Eufisky - The lost book

## 与调和数列有关的级数计算

\begin{align*}\sum_{n=1}^{\infty}{\frac{H_n-H_{2n}}{n\left( 2n+1 \right)}}=&2\sum_{n=1}^{\infty}{\left( H_n-H_{2n} \right) \left( \frac{1}{2n}-\frac{1}{2n+1} \right)}\\=&2\sum_{n=1}^{\infty}{\left( \frac{1}{2n}-\frac{1}{2n+1} \right) \int_0^1{\frac{x^{2n}-x^n}{1-x}\text{d}x}}\\=&\int_0^1{\frac{\sqrt{x}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-\ln \frac{1+x}{1-x}-\ln \left( 1+x \right)}{1-x}\text{d}x}\\&+\int_0^1{\left( \frac{1}{\sqrt{x}}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-\frac{1}{x}\ln \frac{1+x}{1-x} \right) \text{d}x},\end{align*}

$$\int_0^1{\frac{1}{\sqrt{x}}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}\text{d}x}=2\int_0^1{\ln \frac{1+t}{1-t}\text{d}t}=4\ln 2.$$
$$\int_0^1{\frac{1}{x}\ln \frac{1+x}{1-x}\text{d}x}=\mathrm{Li}_2\left( 1 \right) -\mathrm{Li}_2\left( -1 \right) =\frac{\pi ^2}{4}.$$
\begin{align*}\int_0^1{\frac{\left( \sqrt{x}-1 \right)}{1-x}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}\text{d}x}&=-2\int_0^1{\frac{t}{1+t}\ln \frac{1+t}{1-t}\text{d}t}\\&=2\int_0^1{\left[ \frac{\ln \left( 1+t \right)}{1+t}-\frac{\ln \left( 1-t \right)}{1+t} \right] \text{d}t}-2\int_0^1{\ln \frac{1+t}{1-t}\text{d}t}\\&=\ln ^22+2\mathrm{Li}_2\left( \frac{1}{2} \right) -4\ln 2=\frac{\pi^2}{6}-4\ln2.\end{align*}
\begin{align*}&\int_0^1{\frac{\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-\ln \frac{1+x}{1-x}-\ln \left( 1+x \right)}{1-x}\text{d}x}=2\int_0^1{\frac{1}{1-x}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}\text{d}x}\\=&2\int_0^1{\frac{\ln \left( 1+\sqrt{x} \right) -\ln 2}{1-x}\text{d}x}-2\int_0^1{\frac{\ln \left( 1+x \right) -\ln 2}{1-x}\text{d}x},\end{align*}

\begin{align*}\int_0^1{\frac{\ln \left( 1+\sqrt{x} \right) -\ln 2}{1-x}\text{d}x}&=2\int_0^1{\frac{t}{1-t^2}\ln \frac{1+t}{2}\text{d}t}\\&=\int_0^1{\frac{1}{1-t}\ln \frac{1+t}{2}\text{d}t}-\int_0^1{\frac{1}{1+t}\ln \frac{1+t}{2}\text{d}t}\\&=-\mathrm{Li}_2\left( \frac{1}{2} \right) +\frac{1}{2}\ln ^22=\ln ^22-\frac{\pi ^2}{12}.\end{align*}
$$\int_0^1{\frac{\ln \left( 1+x \right) -\ln 2}{1-x}\text{d}x}=-\mathrm{Li}_2\left( \frac{1}{2} \right) =\frac{\ln ^22}{2}-\frac{\pi ^2}{12}.$$

\begin{align*}\sum_{n=1}^{\infty}{\frac{H_n-H_{2n}}{n\left( 2n+1 \right)}}=4\ln 2-\frac{\pi ^2}{4}+\frac{\pi ^2}{6}-4\ln 2+2\left( \ln ^22-\frac{\pi ^2}{12} \right) -2\left( \frac{\ln ^22}{2}-\frac{\pi ^2}{12} \right) =\ln ^22-\frac{\pi ^2}{6}.\end{align*}

## 关于$\pi$的级数

$$\sum\limits_{n = 0}^\infty {\frac{{1 + 14n + 76{n^2} + 168{n^3}}}{{{2^{20n}}}}{{\left( \begin{array}{c}2n\\n\end{array} \right)}^7}} = \frac{{32}}{{{\pi ^3}}}.$$

