阿贝尔分部求和法的应用(二)

14.(阿贝耳定理) 设$\sum\limits_{n=0}^{\infty}a_{n}=s$. 则$\lim_{x\to 1-}\sum\limits_{n=0}^{\infty}a_{n}x^{n}=s$.
证明: 容易看出$f(x)=\sum\limits_{n=0}^{\infty}a_{n}x^{n}$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
$$\left|\sum_{k=n}^{n+p}a_{k}\right|<\varepsilon$$
由Abel引理(命题4)知
$$\left|\sum_{k=n}^{n+p}a_{k}x^{k}\right|<x^{n}\left|\sum_{k=n}^{n+p}a_{k}\right|\leq \varepsilon$$
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
$$\lim_{x\to 1-}\sum_{n=0}^{\infty}a_{n}x^{n}=s$$
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
$$\sum_{n=0}^{\infty}c_{n}=\left(\sum_{n=0}^{\infty}a_{n}\right)\left(\sum_{n=0}^{\infty}b_{n}\right)$$
证明: 因为绝对收敛的级数可以相乘,因此
$$\sum_{n=0}^{\infty}c_{n}x^{n}=\left(\sum_{n=0}^{\infty}a_{n}x^{n}\right)\left(\sum_{n=0}^{\infty}b_{n}x^{n}\right)=s_{1}(x)s_{2}(x)$$
由阿贝尔第二定理(命题14)知
$$\sum_{n=0}^{\infty}c_{n}=\lim_{x\to 1-}\sum_{n=0}^{\infty}c_{n}x^{n}=\lim_{x\to 1-}s_{1}(x)s_{2}(x)=\left(\sum_{n=0}^{\infty}a_{n}\right)\left(\sum_{n=0}^{\infty}b_{n}\right)$$
16. 试证
$$\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)^{2}=\frac{1}{2}-\frac{1}{3}
\left(1+\frac{1}{2}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{5}\left(1+\frac{1}{2}
+\frac{1}{3}+\frac{1}{4}\right)+\cdots$$
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
$$c_{n}=\sum_{k=0}^{n}a_{k}b_{n-k}=(-1)^{n}\sum_{k=1}^{n}\frac{1}{k(n-k)}=\frac{(-1)^{n}}{n}\sum_{k=1}^{n}\left(
\frac{1}{k}+\frac{1}{n-k}\right)=(-1)^{n}\frac{2}{n}\sum_{k=1}^{n}\frac{1}{k}$$
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
$$\lim_{n\to\infty}\frac{1}{n}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)=0$$
且
\begin{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}\right)\cdot \frac{1}{1+\frac{\frac{1}{n+1}}{1+\frac{1}{2}+\cdots+\frac{1}{n}}}\\
\geq&\left(1+\frac{1}{n}\right)\left(\frac{1}{1+\frac{1}{n+1}}\right)\\
\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
$$c_{2n}=\sum_{k=1}^{2n}\frac{1}{\sqrt{k(2n-k)}}\geq \sum_{k=1}^{2n}\frac{1}{n}=2\nrightarrow 0$$
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$\sum\limits_{n=1}^{\infty}(-1)^{n-1}u_{n}$与$\sum\limits_{n=1}^{\infty}(-1)^{n-1}v_{n}$的乘积级数
$\sum\limits_{n=1}^{\infty}(-1)^{n-1}\omega_{n}$收敛的充分必要条件是
$$\omega_{n}=u_{1}v_{n}+u_{2}v_{n-1}+\cdots+u_{n}v_{1}\to 0$$
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
$$U_{n}\to u \,\, ;V_{n}\to v \,\, (n\to \infty)$$
我们有
\begin{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
\begin{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} \to 0
\end{align*}
所以
$$\lim_{n\to\infty}\Omega_{n}=UV$$
19. (Pringsheim) 设$u_{n}\downarrow 0,v_{n}\downarrow 0$.则级数$\sum\limits_{n=1}^{\infty}(-1)^{n-1}u_{n}$与$\sum\limits_{n=1}^{\infty}(-1)^{n-1}v_{n}$的乘积级数收敛的充分必要条件是$U_{n}v_{n}\to 0$ 并且 $V_{n}u_{n}\to 0$,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$U_{n}v_{n}\to 0$ 并且 $V_{n}u_{n}\to 0$则
\begin{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})\to 0
\end{align*}
对 $\omega_{2n+1}$ 类似处理可得 $\omega_{2n+1}\to 0$. 从而 $\omega_{n}\to 0$. 由命题18知乘积级数收敛.
