14.(阿贝耳定理) 设$sumlimits_{n=0}^{infty}a_{n}=s$. 则$lim_{x o 1-}sumlimits_{n=0}^{infty}a_{n}x^{n}=s$.
证明: 容易看出$f(x)=sumlimits_{n=0}^{infty}a_{n}x^{n}$在$0 leq x leq 1$上一致收敛.
由Cauchy收敛准则知,任意$varepsilon>0$,存在$n$任意$p>0$有
$$left|sum_{k=n}^{n+p}a_{k}
ight|<varepsilon$$
由Abel引理(命题4)知
$$left|sum_{k=n}^{n+p}a_{k}x^{k}
ight|<x^{n}left|sum_{k=n}^{n+p}a_{k}
ight|leq varepsilon$$
由Cauchy收敛准则知$f(x)$一致收敛. 一致收敛的级数和保持连续性故
$$lim_{x o 1-}sum_{n=0}^{infty}a_{n}x^{n}=s$$
15. (Abel 级数乘法定理)令$c_{n}=a_{0}b_{n}+a_{1}b_{n-1}+cdots+a_{n}b_{0}$.又设$sum a_{n},sum b_{n}, sum c_{n}$都收敛. 则
$$sum_{n=0}^{infty}c_{n}=left(sum_{n=0}^{infty}a_{n}
ight)left(sum_{n=0}^{infty}b_{n}
ight)$$
证明: 因为绝对收敛的级数可以相乘,因此
$$sum_{n=0}^{infty}c_{n}x^{n}=left(sum_{n=0}^{infty}a_{n}x^{n}
ight)left(sum_{n=0}^{infty}b_{n}x^{n}
ight)=s_{1}(x)s_{2}(x)$$
由阿贝尔第二定理(命题14)知
$$sum_{n=0}^{infty}c_{n}=lim_{x o 1-}sum_{n=0}^{infty}c_{n}x^{n}=lim_{x o 1-}s_{1}(x)s_{2}(x)=left(sum_{n=0}^{infty}a_{n}
ight)left(sum_{n=0}^{infty}b_{n}
ight)$$
16. 试证
$$frac{1}{2}left(1-frac{1}{2}+frac{1}{3}-frac{1}{4}+cdots
ight)^{2}=frac{1}{2}-frac{1}{3}
left(1+frac{1}{2}
ight)+frac{1}{4}left(1+frac{1}{2}+frac{1}{3}
ight)-frac{1}{5}left(1+frac{1}{2}
+frac{1}{3}+frac{1}{4}
ight)+cdots$$
证明: 从左往右证是平凡的,从右式归为左式是非常考察眼力的.
取
$$a_{n}=b_{n}=(-1)^{n-1}frac{1}{n},\, (n=0,1,2,cdots),a_{0}=b_{0}=0$$
则
$$c_{n}=sum_{k=0}^{n}a_{k}b_{n-k}=(-1)^{n}sum_{k=1}^{n}frac{1}{k(n-k)}=frac{(-1)^{n}}{n}sum_{k=1}^{n}left(
frac{1}{k}+frac{1}{n-k}
ight)=(-1)^{n}frac{2}{n}sum_{k=1}^{n}frac{1}{k}$$
只需证明级数$sum c_{n}$收敛, 那么由命题15自然得到命题16成立.
由Euler极限(或Stolz定理)知
$$lim_{n oinfty}frac{1}{n}left(1+frac{1}{2}+cdots+frac{1}{n}
ight)=0$$
且
egin{align*}
&frac{1+frac{1}{2}+cdots+frac{1}{n}}{n}cdot frac{n+1}{1+frac{1}{2}+cdots+frac{1}{n+1}}\
=&left(1+frac{1}{n}
ight)cdot frac{1}{1+frac{frac{1}{n+1}}{1+frac{1}{2}+cdots+frac{1}{n}}}\
geq&left(1+frac{1}{n}
ight)left(frac{1}{1+frac{1}{n+1}}
ight)\
geq& 1
end{align*}
由交错级数的Leibniz判别法知 $sum c_{n}$收敛. 命题证毕.