1.若$\alpha\beta=\pi$,则$$\sqrt \alpha \int_0^\infty {\frac{{{e^{ - {x^2}}}dx}}{{\cosh \alpha x}}} = \sqrt \beta \int_0^\infty {\frac{{{e^{ - {x^2}}}dx}}{{\cosh \beta x}}} .$$
2.求证$\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx} = \frac{\pi }{8} + \frac{{4\left( {837\sqrt 3 + 5\pi \left( {161 - 75\sqrt 3 \pi } \right)} \right)}}{{3375{\pi ^2}}}.$
3.2016.12.30
$\int_{ - \infty }^\infty {\frac{{{x^3}\sin x}}{{{x^4} + 2{x^2} + 1}}dx} = \frac{\pi }{{2e}}.$
4.求$\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{k - 1}}\ln \left( {2k - 1} \right)}}{{2k - 1}}} .$

## 与$\sum \arctan$有关的一些问题

\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*}

\begin{align*}S&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4}\right).\end{align*}

$$(3n^2+2)(9n^2-1)+10in = (n-i)(3n+i)(3n+i+1)(3n+i-1).$$

$$S = \arg \prod\limits_{n=1}^{\infty}\frac{\left(1+\frac{i}{3n}\right)\left(1+\frac{i+1}{3n}\right)\left(1+\frac{i-1}{3n}\right)}{\left(1+\frac{i}{n}\right)}.$$

$$S = \arg \frac{-\Gamma(i)}{\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right)\Gamma\left(\frac{i-1}{3}\right)}.$$

$$\Gamma(3z) = \frac{1}{2\pi}3^{3z - \frac{1}{2}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right).$$

$$\Gamma\left(\frac{i-1}{3}\right)\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right) = 2\pi 3^{-i+\frac{3}{2}}\Gamma(i-1)$$

\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*}

\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*}

\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

$\sqrt[p]{{n + 1}} - \sqrt[p]{n} = \frac{1}{p}{\xi ^{1/p - 1}} \ge \frac{1}{p}{\left( {n + 1} \right)^{1/p - 1}},\quad \xi \in \left( {n,n + 1} \right).$

$\sum\limits_{n = 1}^\infty {\frac{1}{{\left( {n + 1} \right)\sqrt[p]{n}}}} \le p\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{\sqrt[p]{n}}} - \frac{1}{{\sqrt[p]{{n + 1}}}}} \right)} = p.$

$$\frac{1}{a_1^p}+\sum\limits_{n=1}^{\infty}\frac{a_1a_2 \cdots a_n}{a_{n+1}^p} \ge \sum\limits_{n=0}^{\infty}(\frac{p}{p+1})^{n-p}.$$

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

(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.

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

\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*}

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

\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*}

$\sum\limits_{k = 1}^N {\frac{1}{{{k^4}}}} \cdot \sum\limits_{k = 1}^N {{a_k}} \ge {\left( {\sum\limits_{k = 1}^N {\frac{{a_k^{1/2}}}{{{k^2}}}} } \right)^2}.$

(1)$\displaystyle\sum\limits_{n=1}^{\infty}\frac{n}{S_n} \le 2 \sum\limits_{n=1}^{\infty}\frac{1}{a_n}$;
(2)$\displaystyle\sum\limits_{n=1}^{\infty}\frac{2n+1}{S_n} \le 4 \sum\limits_{n=1}^{\infty}\frac{1}{a_n}$.

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

\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*}

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

(2)如法炮制.由柯西不等式我们得

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

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

\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*}

$$\sum\limits_{n=1}^{\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}} \le e \sum\limits_{n=1}^{\infty}a_n$$,且证明$e$是最优值.

## 与$\sin n^2$类似的一些问题

1.证明: $\sum_{k=1}^n\sin k^2$无界.

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

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## 一个与多项式分拆有关的级数题

\begin{align*}&x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k + 1} \right) = {x^k} - {a_{k - 2}}x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k + 2} \right) \\&-  \cdots  - {a_2}x\left( {x - 1} \right)\left( {x - 2} \right) - {a_1}x\left( {x - 1} \right) - {a_0}x.\end{align*}

