哆嗒数学网里代数龙发的一系列级数题

练习题4.设$(0,+\infty)$上的函数列$f_n$由下式定义：$$f_1(x)=x,f_{n+1}(x)=(f_n(x)+\frac{1}{n})f_n(x).$$证明：存在唯一的正数$a$，使得对于所有$n$，$$0<f_n(x)<f_{n+1}(a)<1.$$

练习题5.$\displaystyle\sum\limits_{n=1}^{\infty}a_n$为正项收敛级数，$\displaystyle r_n=\sum\limits_{k=n}^{\infty}a_k，0<p<1$，证明：$$\sum\limits_{n=1}^{\infty}\frac{a_n}{r_n^p}<\frac{1}{1-p}\left(\sum\limits_{n=1}^{\infty}a_n \right)^{1-p}.$$

练习题6.设$a>0,a_n$是一个数列，并且$a_n>0,a_{n+1}\ge a_n$，证明：$$\sum\limits_{n=1}^{\infty}\frac{a_n-a_{n-1}}{a_na_{n-1}^a}$$收敛.

证:首先可以确定给定的级数是正项级数.

(1)当$0<a<1$时,我们利用Lagrange中值定理,有\[\frac{{a_n^a - a_{n - 1}^a}}{{{a_n} - {a_{n - 1}}}} = a{\xi ^{a - 1}} \ge aa_n^{a - 1},\quad \xi  \in \left( {{a_{n - 1}},{a_n}} \right).\]

因此\[\frac{{{a_n} - {a_{n - 1}}}}{{{a_n}a_{n - 1}^a}} = \frac{{a_n^a - a_{n - 1}^a}}{{a_n^aa_{n - 1}^a}} \cdot \left( {\frac{{{a_n} - {a_{n - 1}}}}{{a_n^a - a_{n - 1}^a}} \cdot a_n^{a - 1}} \right) \le \frac{1}{a}\frac{{a_n^a - a_{n - 1}^a}}{{a_n^aa_{n - 1}^a}} = \frac{1}{a}\left( {\frac{1}{{a_{n - 1}^a}} - \frac{1}{{a_n^a}}} \right).\]

故\[\sum\limits_{n = 1}^\infty  {\frac{{{a_n} - {a_{n - 1}}}}{{{a_n}a_{n - 1}^a}}}  \le \frac{1}{a}\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{a_{n - 1}^a}} - \frac{1}{{a_n^a}}} \right)}  = \frac{1}{a}\left( {\frac{1}{{a_0^a}} - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^a}}} \right).\]

由于$\{a_n\}$是单增的正数列,则${\mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^a}}}$必定存在,由此可知原正项级数收敛;

(2)当$a\geq1$时,由\[\sum\limits_{n = 1}^\infty  {\frac{{{a_n} - {a_{n - 1}}}}{{{a_n}a_{n - 1}^a}}}  = \sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{a_{n - 1}^a}} - \frac{{a_{n - 1}^{1 - a}}}{{{a_n}}}} \right)}  \le \sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{a_{n - 1}^a}} - \frac{1}{{a_n^a}}} \right)}  = \frac{1}{a}\left( {\frac{1}{{a_0^a}} - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^a}}} \right)\]同样可知原正项级数收敛.

综上,级数$\sum\limits_{n=1}^{\infty}\frac{a_n-a_{n-1}}{a_na_{n-1}^a}$收敛.

练习题7.设$\displaystyle S(x)=\sum\limits_{n=1}^{\infty}\frac{2n}{(n^2 +x^2)^2}$，证明：$$\frac{1}{x^2 +\frac{1}{2\zeta(3)}}<S(x)<\frac{1}{x^2 +\frac{1}{6}},$$其中$\displaystyle \zeta(3)=\sum\limits_{n=1}^{\infty}\frac{1}{n^3}.$

练习题8.给定序列$\{a_n\}$，且$a_n$满足$a_1=2,a_2=8,a_n=4a_{n-1}-a_{n-2}(n=3,4,\ldots)$，证明：$$\sum\limits_{n=1}^{\infty}\text{arccot}\,\,a_n^2=\frac{\pi}{12}.$$

