与调和数列有关的级数计算
关于$\pi$的级数
与$\sum \arctan$有关的一些问题
证明\[\sum\limits_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}} = \ln 3 - \frac{\pi }{4}.\]
解.(r9m)首先有
\begin{align*}S&=\sum\limits_{n=1}^{\infty} \arctan \frac{10n}{(3n^2+2)(9n^2-1)} \\&= \sum\limits_{n=1}^{\infty} \arg \left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4\left(1+\frac{2}{3n^2}\right)\left(1-\frac{1}{9n^2}\right)}\right).\end{align*}
分母里的无穷乘积$\displaystyle \prod\limits_{n=1}^{\infty}\left(1+\frac{2}{3n^2}\right)$和$\displaystyle \prod\limits_{n=1}^{\infty}\left(1-\frac{1}{9n^2}\right)$ 可以被忽略,当它们收敛于实数时.
因此
\begin{align*}S&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4}\right).\end{align*}
对分子进行分解
$$(3n^2+2)(9n^2-1)+10in = (n-i)(3n+i)(3n+i+1)(3n+i-1).$$
我们得
$$S = \arg \prod\limits_{n=1}^{\infty}\frac{\left(1+\frac{i}{3n}\right)\left(1+\frac{i+1}{3n}\right)\left(1+\frac{i-1}{3n}\right)}{\left(1+\frac{i}{n}\right)}.$$
在$\displaystyle z = i,\frac{i}{3},\frac{i+1}{3},\frac{i-1}{3}$ 处运用$\displaystyle \frac{1}{\Gamma(z)} = ze^{\gamma z}\prod\limits_{n=1}^{\infty}\left(1+\frac{z}{n}\right)e^{-z/n}$ .
我们可以改写为
$$S = \arg \frac{-\Gamma(i)}{\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right)\Gamma\left(\frac{i-1}{3}\right)}.$$
另一方面运用Gauss-Legendre Triplication Formula
$$ \Gamma(3z) = \frac{1}{2\pi}3^{3z - \frac{1}{2}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right).$$
令$z = \dfrac{i-1}{3}$我们有
$$\Gamma\left(\frac{i-1}{3}\right)\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right) = 2\pi 3^{-i+\frac{3}{2}}\Gamma(i-1)$$
因此$$S = \arg \frac{-3^{i}\Gamma(i)}{\Gamma(i-1)} = \arg (3^{i}(1-i)) = \log 3 - \frac{\pi}{4}.$$
解法二.(robjohn)运用$\arctan(x)=\arg(1+ix)$,分解可知
\begin{align*}&1+\frac{10in}{\left(3n^2+2\right)\left( 9n^2-1\right)}\\=&\frac{\left(1-\frac in\right)\left(1+\frac i{3n-1}\right)\left(1+\frac i{3n+1}\right)\left(1+\frac i{3n}\right)}{1+\frac2{3n^2}}.\end{align*}
因此有
\begin{align*}&\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\=&\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right).\end{align*}
裂项可知
\begin{align*}&\sum_{n=1}^\infty\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\=&\lim_{m\to\infty}\sum_{n=1}^m\left[\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right)\right]\\=&-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\arctan\left(\frac1n\right)\\=&-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\left[\frac1n+O\left(\frac1{n^3}\right)\right]\\=&\log(3)-\frac\pi4.\end{align*}
哆嗒数学网里代数龙发的一系列级数题
练习题1.证明：$$\sum\limits_{n=1}^{\infty}\frac{1}{(n+1)\sqrt[p]{n}}\leq p,\,\,(p\ge1).$$
证:由Lagrange中值定理,我们有
\[\sqrt[p]{{n + 1}} - \sqrt[p]{n} = \frac{1}{p}{\xi ^{1/p - 1}} \ge \frac{1}{p}{\left( {n + 1} \right)^{1/p - 1}},\quad \xi \in \left( {n,n + 1} \right).\]
因此\[\frac{1}{{\left( {n + 1} \right)\sqrt[p]{n}}} = \frac{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}}{{\sqrt[p]{n} \cdot \sqrt[p]{{n + 1}}}} \cdot \frac{{{{\left( {n + 1} \right)}^{1/p - 1}}}}{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}} \le p\frac{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}}{{\sqrt[p]{n} \cdot \sqrt[p]{{n + 1}}}} = p\left( {\frac{1}{{\sqrt[p]{n}}} - \frac{1}{{\sqrt[p]{{n + 1}}}}} \right).\]
立即有
\[\sum\limits_{n = 1}^\infty {\frac{1}{{\left( {n + 1} \right)\sqrt[p]{n}}}} \le p\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{\sqrt[p]{n}}} - \frac{1}{{\sqrt[p]{{n + 1}}}}} \right)} = p.\]
练习题2.设$\displaystyle S_n=\sum\limits_{k=1}^{n}a_k, p>1,c>1$,证明：$$\sum\limits_{n=1}^{\infty}\frac{S_n^p}{n^c}\le K\sum\limits_{n=1}^{\infty}\frac{(na_n)^p}{n^c},$$并求出$K$的最优值.
