计算
\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{2}{n+1}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)=\ln^2 2\]
解:
\[ H_{n}=H_{n+1}-\frac{1}{n+1} \]
\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{2}{n+1}\left(H_{n+1}-\frac{1}{n+1}\right)=\sum_{n=1}^{\infty}\frac{2(-1)^{n-1}}{n+1}H_{n+1}-\sum_{n=1}^{\infty}\frac{2(-1)^{n-1}}{(n+1)^{2}}\]
\begin{align*}
H_{n+1}&=\int_{0}^{1}(1+x+\cdots+x^{n})dx\\
&=\int_{0}^{1}\frac{1-x^{n+1}}{1-x}dx\\
&=\int_{0}^{1}\frac{1-(1-t)^{n+1}}{t}dt\qquad (t=1-x)\\
&=[1-(1-t)^{n+1}]\ln{t}\bigg|_{0}^{1}-\int_{0}^{1}(n+1)(1-t)^{n}\ln{t}dt\\
&=-(n+1)\int_{0}^{1}(1-t)^{n}\ln{t}dt
\end{align*}
\begin{align*}
 I_{1}&=\sum_{n=1}^{\infty}\frac{2(-1)^{n-1}}{n+1}H_{n+1}\\
&=\sum_{n=1}^{\infty}2\int_{0}^{1}(t-1)^{n}\ln{t}dt\\
&=2\int_{0}^{1}\frac{t-1}{2-t}\ln{t}dt\\
&=-2\int_{0}^{1}\ln{t}dt+2\int_{0}^{1}\frac{1}{2-t}\ln{t}dt\\
&=2+2\int_{0}^{1}\frac{\ln(1-x)}{1+x}\qquad (x=1-t)\\
&=2+2\int_{0}^{1}\frac{1}{1+z}\cdot \ln\left(\frac{2z}{1+z}\right)dz\qquad (x=\frac{1-z}{1+z})\\
&=2+\ln^2{2}-2\int_{0}^{1}\frac{\ln(1+z)}{z}dz\\
&= 2+\ln^2{2}+2\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^2}
\end{align*}
于是
\[ I=2+\ln^2{2}+2\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^2}-2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+1)^2}=\ln^2 2 \]

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