必要性. 若乘积级数收敛,则$\omega_{n}\to 0$. 利用单调性知
$$U_{n}v_{n}\leq u_{1}v_{n}+u_{2}v_{n-1}+\cdots+u_{n}v_{1}\to 0$$
20. 试讨论$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\alpha}}$和$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\beta}}$的乘积级数的收敛性,这里 $0<\alpha<1,0<\beta<1$.
解答: 利用命题19 以及Stolz定理
$$\left(\frac{1}{1^{\alpha}}+\frac{1}{2^{\alpha}}+\cdots+\frac{1}{n^{\alpha}}\right)\cdot \frac{1}{n^{\beta}}\sim \frac{1}{(n+1)^{\alpha}}\cdot \frac{1}{(n+1)^{\beta}-n^{\beta}}\sim \frac{1}{\beta n^{\alpha+\beta-1}}, n\to \infty$$
同理
$$\left(\frac{1}{1^{\beta}}+\frac{1}{2^{\beta}}+\cdots+\frac{1}{n^{\beta}}\right)\cdot \frac{1}{n^{\alpha}}\sim \frac{1}{\alpha n^{\alpha+\beta-1}}, n\to \infty$$
故当$\alpha+\beta>1$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
$$\left(\sum_{n=0}^{\infty}a_{n}\right)\left(\sum_{n=0}^{\infty}b_{n}\right)=\sum_{n=0}^{\infty}c_{n}$$
证明: 证明方法类似命题18, 设
$$A_{n}=\sum_{k=0}^{n}a_{n}\to A$$
$$B_{n}=\sum_{k=0}^{n}b_{n}\to B$$
又设$\sum a_{n}$绝对收敛, 则对$\sum c_{n}$进行估计
\begin{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
\begin{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|+M\cdot \varepsilon
\end{align*}
其中$M$为$|B_{n}-B|$的一个上界, 两边令$n\to \infty$
$$\limsup_{n\to \infty}|a_{0}(B_{n}-B)+a_{1}(B_{n-1}-B)+\cdots+a_{n}(B_{0}-B)|=0$$
所以
$$\sum_{n=0}^{\infty}c_{n}=AB=\left(\sum_{n=0}^{\infty}a_{n}\right)\left(\sum_{n=0}^{\infty}b_{n}\right)$$
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$A_{n}\to A$,由于$\sum (b_{n}-b_{n+1})$绝对收敛,设$b_{n}\to b$. 只需要证明
$$\sum_{k=1}^{n-1}A_{k}(b_{k}-b_{k+1})$$
收敛即可.
$$\sum_{k=1}^{n-1}A_{k}(b_{k}-b_{k+1})=\sum_{k=1}^{n-1}(A_{k}-A)(b_{k}-b_{k+1})+A\sum_{k=1}^{n-1}(b_{k}-b_{k+1})$$
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$\sum\limits_{k=1}^{n}a_{n}$为有界而$\sum (b_{n}-b_{n+1})$为绝对收敛且$b_{n} \to 0$(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$M$,$b_{n}\to 0$, 故$A_{n}b_{n}\to 0$. 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$\sum\limits_{k=1}^{n-1}A_{k}(b_{k}-b_{k+1})$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$b_{n}\to b$,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$b_{n}\downarrow 0 (n\to \infty)$而$\sum\limits_{1}^{n}a_{k}$有界. 则级数$\sum a_{n}b_{n}$必收敛.
证明: 由$b_{n}\downarrow 0 (n\to \infty)$知$\sum (b_{n}-b_{n+1})$为绝对收敛且$b_{n} \to 0$, 利用命题 23 变得结论.
