Σ[n=1,∞]1/n(n+1)(n+2)(n+3)などの級数

\begin{alignat}{2}
&(1)  \displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}=\frac{1}{4}\\
&(2)  \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)(n+2)}=2 \log 2 -\frac{5}{4}\\
&(3)  \displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{18}\\
&(4)  \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)(n+2)(n+3)}=\frac{4}{3}\log 2-\frac{8}{9}\\
\end{alignat}








<証明>

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






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







\begin{alignat}{2}
&(3)  \displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}\displaystyle\sum_{n=1}^{\infty}\left\{\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right\}\\
&=\frac{1}{3}\left\{\left(\frac{1}{1 \cdot 2 \cdot 3}-\frac{1}{2 \cdot 3 \cdot 4}\right)+\left(\frac{1}{2 \cdot 3 \cdot 4}-\frac{1}{3 \cdot 4 \cdot 5}\right)+\left(\frac{1}{3 \cdot 4 \cdot 5}-\frac{1}{4 \cdot 5 \cdot 6}\right)+ \cdots \right\}=\frac{1}{18}
\end{alignat}以上より$$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{18}$$







\((4)\) 部分分数分解後、\((2)\) の結果を用います。
\begin{alignat}{2}
&  \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)(n+2)(n+3)}\\
&=\frac{1}{3}\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\left\{\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right\}\\
&=\frac{1}{3}\left\{\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)(n+2)}-\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n+1)(n+2)(n+3)}\right\}\\
\end{alignat}右のシグマについて
\begin{alignat}{2}
&  \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)(n+2)(n+3)}=\frac{1}{6}+\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{(n+1)(n+2)(n+3)}\\
&=\frac{1}{6}+\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n(n+1)(n+2)}=\frac{1}{6}-\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)(n+2)}\\
\end{alignat}となるので、元の級数の計算に戻ります。
\begin{alignat}{2}
&=\frac{1}{3}\left\{\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)(n+2)}-\frac{1}{6}+\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)(n+2)}\right\}\\
&=\frac{1}{3}\left\{-\frac{1}{6}+2\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)(n+2)}\right\}\\
&=-\frac{1}{18}+\frac{2}{3}\left(2 \log 2 -\frac{5}{4}\right)=\frac{4}{3}\log 2-\frac{8}{9}
\end{alignat}以上より$$\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)(n+2)(n+3)}=\frac{4}{3}\log 2-\frac{8}{9}$$

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