17. 试证级数
$$1-frac{1}{sqrt{2}}+frac{1}{sqrt{3}}-frac{1}{sqrt{4}}+cdots$$
的自乘级数为发散.
证明: 设
$$a_{k}=b_{k}=frac{(-1)^{k-1}}{sqrt{k}},(k=1,2,3,cdots),a_{0}=b_{0}=0$$
则
$$c_{n}=sum_{k=0}^{n}a_{k}b_{n-k}=(-1)^{n}sum_{k=1}^{n}frac{1}{sqrt{k(n-k)}}$$
利用基本不等式
$$sqrt{ab}leq frac{a+b}{2}$$
有
$$c_{2n}=sum_{k=1}^{2n}frac{1}{sqrt{k(2n-k)}}geq sum_{k=1}^{2n}frac{1}{n}=2
rightarrow 0$$
因此级数$sum c_{n}$必不收敛(这是因为收敛的必要性通项趋于零这一条件的不满足).
18. (Pringsheim) 设$u_{n}downarrow 0,v_{n}downarrow 0$.则级数$sumlimits_{n=1}^{infty}(-1)^{n-1}u_{n}$与$sumlimits_{n=1}^{infty}(-1)^{n-1}v_{n}$的乘积级数
$sumlimits_{n=1}^{infty}(-1)^{n-1}omega_{n}$收敛的充分必要条件是
$$omega_{n}=u_{1}v_{n}+u_{2}v_{n-1}+cdots+u_{n}v_{1} o 0$$
证明: 必要性是平凡的. 只证充分性即可.由Abel级数乘法定理只需要证明级数$sum (-1)^{n}omega_{n}$收敛即可(未必为交错级数).
考察部分和
$$Omega_{n}=omega_{1}-omega_{2}+cdots+(-1)^{n}omega_{n}$$
$$U_{n}=u_{1}-u_{2}+cdots+(-1)^{n}u_{n}$$
$$V_{n}=v_{1}-v_{2}+cdots+(-1)^{n}v_{n}$$
且
$$U_{n} o u \,\, ;V_{n} o v \,\, (n o infty)$$
我们有
egin{align*}
Omega_{n}&=u_{1}V_{n}+u_{2}V_{n-1}+cdots+u_{n}V_{1}\
&=u_{1}(V_{n}-V)+u_{2}(V_{n-1}-V)+cdots+u_{n}(V_{1}-V)+U_{n}V
end{align*}
而
egin{align*}
&|u_{1}(V_{n}-V)+u_{2}(V_{n-1}-V)+cdots+u_{n}(V_{1}-V)|\
leq &u_{1}|V_{n}-V|+u_{2}(V_{n-1}-V)+cdots+u_{n}|V_{1}-V|\
leq &u_{1}v_{n+1}+u_{2}v_{n}+cdots+u_{n}v_{2} \,\,\,(since \, |V_{n}-V|leq v_{n+1},\, Lebnizt\, series)\
leq &u_{1}v_{n}+u_{2}v_{n-1}+cdots+u_{n}v_{1} o 0
end{align*}
所以
$$lim_{n oinfty}Omega_{n}=UV$$
19. (Pringsheim) 设$u_{n}downarrow 0,v_{n}downarrow 0$.则级数$sumlimits_{n=1}^{infty}(-1)^{n-1}u_{n}$与$sumlimits_{n=1}^{infty}(-1)^{n-1}v_{n}$的乘积级数收敛的充分必要条件是$U_{n}v_{n} o 0$ 并且 $V_{n}u_{n} o 0$,这里
$$U_{n}=u_{1}+u_{2}+cdots+u_{n};V_{n}=v_{1}+v_{2}+cdots+v_{n}$$
证明: 充分性.若$U_{n}v_{n} o 0$ 并且 $V_{n}u_{n} o 0$则
egin{align*}
omega_{2n}&=u_{1}v_{2n}+u_{2}v_{2n-1}+cdots+u_{n}v_{n+1}+u_{n+1}v_{n}+u_{n+2}v_{n-1}+cdots+u_{2n}v_{1}\
&leq (u_{1}+u_{2}+cdots+u_{n})v_{n+1}+u_{n+1}(v_{1}+v_{2}+cdots+v_{n}) o 0
end{align*}
对 $omega_{2n+1}$ 类似处理可得 $omega_{2n+1} o 0$. 从而 $omega_{n} o 0$. 由命题18知乘积级数收敛.