证.由${a_n} + {a_{n - 2}} = 4{a_{n - 1}}$可知\[{a_n}\left( {{a_n} + {a_{n - 2}}} \right) = 4{a_{n - 1}}{a_n} = {a_{n - 1}}\left( {{a_{n + 1}} + {a_{n - 1}}} \right),\]递推得\[a_n^2 - {a_{n + 1}}{a_{n - 1}} = a_{n - 1}^2 - {a_n}{a_{n - 2}} = \cdots = a_2^2 - {a_3}{a_1} = 4.\]

注意到$\mathrm{arccot\,} x$的一个公式

\[\mathrm{arccot\,} x-\mathrm{arccot\,} y=\mathrm{arccot\,}\left( \frac{1+xy}{y-x}\right).\]

因此有

\begin{align*}\mathrm{arccot\,} a_n^2 &= \mathrm{arccot\,} \frac{{{a_n} \cdot 4{a_n}}}{4} = \mathrm{arccot\,} \frac{{{a_n}\left( {{a_{n + 1}} + {a_{n - 1}}} \right)}}{4} =\mathrm{arccot\,} \frac{{{a_n}\left( {{a_{n + 1}} + {a_{n - 1}}} \right)}}{{a_n^2 - {a_{n + 1}}{a_{n - 1}}}}\\& = \mathrm{arccot\,} \frac{{1 + \frac{{{a_{n + 1}}}}{{{a_{n - 1}}}}}}{{\frac{{{a_n}}}{{{a_{n - 1}}}} - \frac{{{a_{n + 1}}}}{{{a_n}}}}} = \mathrm{arccot\,} \frac{{{a_{n + 1}}}}{{{a_n}}} -\mathrm{arccot\,} \frac{{{a_n}}}{{{a_{n - 1}}}}.\end{align*}

易得\[\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = 2 + \sqrt 3 .\]

故

\[\sum\limits_{n = 1}^\infty {\mathrm{arccot\,} a_n^2} = \mathrm{arccot\,} a_1^2 + \sum\limits_{n = 2}^\infty {\mathrm{arccot\,} a_n^2} = \mathop {\lim }\limits_{n \to \infty } \mathrm{arccot\,} \frac{{{a_{n + 1}}}}{{{a_n}}} - \mathrm{arccot\,} \frac{{{a_2}}}{{{a_1}}} + \mathrm{arccot\,} a_1^2 = \frac{\pi }{{12}}.\]

练习题9.设$\displaystyle a_n=\arctan \frac{1}{n^2 +n +1}$,证明: $$\sum\limits_{k=1}^{\infty}\frac{a_k^{1/2}}{k^2} \le \sqrt{\frac{\pi}{3}}.$$

证.(1)由柯西不等式我们得

\[\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} \sum\limits_{m = 1}^n {{a_m}}  \ge {\left( {1 + 2 +  \cdots  + n} \right)^2} = \frac{1}{4}{n^2}{\left( {n + 1} \right)^2},\]

即\[\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}} \le \frac{4}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} .\]

因此

\begin{align*}\sum\limits_{n = 1}^\infty  {\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}}}  &\le 4\sum\limits_{n = 1}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} }  = 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}} } \\&\le 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right]} }  = 2\sum\limits_{m = 1}^\infty  {\frac{1}{{{a_m}}}} .\end{align*}

这里用到了\[\frac{1}{{n{{\left( {n + 1} \right)}^2}}} \le \frac{1}{2}\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}} = \frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right].\]

注意到$a_n=n^\alpha,\alpha>1$时有

\[\mathop {\lim }\limits_{\alpha \to 1} \frac{{\sum\limits_{n = 1}^\infty {\frac{n}{{{a_1} + {a_2} + \cdots + {a_n}}}} }}{{\sum\limits_{j = 1}^\infty {\frac{1}{{{a_j}}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\sum\limits_{n = 1}^N {\frac{n}{{{1^\alpha } + {2^\alpha } + \cdots + {n^\alpha }}}} }}{{\sum\limits_{n = 1}^N {\frac{1}{{{n^\alpha }}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\frac{N}{{\frac{1}{{\alpha + 1}}{N^{\alpha + 1}} + O\left( {{N^\alpha }} \right)}}}}{{\frac{1}{{{N^\alpha }}}}} = 2.\]