Σ[n=0,∞](-1)^n/(3n+1)(3n+2)(3n+3)(3n+4)などの級数

\begin{alignat}{2}
&(1)  \displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)(3n+3)(3n+4)}=\frac{1}{6}-\frac{1}{4}\log 3+\frac{\sqrt{3}}{36}π\\
&(2)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(3n+1)(3n+2)(3n+3)(3n+4)}=-\frac{1}{6}+\frac{4}{9}\log 3+\frac{\sqrt{3}}{54}π\\
\end{alignat}








<証明>

次の等式を用います。[詳細はこちらです。(A)(B)(C)]
\begin{alignat}{2}
&(A)  ψ\left(\frac{2}{3}\right)=-γ-\frac{3}{2} \log 3+\frac{π}{2\sqrt{3}}\\
&(B)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+1}=\frac{\sqrt{3}π+3 \log 2}{9}\\
&(C)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+2}=\frac{\sqrt{3}π-3 \log 2}{9}\\
\end{alignat}




\((1)\) 部分分数分解を行います。
\begin{alignat}{2}
&  \displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)(3n+3)(3n+4)}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty} \left\{\frac{1}{(3n+1)(3n+4)}-\frac{1}{(3n+2)(3n+3)}\right\}\\
\end{alignat}それぞれのシグマの計算をします。

\begin{alignat}{2}
&(A) \displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+4)}=\frac{1}{3}\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{3n+1} -\frac{1}{3n+4}\right)\\
&=\frac{1}{3}\left\{\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+ \cdots \right\}=\frac{1}{3}
\end{alignat}
\begin{alignat}{2}
&(B) \displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+2)(3n+3)}=\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{3n+2} -\frac{1}{3n+3}\right)\\
&=\frac{1}{3}\left\{ψ(1)-ψ\left(\frac{2}{3}\right)\right\}=\frac{1}{3}\left\{-γ-\left(-γ-\frac{3}{2}\log 3+\frac{π}{2\sqrt{3}}\right)\right\}\\
&=\frac{1}{3}\left(\frac{3}{2}\log 3-\frac{π}{2\sqrt{3}}\right)=\frac{1}{2}\log 3 -\frac{π}{6\sqrt{3}}\\
\end{alignat}となるので、元の級数の計算は$$=\frac{1}{2}\left\{\frac{1}{3}-\left(\frac{1}{2}\log 3 -\frac{π}{6\sqrt{3}}\right)\right\}=\frac{1}{6}-\frac{1}{4}\log 3+\frac{\sqrt{3}}{36}π$$以上より$$\displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)(3n+3)(3n+4)}=\frac{1}{6}-\frac{1}{4}\log 3+\frac{\sqrt{3}}{36}π$$








\((2)\) 部分分数分解を行います。
\begin{alignat}{2}
&  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(3n+1)(3n+2)(3n+3)(3n+4)}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty} (-1)^n\left\{\frac{1}{(3n+1)(3n+4)}-\frac{1}{(3n+2)(3n+3)}\right\}\\
\end{alignat}それぞれのシグマの計算をします。

\begin{alignat}{2}
&(A) \displaystyle\sum_{n=0}^{\infty}(-1)^n \frac{1}{(3n+1)(3n+4)}=\frac{1}{3}\displaystyle\sum_{n=0}^{\infty}(-1)^n \left(\frac{1}{3n+1} -\frac{1}{3n+4}\right)\\
&=\frac{1}{3}\left(1-\frac{2}{4}+\frac{2}{7}-\frac{2}{10}+ \cdots \right)=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+ \cdots \right)\\
&=\frac{2}{3}\left\{-\frac{1}{2}+\left(1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+ \cdots \right)\right\}=-\frac{1}{3}+\frac{2}{3} \cdot \frac{\sqrt{3}π+3 \log 2}{9}
\end{alignat}
\begin{alignat}{2}
&(B) \displaystyle\sum_{n=0}^{\infty} (-1)^n\frac{1}{(3n+2)(3n+3)}=\displaystyle\sum_{n=0}^{\infty} (-1)^n\left(\frac{1}{3n+2} -\frac{1}{3n+3}\right)\\
&=\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+2}-\frac{1}{3}\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}=\frac{\sqrt{3}π-3 \log 2}{9}-\frac{1}{3}\log 2\\
\end{alignat}となるので、元の級数の計算は
\begin{alignat}{2}
&=\frac{1}{2}\left\{-\frac{1}{3}+\frac{2}{3} \cdot \frac{\sqrt{3}π+3 \log 2}{9}-\left(\frac{\sqrt{3}π-3 \log 2}{9}-\frac{1}{3}\log 2\right)\right\}\\
&=-\frac{1}{6}+\frac{\sqrt{3}}{27}π+\frac{1}{9}\log 2-\frac{\sqrt{3}}{18}π+\frac{1}{6}\log 2+\frac{1}{6}\log 2=-\frac{1}{6}+\frac{4}{9}\log 3+\frac{\sqrt{3}}{54}π
\end{alignat}以上より$$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(3n+1)(3n+2)(3n+3)(3n+4)}=-\frac{1}{6}+\frac{4}{9}\log 3+\frac{\sqrt{3}}{54}π$$


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