必要性. 若乘积级数收敛,则$omega_{n} o 0$. 利用单调性知
$$U_{n}v_{n}leq u_{1}v_{n}+u_{2}v_{n-1}+cdots+u_{n}v_{1} o 0$$
20. 试讨论$sumlimits_{n=1}^{infty}frac{(-1)^{n-1}}{n^{alpha}}$和$sumlimits_{n=1}^{infty}frac{(-1)^{n-1}}{n^{eta}}$的乘积级数的收敛性,这里 $0<alpha<1,0<eta<1$.
解答: 利用命题19 以及Stolz定理
$$left(frac{1}{1^{alpha}}+frac{1}{2^{alpha}}+cdots+frac{1}{n^{alpha}}
ight)cdot frac{1}{n^{eta}}sim frac{1}{(n+1)^{alpha}}cdot frac{1}{(n+1)^{eta}-n^{eta}}sim frac{1}{eta n^{alpha+eta-1}}, n o infty$$
同理
$$left(frac{1}{1^{eta}}+frac{1}{2^{eta}}+cdots+frac{1}{n^{eta}}
ight)cdot frac{1}{n^{alpha}}sim frac{1}{alpha n^{alpha+eta-1}}, n o infty$$
故当$alpha+eta>1$时乘积级数收敛.
21. (Mertens) 设级数$sum a_{n}$与级数$sum b_{n}$二收敛级数中至少有一个绝对收敛,又设
$$c_{n}=a_{0}b_{n}+a_{1}b_{n-1}+cdots+a_{n}b_{0}$$
则$sum c_{n}$必收敛,且
$$left(sum_{n=0}^{infty}a_{n}
ight)left(sum_{n=0}^{infty}b_{n}
ight)=sum_{n=0}^{infty}c_{n}$$
证明: 证明方法类似命题18, 设
$$A_{n}=sum_{k=0}^{n}a_{n} o A$$
$$B_{n}=sum_{k=0}^{n}b_{n} o B$$
又设$sum a_{n}$绝对收敛, 则对$sum c_{n}$进行估计
egin{align*}
sum_{k=0}^{n}c_{k}&=a_{0}B_{n}+a_{1}B_{n-1}+cdots+a_{n}B_{0}\
&=a_{0}(B_{n}-B)+a_{1}(B_{n-1}-B)+cdots+a_{n}(B_{0}-B)+A_{n}B
end{align*}
由收敛级数的Cauchy准则有 $forall varepsilon>0, exists N,n>N$
$$sum_{N}^{n}|a_{n}|<varepsilon$$
故
egin{align*}
&|a_{0}(B_{n}-B)+a_{1}(B_{n-1}-B)+cdots+a_{n}(B_{0}-B)|\
leq &|a_{0}|cdot|B_{n}-B|+|a_{1}|cdot |B_{n-1}-B|+cdots+|a_{N}|cdot |B_{n-N}-B|+cdots+|a_{n}|cdot|B_{0}-B|\
leq &|a_{0}|cdot|B_{n}-B|+|a_{1}|cdot |B_{n-1}-B|+cdots+|a_{N}|cdot |B_{n-N}-B|+Mcdot varepsilon
end{align*}
其中$M$为$|B_{n}-B|$的一个上界, 两边令$n o infty$
$$limsup_{n o infty}|a_{0}(B_{n}-B)+a_{1}(B_{n-1}-B)+cdots+a_{n}(B_{0}-B)|=0$$
所以
$$sum_{n=0}^{infty}c_{n}=AB=left(sum_{n=0}^{infty}a_{n}
ight)left(sum_{n=0}^{infty}b_{n}
ight)$$
22. (Abel判别法) 设级数$sum a_{n}$收敛而$sum (b_{n}-b_{n+1})$绝对收敛(其中$a_{n},b_{n}$可以是复数),则$sum a_{n}b_{n}$必收敛.
