# 与调和数列有关的级数计算

Eufisky posted @ 2017年8月11日 16:49 in 数学分析 with tags 级数 , 1061 阅读

\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*}

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