$\frac{{{{\left( {1 + 2 + \cdots + \left( {k - 1} \right)} \right)}^2} - \left( {{1^2} + {2^2} + \cdots + {{\left( {k - 1} \right)}^2}} \right)}}{2} = \left( {1 + 2 + \cdots + \left( {k - 2} \right)} \right){a_{k - 2}} - {a_{k - 3}}.$故${a_{k - 3}} = \frac{{\left( {3k - 5} \right)\left( {k - 2} \right)\left( {k - 1} \right)k}}{{24}}.$

\begin{align*}&{\left( { - 1} \right)^{k - n}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n}} \le k - 1} {{k_1}{k_2} \cdots {k_{k - n}}}  = {\left( { - 1} \right)^{k - n}}{a_{k - 2}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n - 1}} \le k - 2} {{k_1}{k_2} \cdots {k_{k - n - 1}}} \\&+ {\left( { - 1} \right)^{k - n - 1}}{a_{k - 3}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n - 2}} \le k - 3} {{k_1}{k_2} \cdots {k_{k - n - 2}}}  +  \cdots  + {\left( { - 1} \right)^2}{a_n}\sum\limits_{1 \le {k_1} \le n} {{k_1}} {x^n} + {\left( { - 1} \right)^1}{a_{n - 1}}.\end{align*}

\begin{align*}\sum\limits_{n = 1}^\infty  {\frac{{{n^k}}}{{n!}}}  &= \sum\limits_{n = 1}^\infty  {\frac{{{a_0}n + {a_1}n\left( {n - 1} \right) + {a_2}n\left( {n - 1} \right)\left( {n - 2} \right) +  \cdots  + {a_{k - 1}}n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k + 1} \right)}}{{n!}}} \\&= {a_0}\sum\limits_{n = 1}^\infty  {\frac{1}{{\left( {n - 1} \right)!}}}  + {a_1}\sum\limits_{n = 2}^\infty  {\frac{1}{{\left( {n - 2} \right)!}}}  + {a_2}\sum\limits_{n = 3}^\infty  {\frac{1}{{\left( {n - 3} \right)!}}}  +  \cdots  + {a_{k - 1}}\sum\limits_{n = k}^\infty  {\frac{1}{{\left( {n - k} \right)!}}} \\&= \left( {{a_0} + {a_1} + {a_2} +  \cdots {a_{k - 1}}} \right)e = Ne.\end{align*}

## $\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解

Proof.What you require here are the Eisenstein series. In particular the evaluation of

$$E_2(\tau) = 1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}},$$

at $\tau = i.$ Rearrange to get

$$\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau} } = \frac{1}{24}(1 – E_2(i) ).$$

See Lambert series for additional information.

The function

$$G_ 2(\tau) = \zeta(2) \left(1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}} \right)=\zeta(2)E_2(\tau)$$

satisfies the quasimodular transformation

$$G_ 2\left( \frac{a\tau+b}{c\tau+d} \right) =(c\tau+d)^2G_ 2(\tau) - \pi i c (c\tau + d).$$

And so with $a=d=0,$ $c=1$ and $b=-1$ we find $G_ 2(i) = \pi/2.$ Therefore

$$E_2(i) = \frac{ G_ 2( i)}{ \zeta(2)} = \frac{\pi}{2}\frac{6}{\pi^2} = \frac{3}{\pi}.$$

Hence we obtain

$$\sum_{n=1}^\infty \frac{n}{e^{2\pi n} – 1} = \frac{1}{24} - \frac{1}{8\pi},$$

as given in the comment to the question by Slowsolver.

There is a very nice generalisation of the sum in the question.

For odd $m > 1$ we have

$$\sum_{n=1}^\infty \frac{n^{2m-1} }{ e^{2\pi n} -1 } = \frac{B_{2m}}{4m},$$

where $B_k$ are the Bernoulli numbers defined by

$\frac{z}{{{e^z} - 1}} = \sum\limits_{k = 0}^\infty {\frac{{{B_k}}}{{k!}}} {z^k}\;\;\; for |z|<2\pi .$

$$\displaystyle\mathcal{M}\Big[\frac{x}{e^{2\pi x}-1}] = \int_{0}^{\infty} \frac{x^{s}}{e^{2 \pi x}-1} \ dx = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)$$

So $$\displaystyle \frac{x}{e^{2\pi x}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) x^{-s}\ ds$$

which implies $$\displaystyle \sum_{n=1}^{\infty}\frac{n}{e^{2\pi n}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s)\ ds$$

The integrand has poles at $s=-1, s=0$, and $s=1$ (and removable singularities at $s= -2, -3, -4, \ldots$)

I'm going to close the contour with a rectangle that has vertices at $$-i \infty, \frac{3}{2} - i \infty, \frac{3}{2} + i \infty, and \;i \infty$$ and is indented at the origin

$\Gamma(s)$ decays rapidly as $\text{Im} (s) \to \pm \infty$. So the integral goes to zero along the top and bottom of the rectangle.

And on the imaginary axis, the integrand is odd.