证明: 利用分部求和公式
$$sum_{k=1}^{n}a_{k}b_{k}=A_{n}b_{n}+sum_{k=1}^{n-1}A_{k}(b_{k}-b_{k+1})$$
设$A_{n} o A$,由于$sum (b_{n}-b_{n+1})$绝对收敛,设$b_{n} o b$. 只需要证明
$$sum_{k=1}^{n-1}A_{k}(b_{k}-b_{k+1})$$
收敛即可.
$$sum_{k=1}^{n-1}A_{k}(b_{k}-b_{k+1})=sum_{k=1}^{n-1}(A_{k}-A)(b_{k}-b_{k+1})+Asum_{k=1}^{n-1}(b_{k}-b_{k+1})$$
而
$$sum_{k=1}^{n-1}|(A_{k}-A)(b_{k}-b_{k+1})|leq M sum_{k=1}^{n-1}|b_{k}-b_{k+1}|$$
其中$M$为$|A_{n}-A|$的一个上界, 由Weirstrass判别法知$sum (A_{k}-A)(b_{k}-b_{k+1})$绝对收敛.运用收敛级数的四则运算法则知级数$sum a_{n}b_{n}$收敛.
23. (Dirichlet判别法) 设$sumlimits_{k=1}^{n}a_{n}$为有界而$sum (b_{n}-b_{n+1})$为绝对收敛且$b_{n} o 0$(其中$a_{n},b_{n}$可以是复数).则$sum a_{n}b_{n}$收敛.
证明: 由分部求和公式
$$sum_{k=1}^{n}a_{k}b_{k}=A_{n}b_{k}+sum_{k=1}^{n-1}A_{k}(b_{k}-b_{k+1})$$
由于$|A_{n}|$有界$M$,$b_{n} o 0$, 故$A_{n}b_{n} o 0$. 且
$$sum_{k=1}^{n-1}|A_{k}(b_{k}-b_{k+1})|leq M sum_{k=1}^{n-1}|b_{k}-b_{k+1}|$$
由Weistrass判别法知$sumlimits_{k=1}^{n-1}A_{k}(b_{k}-b_{k+1})$收敛. 从而 $sum a_{n}b_{n}$收敛.
24. (Abel)设${b_{n}}_{1}^{infty}$为一有界单调实数列而级数$sum a_{n}$收敛. 则级数$sum a_{n}b_{n}$必收敛.
证明: 不妨设$b_{n}$单调递减, 由单调有界收敛准则知$b_{n} o b$,且级数
$$sum_{k=1}^{infty}(b_{k}-b_{k+1})=b_{1}-b$$
由命题 22 知$sum a_{n}b_{n}$收敛.
25. (Dirichlet) 设$b_{n}downarrow 0 (n o infty)$而$sumlimits_{1}^{n}a_{k}$有界. 则级数$sum a_{n}b_{n}$必收敛.
证明: 由$b_{n}downarrow 0 (n o infty)$知$sum (b_{n}-b_{n+1})$为绝对收敛且$b_{n} o 0$, 利用命题 23 变得结论.
26. 令$t=cos heta+isin heta=e^{i heta}$. 则当$0< heta<2pi$且$a_{n}downarrow 0$时级数$sum a_{n}t^{n}$收敛,同时
$$ extcolor{red}{left|sum_{k=n}^{n+p}a_{k}t^{k}
ight|leq frac{4\, a_{n}}{|1-t|}}$$
证明: 设 $b_{n}=t^{n}$.
$$left|sum_{k=0}^{n}t^{k}
ight|=left|frac{1-t^{n+1}}{1-t}
ight|leq frac{2}{|1-t|}$$
又$a_{n}downarrow 0$由Dirichlet收敛原理知级数$sum a_{n}t^{n}$收敛.
