( { - 1} \right) \cdot \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} \cdot \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} + \prod\limits_{k = 1}^n {{y_k}} \cdot \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \\& = \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \prod\limits_{k = 1}^n {{y_k}} } \right].\end{align*}

因此

\begin{align*}{\Delta _n} &= \det A + x_1^{n - 1}\prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \prod\limits_{k = 1}^n {{y_k}} } \right]\\& = x_1^n \cdot \prod\limits_{k = 1}^n {{y_k}} \cdot \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} + x_1^{n - 1}\prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \prod\limits_{k = 1}^n {{y_k}} } \right]\\& = x_1^{n - 1}\prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \left( {{x_1} + 1} \right)\prod\limits_{k = 1}^n {{y_k}} } \right].\end{align*}

1.证明: $\sum_{k=1}^n\sin k^2$无界.

参看: http://www.zhihu.com/question/29094450

2.证明\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\sin \sqrt k }  = 0.\]

TeX数学公式网页预览版

http://jaxedit.com/note/

“数学是你们的选择，你们随时都可以放弃。但当数学仍是你们的选择时，就必须为此负责。”

——S.Lang对他学生上课前说的话

大风起兮云飞扬，一生挚爱美娇娘.

爱情，本来就是勇敢者的游戏！

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} = \int_0^1 {\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)dx} \\=& \left. {\left[ {\frac{{5 - \sqrt 5 }}{{10}}\sqrt {\frac{{10 + 2\sqrt 5 }}{5}} \arctan \frac{{4x + \sqrt 5 + 1}}{{\sqrt {10 - 2\sqrt 5 } }} + \frac{{5 + \sqrt 5 }}{5}\sqrt {\frac{2}{{5 + \sqrt 5 }}} \arctan \frac{{4x - \sqrt 5 + 1}}{{\sqrt {10 + 2\sqrt 5 } }}} \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*}

1.Catalan数

设数列$\{a_n\}$满足$a_0=1$,以及

\[a_n=a_0a_{n-1}+a_1a_{n-2}+\cdots+a_{n-1}a_0,\quad n\geq1,\]

试求$a_n$的表达式.(详见周民强第二册P378)

解.令$f\left( x \right) = {a_0} + {a_1}x +  \cdots  + {a_n}{x^n} +  \cdots $,以及

证.利用幂级数展开式如下:

今天13级数院的JJ发来提问:

求对称矩阵

\[A = \left( {\begin{array}{*{20}{c}}{a_{11}^2}&{{a_{11}}{a_{12}} + 1}&{{a_{11}}{a_{13}} + 1}& \cdots &{{a_{11}}{a_{1n}} + 1}\\{{a_{11}}{a_{12}} + 1}&{a_{22}^2}&{{a_{22}}{a_{23}} + 1}& \cdots &{{a_{22}}{a_{2n}} + 1}\\{{a_{11}}{a_{13}} + 1}&{{a_{22}}{a_{23}} + 1}& \ddots & \ddots & \vdots \\\vdots & \vdots & \ddots &{a_{n - 1,n - 1}^2}&{{a_{n - 1,n - 1}}{a_{n - 1,n}} + 1}\\{{a_{11}}{a_{1n}} + 1}&{{a_{22}}{a_{2n}} + 1}& \cdots &{{a_{n - 1,n - 1}}{a_{n - 1,n}} + 1}&{a_{nn}^2}\end{array}} \right)\]的特征值.

西西在大学群里的一道题,也是2014年第六届非数竞赛预赛最后一题的推广:

设${A_n} = \frac{n}{{{n^2} + 1}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}}$,求极限

\[\mathop {\lim }\limits_{n \to \infty } {n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right).\]

这里提供个一般的方法.