练习题3.设$a_n$是有界的正数列，$p>0$，证明：
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.注意到
\begin{align*}\sum\limits_{k = 1}^\infty {{a_k}} &= \sum\limits_{k = 1}^\infty {\arctan \frac{1}{{{k^2} + k + 1}}} = \sum\limits_{k = 1}^\infty {\left( {\arctan \frac{1}{k} - \arctan \frac{1}{{k + 1}}} \right)} = \frac{\pi }{4}\\\sum\limits_{k = 1}^\infty {\frac{1}{{{k^4}}}} &= \zeta \left( 4 \right) = \frac{{{\pi ^4}}}{{90}}.\end{align*}
由Cauchy-Schwarz不等式可知
\[\sum\limits_{k = 1}^N {\frac{1}{{{k^4}}}} \cdot \sum\limits_{k = 1}^N {{a_k}} \ge {\left( {\sum\limits_{k = 1}^N {\frac{{a_k^{1/2}}}{{{k^2}}}} } \right)^2}.\]
令$N\to\infty$,我们有\[\sum\limits_{k = 1}^\infty {\frac{{a_k^{1/2}}}{{{k^2}}}} \le \sqrt {\frac{{{\pi ^4}}}{{90}} \cdot \frac{\pi }{4}} = \sqrt {\frac{{{\pi ^5}}}{{360}}} < \sqrt {\frac{\pi }{3}} .\]
也可通过放缩实现\[\sum\limits_{k = 1}^\infty {\frac{1}{{{k^4}}}} = 1 + \sum\limits_{k = 2}^\infty {\frac{1}{{{k^4}}}} < 1 + \sum\limits_{k = 2}^\infty {\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}} = \frac{4}{3}.\]
练习题10.设$\displaystyle a_n > 0, S_n=\sum\limits_{k=1}^na_k$，证明：
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
\[\mathop {\lim }\limits_{\alpha \to 1} \frac{{\sum\limits_{n = 1}^\infty {\frac{n}{{{a_1} + {a_2} + \cdots + {a_n}}}} }}{{\sum\limits_{j = 1}^\infty {\frac{1}{{{a_j}}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\sum\limits_{n = 1}^N {\frac{n}{{{1^\alpha } + {2^\alpha } + \cdots + {n^\alpha }}}} }}{{\sum\limits_{n = 1}^N {\frac{1}{{{n^\alpha }}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\frac{N}{{\frac{1}{{\alpha + 1}}{N^{\alpha + 1}} + O\left( {{N^\alpha }} \right)}}}}{{\frac{1}{{{N^\alpha }}}}} = 2.\]
(2)如法炮制.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
\[\frac{{2n + 1}}{{{a_1} + {a_2} + \cdots + {a_n}}} \le \frac{{4\left( {2n + 1} \right)}}{{{n^2}{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} .\]
因此
\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{{a_1} + {a_2} + \cdots + {a_n}}}} &\le 4\sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{{n^2}{{\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{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}}} } \\&= 4\sum\limits_{m = 1}^\infty {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty {\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right]} } = 4\sum\limits_{m = 1}^\infty {\frac{1}{{{a_m}}}}.\end{align*}
练习题11.设$\displaystyle a_n \ge 0, n=1,2,\ldots,\sum\limits_{n=1}^{\infty}a_n < \infty$,证明:
$$\sum\limits_{n=1}^{\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}} \le e \sum\limits_{n=1}^{\infty}a_n$$,且证明$e$是最优值.
此题再拓展下求证:$$\sum\limits_{n=1}^{\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}} \le e \sum\limits_{n=1}^{\infty}[1-\frac{1}{2(n+1)}]a_n.$$
练习题12.如果正项级数$\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{p_n}$收敛,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}\frac{n^2}{(p_1+p_2+\cdots+p_n)^2}p_n$也收敛.
练习题13.设$\displaystyle \sum\limits_{n=1}^{\infty}a_n$为正项级数,且$\displaystyle \sum\limits_{k=1}^{n}(a_k-a_n)$对$n$有界,$a_n$单调递减趋于$0$,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}a_n$收敛.
练习题14.设级数$\displaystyle \sum\limits_{n=1}^{\infty}a_n$收敛, $\displaystyle \sum\limits_{n=1}^{\infty}(b_{n+1}-b_n)$绝对收敛,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}a_nb_n$收敛.
练习题15.设$a_n>0,\left\{ a_n-a_{n+1}\right\}$为一个严格递减的数列.如果$\sum_{n=1}^{\infty}a_n$收敛。试证：$$\lim\limits_{n \to \infty}\left( \cfrac{1}{a_{n+1}}-\cfrac{1}{a_n}\right)=+\infty.$$
练习题16.能否构造一个收敛数列$\sum\limits_{n=1}^{\infty}a_n$,使得级数$\sum\limits_{n=1}^{\infty}a_n^3$发散.
练习题17.设$\lim \limits_{n\rightarrow +\infty}x_n=+\infty$,正项级数$\sum\limits_{n=1}^{\infty}y_n$收敛,设$n_0$是某一自然数,
练习题18.设$\sum\limits_{n=1}^{\infty}a_n$是一正项收敛级数,且有$a_{n+1}< \frac{1}{2}(a_n+a_{n+2}),\,\frac{1}{a_{n+2}}-\frac{1}{a_{n+1}}\le \frac{1}{3}(\frac{1}{a_{n+3}}-\frac{1}{a_{n}})$,
与$\sin n^2$类似的一些问题
1.证明: $\sum_{k=1}^n\sin k^2$无界.
参看: http://www.zhihu.com/question/29094450
2.证明\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\sin \sqrt k } = 0.\]
一个与多项式分拆有关的级数题
$\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解
Proof.What you require here are the Eisenstein series. In particular the evaluation of
Euler Sum的若干研究
本文主要展示并分享有关Euler Sum的若干求解问题。
问题一：求\[\sum_{n=1}^\infty{\frac{H_n}{n^q}}\]
Solution.
参考文献
[1]
[2]
一个级数求解