26. 令$t=\cos \theta+i\sin \theta=e^{i\theta}$. 则当$0<\theta<2\pi$且$a_{n}\downarrow 0$时级数$\sum a_{n}t^{n}$收敛,同时
$$\textcolor{red}{\left|\sum_{k=n}^{n+p}a_{k}t^{k}\right|\leq \frac{4\, a_{n}}{|1-t|}}$$
证明: 设 $b_{n}=t^{n}$.
$$\left|\sum_{k=0}^{n}t^{k}\right|=\left|\frac{1-t^{n+1}}{1-t}\right|\leq \frac{2}{|1-t|}$$
又$a_{n}\downarrow 0$由Dirichlet收敛原理知级数$\sum a_{n}t^{n}$收敛.
同理
$$\left|\sum_{k=n}^{n+p}t^{k}\right|=\left|\frac{t^{n}(1-t^{p+1})}{1-t}\right|\leq \frac{2}{|1-t|}$$
利用分布求和法估计和式 $\sum\limits_{k=n}^{n+p}a_{k}t^{k}$
\begin{align*}
\left|\sum_{k=n}^{n+p}a_{k}t^{k}\right|&=\left|\sum_{k=n}^{n+p}a_{k}(T_{k}-T_{k-1})\right|\\
&=\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}\right|\\
&\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$0<\theta<2\pi$时级数$\sum a_{n}\cos n\theta$以及$\sum a_{n}\sin n\theta$ 为收敛且有
$$\textcolor{red}{\left|\sum_{k=n}^{n+p}a_{k}\cos k\theta\right|\leq \frac{2\, a_{n}}{\sin \frac{\theta}{2}};\,\,\left|\sum_{k=n}^{n+p}a_{k}\sin k\theta\right|\leq \frac{2\, a_{n}}{\sin \frac{\theta}{2}}}$$
证明: 收敛性可由Dirichlet判别法得知.由命题 26 有
$$\left|\sum_{k=n}^{n+p}\cos k\theta+i\sum_{k=n}^{n+p}\sin k\theta\right|\leq \frac{2}{|1-t|}=\frac{1}{\sin\frac{\theta}{2}}$$
$$\left|\sum_{k=n}^{n+p}a_{k}\cos k\theta+i\sum_{k=n}^{n+p}a_{k}\sin k\theta\right|\leq \frac{4\, a_{n}}{|1-t|}=\frac{2\,a_{n}}{\sin\frac{\theta}{2}}$$
复数的实部与虚部均小于其模长
$$\textcolor{red}{\left|\sum_{k=n}^{n+p}a_{k}\cos k\theta\right|\leq \frac{2\, a_{n}}{\sin \frac{\theta}{2}};\,\,\left|\sum_{k=n}^{n+p}a_{k}\sin k\theta\right|\leq \frac{2\, a_{n}}{\sin \frac{\theta}{2}}}$$
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
$$\sum_{1}^{n}u_{k}=\sin\left(n+\frac{1}{2}\right)^{2}\theta,\,\, \sum_{1}^{n}v_{k}=\cos\left(n+\frac{1}{2}\right)^{2}\theta$$
并从而推断$\sum a_{n}\sin n\theta\cdot \cos n^{2}\theta,\,\, \sum a_{n}\sin n\theta\cdot \sin n^{2}\theta$
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
\begin{align*}
\sum_{1}^{n}\sin k\theta\cdot \cos k^{2}\theta&=\frac{1}{2}\sum_{1}^{n}\left[\sin(k+k^{2})\theta+\sin(k-k^{2})\theta\right]\\
&=\frac{1}{2}\sum_{1}^{n}\left\{\sin\left[\left(k+\frac{1}{2}\right)^{2}-\frac{1}{4}\right]\theta-\sin\left[\left(k-\frac{1}{2}\right)^{2}-\frac{1}{4}\right]\theta\right\}\\
&=\frac{1}{2}\cos\frac{\theta}{4}\sum_{1}^{n}\left[\sin\left(k+\frac{1}{2}\right)^{2}\theta-\sin\left(k-\frac{1}{2}\right)^{2}\theta\right]\\