Euler-Maclaurin求和公式

设函数$f\in C^{(2m+2)}[a,b],h=\frac{b-a}{n},x_i=a+ih,i=0,1,\cdots,n$,则

\begin{align*}\frac{{b - a}}{n}\sum\limits_{i = 1}^n {\frac{1}{2}\left[ {f\left( {{x_{i - 1}}} \right) + f\left( {{x_i}} \right)} \right]} - \int_a^b {f\left( x \right)dx} = &\sum\limits_{k = 1}^m {\frac{{{B_{2k}}}}{{\left( {2k} \right)!}}{h^{2k}}\left[ {{f^{\left( {2k - 1} \right)}}\left( b \right) - {f^{\left( {2k - 1} \right)}}\left( a \right)} \right]} \\&+ \frac{{{B_{2m + 2}}}}{{\left( {2m + 2} \right)!}}{h^{2m + 2}}{f^{\left( {2m + 2} \right)}}\left( \xi \right)\left( {b - a} \right),\end{align*}

其中$\xi\in [a,b]$, $B_{2k}(k=1,2,\cdots,m+1)$是Bernoulli数且$B_2=\frac16,B_4=-\frac{1}{30},B_6=\frac{1}{42}$.

解:取$a=0,b=1,f(x)=\frac{1}{1+x^2}$,则$h=\frac1n,x_i=\frac{i}{n},A_n=\frac{1}{n}\sum\limits_{i = 1}^n {f\left( {{x_i}} \right)}$,则

\begin{align*}&{A_n} + \frac{1}{{4n}} - \frac{\pi }{4} = \frac{1}{2}\left[ {\left( {{A_n} - \frac{1}{{2n}} + \frac{1}{n}} \right) + {A_n}} \right] - \frac{\pi }{4} = \frac{{{B_2}}}{{2!}} \cdot \frac{1}{{{n^2}}}\left[ {f'\left( 1 \right) - f'\left( 0 \right)} \right]\\+ &\frac{{{B_4}}}{{4!}} \cdot \frac{1}{{{n^4}}}\left[ {f'''\left( 1 \right) - f'''\left( 0 \right)} \right] + \frac{{{B_6}}}{{6!}} \cdot \frac{1}{{{n^6}}}\left[ {{f^{\left( 5 \right)}}\left( 1 \right) - {f^{\left( 5 \right)}}\left( 0 \right)} \right] + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^8}}}{f^{\left( 8 \right)}}\left( \xi \right),\end{align*}

其中$\xi\in [0,1]$,也即

\[{n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right) = \frac{1}{{2016}} + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^2}}}{f^{\left( 8 \right)}}\left( \xi \right),\]

注意到${f^{\left( 8 \right)}}\left( \xi \right)$有界,因此$n\to\infty$时,所求极限为$\frac{1}{{2016}}$.

6.设空间体积为$V$的任意$\Omega,X_0\in \Omega ,0<\alpha<3$.证明

\[\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \le C{V^{\alpha /3}}, \text{其中$C$只与$\alpha$有关}.\]

证:(Veer)由于$\alpha-3>-3$且$\displaystyle \int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX}$为三重积分,故积分广义可积.

在$X_0$处作一以$X_0$为圆心的球$D$,使其体积为$V_D=V$.设$D$的半径为$R$.记$D_1=D\cap \Omega,D_2=D/D_1,\Omega_2=\Omega/D_1$,则易知$V_{D_2}=V_{\Omega_2}$.显然

\begin{align*}\int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= \int_{{D_1}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} + \int_{{D_2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \\\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= \int_{{D_1}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} + \int_{{\Omega _2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} .\end{align*}

由积分中值定理有

\begin{align*}\int_{{D_2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= {\left| {\xi - {X_0}} \right|^{\alpha - 3}}{V_{{D_2}}},\xi \in {D_2}\\\int_{{\Omega _2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= {\left| {\eta - {X_0}} \right|^{\alpha - 3}}{V_{{\Omega_2}}},\eta \in {\Omega _2}.\end{align*}