So $$\displaystyle \int_{\frac{3}{2}-i\infty}^{\frac{3}{2}+i\infty}(2\pi)^{-s}\Gamma(s+1)\zeta(s+1)\zeta(s)\ ds-\pi i \text{Res}[f,0] = 2\pi i\text{Res}[f,1]$$

where $$\displaystyle f(s) = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s)$$.

$$\displaystyle \text{Res}[f,0] = \lim_{s \to 0} s (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) = \lim_{s\to 0} s\zeta(s+1) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s)$$

$$= 1\Big(\frac{1}{2 \pi} \Big)(1)\zeta(0)=-\frac{1}{4 \pi}$$

$$\displaystyle\text{Res}[f,1] = \lim_{s \to 1} (s-1) (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s)$$

$$= \displaystyle \lim_{s\to 1}(s-1)\zeta(s) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)= 1\Big(\frac{1}{4 \pi^{2}}\Big)(1)\Big(\frac{\pi^{2}}{6}\Big) =\frac{1}{24}$$

Therefore,  $$\displaystyle \sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1} = \frac{1}{2 \pi i} \Big( 2 \pi i (\frac{1}{24}) + \pi i (\frac{-1}{4 \pi}) \Big) = \frac{1}{24} - \frac{1}{8 \pi}.$$

Poof.We will use the Mellin transform technique. Recalling the Mellin transform and its inverse

$$F(s) =\int_0^{\infty} x^{s-1} f(x)dx, \quad\quad f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} F(s)\, ds.$$

Now, let's consider the function

$$f(x)= \frac{x}{e^{\pi x}+1}.$$

Taking the Mellin transform of $f(x)$, we get

$$F(s)={\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1- {2}^{-s} \right) \zeta \left( s+1 \right),$$

where $\zeta(s)$ is the zeta function. Representing the function in terms of the inverse Mellin Transform, we have

$$\frac{x}{e^{\pi x}+1}=\frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left( 1-{2}^{-s} \right) \zeta \left( s+1 \right) x^{-s}ds.$$

Substituting $x=2n+1$ and summing yields

$$\sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{2\pi i}\int_{C}{\pi}^{-s-1}\Gamma \left( s+1 \right)\left(1-{2}^{-s} \right) \zeta\left( s+1 \right) \sum_{n=0}^{\infty}(2n+1)^{-s}ds$$

$$= \frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1-{2}^{-s} \right)^2\zeta\left( s+1 \right) \zeta(s)ds.$$

Now, the only contribution of the poles comes from the simple pole $s=1$ of $\zeta(s)$ and the residue equals to $\frac{1}{24}$. So, the sum is given by

$$\sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{24}$$

Notes: 1)

$$\sum_{n=0}^{\infty}(2n+1)^{-s}= \left(1- {2}^{-s} \right) \zeta \left( s \right).$$

2) The residue of the simple pole $s=1$, which is the pole of the zeta function, can be calculated as

$$r = \lim_{s=1}(s-1)({\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) \zeta(s))$$

$$= \lim_{s\to 1}(s-1)\zeta(s)\lim_{s\to 1} {\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) = \frac{1}{24}.$$

For calculating the above limit, we used the facts

$$\lim_{s\to 1}(s-1)\zeta(s)=1, \quad \zeta(2)=\frac{\pi^2}{6}.$$

3) Here is the technique for computing the Mellin transform of $f(x)$.
Using the change of variables $u=-\ln(x)$ and the identity

$$\int_{0}^{\infty}\frac{u^{s-1}}{e^u -1}=\zeta{(s)}\Gamma{(s)}$$

we reach to the deisred result

$$\int_0^1 \frac{\ln x }{x-1}= \int_{0}^{\infty}\frac{u}{e^u -1}=\zeta{(2)}\Gamma{(2)} =\sum_{n=1}^\infty \frac{1}{n^2}.$$

Note that,

$$\int_{0}^{\infty}\frac{u^{s-1}}{e^u - 1}=\int_{0}^{\infty}\frac{u^{s-1}}{e^u}(1-e^{-u})^{-1}= \sum_{n=0}^{\infty} \int_{0}^{\infty}{u^{s-1}e^{-(n+1)u}}$$

$$= \sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \int_{0}^{\infty}{y^{s-1}e^{-y}}= \sum_{n=1}^{\infty}\frac{1}{n^s} \Gamma(s)= \zeta(s) \Gamma(s).$$

If $Re(s)>1,Re(q)>0$,define$\zeta(s,q)=\sum_{n=0}^{\infty}\frac1{(q+n)^s}.$,then the function has an integral representation in terms of the Mellin transform as $\zeta(s,q)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}dt.$

