log絡みの級数

\begin{alignat}{2}
&(1) \log \frac{x+1}{x-1}=2\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2n-1)x^{2n-1}}  (|x| \gt 1)\\
&(2) \log \frac{x}{x-1}=\displaystyle\sum_{n=1}^{\infty} \frac{1}{nx^n}  (|x| \gt 1)\\
&(3) \frac{1-x}{x}\log \frac{1}{1-x}=1-\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n(n+1)}  (|x| \lt 1)\\
&(4) \frac{1}{1-x}\log \frac{1}{1-x}=\displaystyle\sum_{n=1}^{\infty} x^n \displaystyle\sum_{k=1}^n \frac{1}{k}  (|x| \lt 1)\\
&(5) \frac{(1-x)^2}{2x^3}\log \frac{1}{1-x}=\frac{1}{2x^2}-\frac{3}{4x}+\displaystyle\sum_{n=1}^{\infty} \frac{x^{n-1}}{n(n+1)(n+2)}  (|x| \lt 1)
\end{alignat}









<証明>

次の級数展開の結果を用います。(詳細はこちらです)
\begin{alignat}{2}
&(A) \log (1-x)=-\displaystyle\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}  ( -1 \lt x \lt 1) \\
&(B) \log \left(\frac{1+x}{1-x}\right)=2\displaystyle\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}  ( -1 \lt x \lt 1) \\
\end{alignat}





\((1)\) \((B)\) の式で \(x\) を \(\displaystyle \frac{1}{x}\) に書き換えます。$$\log \frac{1+\frac{1}{x}}{1-\frac{1}{x}}=2\displaystyle\sum_{n=1}^{\infty} \frac{1}{2n-1}\left(\frac{1}{x}\right)^{2n-1}  (|x| \gt 1)$$以上より$$\log \frac{x+1}{x-1}=2\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2n-1)x^{2n-1}}  (|x| \gt 1)$$





\((2)\) \((A)\) の式で \(x\) を \(\displaystyle \frac{1}{x}\) に書き換えます。$$\log \left(1-\frac{1}{x}\right)=-\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{1}{x}\right)^n,  -\log \frac{x-1}{x}=\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{1}{x}\right)^n$$以上より$$\log \frac{x}{x-1}=\displaystyle\sum_{n=1}^{\infty} \frac{1}{nx^n}  (|x| \gt 1)\\$$






\((3)(4)(5)\) は \(\log \) を級数で表してから \(x\) の次数ごとにまとめます。

\begin{alignat}{2}
&(3)  \frac{1-x}{x}\log \frac{1}{1-x}=\frac{1-x}{x}\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}\\
&=-\left(1-\frac{1}{x}\right)\left(x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots \right)\\
&=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots\right)+\frac{1}{x}\left(x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots\right)\\
&=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots\right)+1+\left(\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+ \cdots\right)\\
&=1-\left\{\left(1-\frac{1}{2}\right)x+\left(\frac{1}{2}-\frac{1}{3}\right)x+\left(\frac{1}{3}-\frac{1}{4}\right)x^2+ \cdots\right\}\\
&=1-\left(\frac{x}{1 \cdot 2}+\frac{x^2}{2 \cdot 3}+\frac{x^3}{3 \cdot 4}+\frac{x^4}{4 \cdot 5}+ \cdots \right)=1-\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n(n+1)}
\end{alignat}以上より$$\frac{1-x}{x}\log \frac{1}{1-x}=1-\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n(n+1)}  (|x| \lt 1)$$







\begin{alignat}{2}
&(4)  \frac{1}{1-x}\log \frac{1}{1-x}=\displaystyle\sum_{n=1}^{\infty} x^{n-1}\displaystyle\sum_{k=1}^{\infty}\frac{x^k}{k}\\
&=(1+x+x^2+x^3+ \cdots)\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+ \cdots\right)\\
&=x+\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3+ \cdots \\
&            \cdots +\left(1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}\right)x^n+ \cdots \\
&=\displaystyle\sum_{n=1}^{\infty} \left(1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}\right)x^n\\
&=\displaystyle\sum_{n=1}^{\infty} \left(\displaystyle\sum_{k=1}^n \frac{1}{k}\right)x^n=\displaystyle\sum_{n=1}^{\infty} x^n \displaystyle\sum_{k=1}^n \frac{1}{k}
\end{alignat}以上より$$\frac{1}{1-x}\log \frac{1}{1-x}=\displaystyle\sum_{n=1}^{\infty} x^n \displaystyle\sum_{k=1}^n \frac{1}{k}  (|x| \lt 1)$$







\begin{alignat}{2}
&(5)  \frac{(1-x)^2}{2x^3}\log \frac{1}{1-x}=\frac{1-2x+x^2}{2x^3}\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}\\
&=\left(\frac{1}{2x^3}-\frac{1}{x^2}+\frac{1}{2x}\right)\left(x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots\right)\\
&=\frac{1}{2x^3}\left(x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots\right)-\frac{1}{x^2}\left(x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots\right)+\frac{1}{2x}\left(x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots\right)\\
&=\frac{1}{2x^2}+\frac{1}{4x}+\frac{1}{2}\left(\frac{1}{3}+\frac{x}{4}+\frac{x^2}{5}+\frac{x^3}{6}+ \cdots\right)-\frac{1}{x}+\left(-\frac{1}{2}-\frac{x}{3}-\frac{x^2}{4}-\frac{x^3}{5}- \cdots \cdots\right)+\frac{1}{2}\left(1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+ \cdots\right)\\
&=\frac{1}{2x^2}-\frac{3}{4x}+\left[\left(\frac{1}{6}-\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{8}-\frac{1}{3}+\frac{1}{4}\right)x+\left(\frac{1}{10}-\frac{1}{4}+\frac{1}{6}\right)x^2+ \cdots +\left\{\frac{1}{2(n+1)}-\frac{1}{n+1}+\frac{1}{2n}\right\}x^{n-1}+ \cdots \right]\\
&=\frac{1}{2x^2}-\frac{3}{4x}+\displaystyle\sum_{n=1}^{\infty} \left\{\frac{1}{2(n+1)}-\frac{1}{n+1}+\frac{1}{2n}\right\}x^{n-1}\\
&=\frac{1}{2x^2}-\frac{3}{4x}+\displaystyle\sum_{n=1}^{\infty} \frac{x^{n-1}}{n(n+1)(n+2)}
\end{alignat}以上より$$\frac{(1-x)^2}{2x^3}\log \frac{1}{1-x}=\frac{1}{2x^2}-\frac{3}{4x}+\displaystyle\sum_{n=1}^{\infty} \frac{x^{n-1}}{n(n+1)(n+2)}  (|x| \lt 1)$$

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