同理
$$left|sum_{k=n}^{n+p}t^{k}
ight|=left|frac{t^{n}(1-t^{p+1})}{1-t}
ight|leq frac{2}{|1-t|}$$
利用分布求和法估计和式 $sumlimits_{k=n}^{n+p}a_{k}t^{k}$
egin{align*}
left|sum_{k=n}^{n+p}a_{k}t^{k}
ight|&=left|sum_{k=n}^{n+p}a_{k}(T_{k}-T_{k-1})
ight|\
&=left|a_{n+p}T_{n+p}+sum_{k=n}^{n+p-1}(a_{k}-a_{k+1})T_{k}-a_{n}T_{n-1}
ight|\
&leq frac{2}{|1-t|}a_{n+p}+frac{2}{|1-t|}sum_{k=n}^{n+p-1}(a_{k}-a_{k+1})+frac{2}{|1-t|}a_{n}\
&=frac{4\, a_{n}}{|1-t|}
end{align*}
27. 设$a_{n}downarrow 0$. 则当$0< heta<2pi$时级数$sum a_{n}cos n heta$以及$sum a_{n}sin n heta$ 为收敛且有
$$ extcolor{red}{left|sum_{k=n}^{n+p}a_{k}cos k heta
ight|leq frac{2\, a_{n}}{sin frac{ heta}{2}};\,\,left|sum_{k=n}^{n+p}a_{k}sin k heta
ight|leq frac{2\, a_{n}}{sin frac{ heta}{2}}}$$
证明: 收敛性可由Dirichlet判别法得知.由命题 26 有
$$left|sum_{k=n}^{n+p}cos k heta+isum_{k=n}^{n+p}sin k heta
ight|leq frac{2}{|1-t|}=frac{1}{sinfrac{ heta}{2}}$$
$$left|sum_{k=n}^{n+p}a_{k}cos k heta+isum_{k=n}^{n+p}a_{k}sin k heta
ight|leq frac{4\, a_{n}}{|1-t|}=frac{2\,a_{n}}{sinfrac{ heta}{2}}$$
复数的实部与虚部均小于其模长
$$ extcolor{red}{left|sum_{k=n}^{n+p}a_{k}cos k heta
ight|leq frac{2\, a_{n}}{sin frac{ heta}{2}};\,\,left|sum_{k=n}^{n+p}a_{k}sin k heta
ight|leq frac{2\, a_{n}}{sin frac{ heta}{2}}}$$
28. (Hardy) 试应用命题 25 来讨论级数$sum a_{n}u_{n},sum a_{n}v_{n}$的收敛性问题, 其中
$$sum_{1}^{n}u_{k}=sinleft(n+frac{1}{2}
ight)^{2} heta,\,\, sum_{1}^{n}v_{k}=cosleft(n+frac{1}{2}
ight)^{2} heta$$
并从而推断$sum a_{n}sin n hetacdot cos n^{2} heta,\,\, sum a_{n}sin n hetacdot sin n^{2} heta$
二级数当$a_{n}downarrow 0$时收敛.