## Euler Sum的若干研究

Solution.

\begin{align}&\sum_{j=0}^k\zeta(k+2-j)\zeta(j+2)\\&=\sum_{m=1}^\infty\sum_{n=1}^\infty\sum_{j=0}^k\frac1{m^{k+2-j}n^{j+2}}\tag{1}\\&=(k+1)\zeta(k+4)+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{m^2n^2}\frac{\frac1{m^{k+1}}-\frac1{n^{k+1}}}{\frac1m-\frac1n}\tag{2}\\&=(k+1)\zeta(k+4)+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{3}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{4}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{(n+m)m^{k+2}n}-\frac1{m(n+m)^{k+2}n}\tag{5}\\&=(k+1)\zeta(k+4)\\&+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^{k+3}n}-\frac1{(m+n)m^{k+3}}\\&-2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}\tag{6}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{n(n+m)^{k+3}}\tag{7}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{nm^{k+3}}\tag{8}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=n}^\infty\frac1{nm^{k+3}}+4\zeta(k+4)\tag{9}\\&=(k+5)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{m=1}^\infty\sum_{n=1}^m\frac1{nm^{k+3}}\tag{10}\\&=(k+5)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{11}\\&=(k+5)\zeta(k+4)-2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{12}\end{align}

Letting $q=k+3$ and reindexing $j\mapsto j-1$ yields
$$\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)=(q+2)\zeta(q+1)-2\sum_{m=1}^\infty\frac{H_m}{m^q}\tag{13}$$
and finally
$$\sum_{m=1}^\infty\frac{H_m}{m^q}=\frac{q+2}{2}\zeta(q+1)-\frac12\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)\tag{14}$$

Explanation

$\hphantom{0}(1)$ expand $\zeta$
$\hphantom{0}(2)$ pull out the terms for $m=n$ and use the formula for finite geometric sums on the rest
$\hphantom{0}(3)$ simplify terms
$\hphantom{0}(4)$ utilize the symmetry of $\frac1{nm^{k+2}(n-m)}+\frac1{mn^{k+2}(m-n)}$
$\hphantom{0}(5)$ $n\mapsto n+m$ and change the order of summation
$\hphantom{0}(6)$ $\frac1{mn}=\frac1{m(m+n)}+\frac1{n(m+n)}$
$\hphantom{0}(7)$ $H_m=\sum_{n=1}^\infty\frac1n-\frac1{n+m}$ and use the symmetry of $\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}$
$\hphantom{0}(8)$ $m\mapsto m-n$
$\hphantom{0}(9)$ subtract and add the terms for $m=n$
$(10)$ combine $\zeta(k+4)$ and change the order of summation
$(11)$ $H_m=\sum_{n=1}^m\frac1n$
$(12)$ combine sums

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## 一个级数求解

$\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^2}}}} = - \frac{{{\pi ^2}}}{6}\left( { - 12\ln A + \gamma + \ln 2 + \ln \pi } \right).$

(Riemann zeta函数的导函数)$\boxed{\zeta '\left( s \right) = - \sum\limits_{k = 1}^\infty {\frac{{\ln k}}{{{k^s}}}} .}$
(Gamma 函数)
$\Gamma \left( s \right) = \int_0^\infty {{x^{s - 1}}{e^{ - x}}dx}.$

$\Gamma '\left( 2 \right) = \int_0^\infty {x{e^{ - x}}\ln xdx} = \int_0^\infty {\left( {{e^{ - x}} + {e^{ - x}}\ln x} \right)dx} = 1 - \gamma .$

$\zeta \left( s \right) = {2^s}{\pi ^{s - 1}}\sin \frac{{\pi s}}{2}\Gamma \left( {1 - s} \right)\zeta \left( {1 - s} \right).$

$\ln \zeta \left( s \right) = s\ln 2 + \left( {s - 1} \right)\ln \pi + \ln \sin \frac{{\pi s}}{2} + \ln \Gamma \left( {1 - s} \right) + \ln \zeta \left( {1 - s} \right).$

$\frac{{\zeta '\left( { - 1} \right)}}{{\zeta \left( { - 1} \right)}} = 12\ln A - 1 = \ln \left( {2\pi } \right) - 1 + \gamma - \frac{{\zeta '\left( 2 \right)}}{{\zeta \left( 2 \right)}}.$

$\Rightarrow \zeta '\left( 2 \right) = \frac{{{\pi ^2}}}{6}\left( {\ln \left( {2\pi } \right) - 12\ln A + \gamma } \right).$