易知${\left| {\xi - {X_0}} \right|^{\alpha - 3}} \ge {\left| {\eta - {X_0}} \right|^{\alpha - 3}}$.又因为${V_{{D_2}}} = {V_{{\Omega _2}}}$,则\[\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \le \int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} .\]由球坐标变换易得\[\int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} = \int_0^\pi {d\varphi } \int_0^{2\pi } {d\theta } \int_0^R {{r^{\alpha - 1}}\sin \varphi dr} = 4\pi \frac{{{R^\alpha }}}{\alpha }.\]又因为$\displaystyle {V_D} = V = \frac{4}{3}\pi {R^3}$,则\[\int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} = \frac{{{3^{\alpha /3}}{{\left( {4\pi } \right)}^{1 - \alpha /3}}}}{\alpha }{V^{\alpha /3}}.\]故\[\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \le \frac{{{3^{\alpha /3}}{{\left( {4\pi } \right)}^{1 - \alpha /3}}}}{\alpha }{V^{\alpha /3}},\]取$\displaystyle C = \frac{{{3^{\alpha /3}}{{\left( {4\pi } \right)}^{1 - \alpha /3}}}}{\alpha }$.

注:从上可以看出当$\Omega=D$时不等式可取等号,故$C$是最佳的,且此题可推广到$n$维上.

7.$f(x)$在$[0,1]$单增,证明:

\[\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right).\]

证:这是Dirichlet引理,菲赫金哥尔茨的《微积分教程》第三卷P358有详细的证明.另外,汪林的《数学分析问题研究与评注》P147上有他的推广及其证明.

对任意给出的$\varepsilon>0$, $\exists 0<\delta<1$,使得对于$0<t\leq \delta$,

\[0 \le g\left( t \right) - g\left( {{0_ + }} \right) < M_1\varepsilon ,\]

其中$M_1$是任意给定的常数.

考察积分

\begin{align*}\int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} &= \left( {\int_0^\delta {} + \int_\delta ^1 {} } \right)\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx\\&= {I_1} + {I_2}.\end{align*}

对于$I_1$,运用积分第二中值定理,我们有

\[{I_1} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_\eta ^\delta {\frac{{\sin xy}}{x}dx} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} ,\]

其中第二个因子对于一切值$y$一致有界.事实上,由反常积分$\displaystyle \int_0^\infty {\frac{{\sin z}}{z}dz}$的收敛性,可见当$z\to \infty$时, $z(z\geq 0)$的连续函数$\displaystyle \int_0^z {\frac{{\sin z}}{z}dz} $有有限的极限,并且对于一切值$z$有界

\[\left| {\int_0^z {\frac{{\sin z}}{z}dz} } \right| \le L\left( L \text{为常数}\right),\]从而

\[\left| {\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} } \right| = \left| {\int_0^{y\delta } {} + \int_0^{y\eta } {} } \right| \le 2L.\]

对于第一个因子,取$M_1=\frac{1 }{{4L}}$,则有$f\left( \delta \right) - f\left( {{0_ + }} \right) < \frac{\varepsilon }{{4L}}$.

因此\[\left| {{I_1}} \right| \le \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\left| {\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} } \right| < \frac{\varepsilon }{{4L}} \cdot 2L = \frac{\varepsilon }{2}.\]

至于$I_2$,由于$\displaystyle \int_\delta ^1 {\frac{{f\left( x \right) - f\left( {{0_ + }} \right)}}{x}dx} $存在,由Riemann-Lebesgue引理可知$\mathop {\lim }\limits_{y \to \infty } {I_2} = 0$,即对$\varepsilon >0,\exists M_2>0$,使得$y>M_2$时,有$\left| {{I_2}} \right| < \frac{\varepsilon }{2}$.

因此\[\left| {\int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} } \right| \le \left| {{I_1}} \right| + \left| {{I_2}} \right| < \varepsilon .\]

即\[\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} = 0.\]

从而

\begin{align*}&\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right)\\=& \mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} + f\left( {{0_ + }} \right)\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\frac{{\sin xy}}{x}dx} \\= &0 + f\left( {{0_ + }} \right)\int_0^{ + \infty } {\frac{{\sin z}}{z}dz} = \frac{\pi }{2}f\left( {{0_ + }} \right).\end{align*}