证明: 关键点在于三角函数的积化和差公式以及二倍角公式
egin{align*}
sum_{1}^{n}sin k hetacdot cos k^{2} heta&=frac{1}{2}sum_{1}^{n}left[sin(k+k^{2}) heta+sin(k-k^{2}) heta
ight]\
&=frac{1}{2}sum_{1}^{n}left{sinleft[left(k+frac{1}{2}
ight)^{2}-frac{1}{4}
ight] heta-sinleft[left(k-frac{1}{2}
ight)^{2}-frac{1}{4}
ight] heta
ight}\
&=frac{1}{2}cosfrac{ heta}{4}sum_{1}^{n}left[sinleft(k+frac{1}{2}
ight)^{2} heta-sinleft(k-frac{1}{2}
ight)^{2} heta
ight]\
&+frac{1}{2}sinfrac{ heta}{4}sum_{1}^{n}left[cosleft(k-frac{1}{2}
ight)^{2} heta-cosleft(k+frac{1}{2}
ight)^{2} heta
ight]\
&=frac{1}{2}left[sinleft(k+frac{1}{2}
ight)^{2} hetacosfrac{ heta}{4}-cosleft(k+frac{1}{2}
ight)^{2} hetasinfrac{ heta}{4}
ight]\
&=frac{1}{2}sin (n+n^{2}) heta
end{align*}
类似地,
egin{align*}
sum_{1}^{n}sin k hetacdot sin k^{2} heta&=frac{1}{2}sum_{1}^{n}left[cos(k-k^{2}) heta-cos(k+k^{2}) heta
ight]\
&=frac{1}{2}sum_{1}^{n}left{cosleft[left(k-frac{1}{2}
ight)^{2}-frac{1}{4}
ight] heta-cosleft[left(k+frac{1}{2}
ight)^{2}-frac{1}{4}
ight] heta
ight}\
&=frac{1}{2}cosfrac{ heta}{4}sum_{1}^{n}left[cosleft(k-frac{1}{2}
ight)^{2} heta-cosleft(k+frac{1}{2}
ight)^{2} heta
ight]\
&+frac{1}{2}sinfrac{ heta}{4}sum_{1}^{n}left[sinleft(k-frac{1}{2}
ight)^{2} heta-sinleft(k+frac{1}{2}
ight)^{2} heta
ight]\
&=frac{1}{2}cosfrac{ heta}{4}left[cosfrac{ heta}{4}-cosleft(n+frac{1}{2}
ight)^{2} heta
ight]\
&+frac{1}{2}sinfrac{ heta}{4}left[sinfrac{ heta}{4}-sinleft(n+frac{1}{2}
ight)^{2} heta
ight]\
&=frac{1}{2}[1-cos(n+n^{2})]\
&=sin frac{(n+n^{2}) heta}{2}
end{align*}
从而
$$left|sum_{1}^{n}sin k hetacdot cos k^{2} heta
ight|leq frac{1}{2},\,left|sum_{1}^{n}sin k hetacdot sin k^{2} heta
ight|leq 1 $$
由Dirichlet判别法知$sum a_{n}sin n hetacdot cos n^{2} heta,\,\, sum a_{n}sin n hetacdot sin n^{2} heta$二级数当$a_{n}downarrow 0$时收敛.
29. 设$a_{n}downarrow 0(n oinfty)$且
$$frac{1}{2}(a_{n}+a_{n+2})geq a_{n+1}$$
则当$0< heta<2pi$时级数
$$frac{a_{0}}{2}+sum_{n=1}^{infty}a_{n}cos n heta$$
收敛于一个非负函数.
证明:核函数
$$D_{n}( heta)=frac{1}{2}+sum_{k=1}^{n}=frac{1}{2}frac{sinleft(n+frac{1}{2}
ight) heta}{sinfrac{ heta}{2}}$$
$$K_{n}( heta)=sum_{k=0}^{n}D_{k}( heta)=frac{1}{2}left(frac{sinleft(n+frac{1}{2}
ight) heta}{sinfrac{ heta}{2}}
ight)^{2}$$
利用两次和差变换可得
egin{align*}
frac{a_{0}}{2}+sum_{k=1}^{n}a_{k}cos k heta&=sum_{k=0}^{n-1}D_{k}( heta)(a_{k}-a_{k+1})+a_{n}D_{n}( heta)\
&=a_{n}D_{n}( heta)+(a_{n-1}-a_{n})K_{n-1}( heta)+sum_{k=0}^{n-2}(a_{k}-2a_{k+1}+a_{k+2})K_{k}( heta)\
&geq 0
end{align*}
由极限的保号性知收敛于一个非负函数.