&+\frac{1}{2}\sin\frac{\theta}{4}\sum_{1}^{n}\left[\cos\left(k-\frac{1}{2}\right)^{2}\theta-\cos\left(k+\frac{1}{2}\right)^{2}\theta\right]\\
&=\frac{1}{2}\left[\sin\left(k+\frac{1}{2}\right)^{2}\theta\cos\frac{\theta}{4}-\cos\left(k+\frac{1}{2}\right)^{2}\theta\sin\frac{\theta}{4}\right]\\
&=\frac{1}{2}\sin (n+n^{2})\theta
\end{align*}
类似地,
\begin{align*}
\sum_{1}^{n}\sin k\theta\cdot \sin k^{2}\theta&=\frac{1}{2}\sum_{1}^{n}\left[\cos(k-k^{2})\theta-\cos(k+k^{2})\theta\right]\\
&=\frac{1}{2}\sum_{1}^{n}\left\{\cos\left[\left(k-\frac{1}{2}\right)^{2}-\frac{1}{4}\right]\theta-\cos\left[\left(k+\frac{1}{2}\right)^{2}-\frac{1}{4}\right]\theta\right\}\\
&=\frac{1}{2}\cos\frac{\theta}{4}\sum_{1}^{n}\left[\cos\left(k-\frac{1}{2}\right)^{2}\theta-\cos\left(k+\frac{1}{2}\right)^{2}\theta\right]\\
&+\frac{1}{2}\sin\frac{\theta}{4}\sum_{1}^{n}\left[\sin\left(k-\frac{1}{2}\right)^{2}\theta-\sin\left(k+\frac{1}{2}\right)^{2}\theta\right]\\
&=\frac{1}{2}\cos\frac{\theta}{4}\left[\cos\frac{\theta}{4}-\cos\left(n+\frac{1}{2}\right)^{2}\theta\right]\\
&+\frac{1}{2}\sin\frac{\theta}{4}\left[\sin\frac{\theta}{4}-\sin\left(n+\frac{1}{2}\right)^{2}\theta\right]\\
&=\frac{1}{2}[1-\cos(n+n^{2})]\\
&=\sin \frac{(n+n^{2})\theta}{2}
\end{align*}
从而
$$\left|\sum_{1}^{n}\sin k\theta\cdot \cos k^{2}\theta\right|\leq \frac{1}{2},\,\left|\sum_{1}^{n}\sin k\theta\cdot \sin k^{2}\theta \right|\leq 1 $$
由Dirichlet判别法知$\sum a_{n}\sin n\theta\cdot \cos n^{2}\theta,\,\, \sum a_{n}\sin n\theta\cdot \sin n^{2}\theta$二级数当$a_{n}\downarrow 0$时收敛.

29. 设$a_{n}\downarrow 0(n\to\infty)$且
$$\frac{1}{2}(a_{n}+a_{n+2})\geq a_{n+1}$$
则当$0<\theta<2\pi$时级数
$$\frac{a_{0}}{2}+\sum_{n=1}^{\infty}a_{n}\cos n\theta$$
收敛于一个非负函数.
证明:核函数
$$D_{n}(\theta)=\frac{1}{2}+\sum_{k=1}^{n}=\frac{1}{2}\frac{\sin\left(n+\frac{1}{2}\right)\theta}{\sin\frac{\theta}{2}}$$
$$K_{n}(\theta)=\sum_{k=0}^{n}D_{k}(\theta)=\frac{1}{2}\left(\frac{\sin\left(n+\frac{1}{2}\right)\theta}{\sin\frac{\theta}{2}}\right)^{2}$$
利用两次和差变换可得
\begin{align*}
\frac{a_{0}}{2}+\sum_{k=1}^{n}a_{k}\cos k\theta&=\sum_{k=0}^{n-1}D_{k}(\theta)(a_{k}-a_{k+1})+a_{n}D_{n}(\theta)\\
&=a_{n}D_{n}(\theta)+(a_{n-1}-a_{n})K_{n-1}(\theta)+\sum_{k=0}^{n-2}(a_{k}-2a_{k+1}+a_{k+2})K_{k}(\theta)\\
&\geq 0
\end{align*}
由极限的保号性知收敛于一个非负函数.