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

\begin{alignat}{2}
&(1)  \displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)(3n+3)}=\frac{\sqrt{3}}{12}π-\frac{1}{4}\log 2\\
&(2)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(3n+1)(3n+2)(3n+3)}=\frac{2}{3}\log 2-\frac{\sqrt{3}}{18}π\\
&(3)  \displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+2)(3n+3)(3n+4)}=\frac{1}{2}\log 3 -\frac{1}{2}\\
&(4)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(3n+2)(3n+3)(3n+4)}=\frac{1}{2}-\frac{2}{3}\log 2\\
\end{alignat}






<証明>

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





\begin{alignat}{2}
&(1)  \displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)(3n+3)}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}\left\{\frac{1}{(3n+1)(3n+2)}-\frac{1}{(3n+2)(3n+3)}\right\}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}\left(\frac{1}{3n+1}-\frac{1}{3n+2}-\frac{1}{3n+2}+\frac{1}{3n+3}\right)\\
&=\frac{1}{6}\left\{ψ\left(\frac{2}{3}\right)-ψ\left(\frac{1}{3}\right)+ψ\left(\frac{2}{3}\right)-ψ(1)\right\}\\
&=\frac{1}{6}\left\{2\left(-γ-\frac{3}{2}\log 3 +\frac{π}{2\sqrt{3}}\right)-\left(-γ-\frac{3}{2}\log 3 -\frac{π}{2\sqrt{3}}\right)+γ\right\}\\
&=\frac{1}{6}\left(-3 \log 3+\frac{π}{\sqrt{3}}+\frac{3}{2}\log 3 +\frac{π}{2\sqrt{3}}\right)\\
&=\frac{1}{6}\left(\frac{\sqrt{3}}{2}π-\frac{3}{2}\log 3\right)=\frac{\sqrt{3}}{12}π-\frac{1}{4}\log 2
\end{alignat}以上より$$\displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)(3n+3)}=\frac{\sqrt{3}}{12}π-\frac{1}{4}\log 2$$







\begin{alignat}{2}
&(2)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(3n+1)(3n+2)(3n+3)}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}(-1)^n\left\{\frac{1}{(3n+1)(3n+2)}-\frac{1}{(3n+2)(3n+3)}\right\}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{3n+1}-\frac{1}{3n+2}-\frac{1}{3n+2}+\frac{1}{3n+3}\right)\\
&=\frac{1}{2}\left\{\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1}-2\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{3n+2}+\frac{1}{3}\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}\right\}\\
&=\frac{1}{2}\left(\frac{\sqrt{3}π+3 \log 2}{9}-2 \cdot \frac{\sqrt{3}π-3 \log 2}{9}+\frac{1}{3} \log 2\right)\\
&=\frac{1}{2}\left(\frac{\sqrt{3}}{9}π+\frac{1}{3}\log 2-\frac{2\sqrt{3}}{9}π+\frac{2}{3}\log 2+\frac{1}{3}\log 2\right)\\
&=\frac{1}{2}\left(\frac{4}{3}\log 2 -\frac{\sqrt{3}}{9}π\right)=\frac{2}{3}\log2 -\frac{\sqrt{3}}{18}π\\
\end{alignat}以上より$$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(3n+1)(3n+2)(3n+3)}=\frac{2}{3}\log 2-\frac{\sqrt{3}}{18}π$$







\begin{alignat}{2}
&(3)  \displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+2)(3n+3)(3n+4)}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}\left\{\frac{1}{(3n+2)(3n+3)}-\frac{1}{(3n+3)(3n+4)}\right\}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}\left(\frac{1}{3n+2}-\frac{1}{3n+3}-\frac{1}{3n+3}+\frac{1}{3n+4}\right)\\
&=\frac{1}{6}\left\{ψ\left(1\right)-ψ\left(\frac{2}{3}\right)+ψ\left(1\right)-ψ\left(\frac{4}{3}\right)\right\}\\
&=\frac{1}{6}\left\{2ψ(1)-ψ\left(\frac{2}{3}\right)-ψ\left(\frac{1}{3}\right)-3\right\}\\
&=\frac{1}{6}\left\{-2γ-\left(-γ-\frac{3}{2}\log 3+\frac{π}{2\sqrt{3}}\right)-\left(-γ-\frac{3}{2}\log 3-\frac{π}{2\sqrt{3}}\right)-3\right\}\\
&=\frac{1}{6}\left(\frac{3}{2}\log 3-\frac{π}{2\sqrt{3}}+\frac{3}{2}\log 3+\frac{π}{2\sqrt{3}}-3\right)\\
&=\frac{1}{6}(3 \log 3-3)=\frac{1}{2}\log 3-\frac{1}{2}\\
\end{alignat}以上より$$\displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+2)(3n+3)(3n+4)}=\frac{1}{2}\log 3 -\frac{1}{2}$$







\begin{alignat}{2}
&(4)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(3n+2)(3n+3)(3n+4)}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}(-1)^n\left\{\frac{1}{(3n+2)(3n+3)}-\frac{1}{(3n+3)(3n+4)}\right\}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{3n+2}-\frac{1}{3n+3}-\frac{1}{3n+3}+\frac{1}{3n+4}\right)\\
&=\frac{1}{2}\left\{\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+2}-\frac{2}{3}\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}+\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+4}\right\}\\
\end{alignat}一番右のシグマについて$$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+4}=\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{3n+1}=1-\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+1}$$となるので
\begin{alignat}{2}
&=\frac{1}{2}\left\{\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+2}-\frac{2}{3}\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}-\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{3n+1}\right\}+\frac{1}{2}\\
&=\frac{1}{2}\left(\frac{\sqrt{3}π-3 \log 2}{9}-\frac{2}{3}\log 2-\frac{\sqrt{3}π+3 \log 2}{9}\right)+\frac{1}{2}\\
&=\frac{1}{2}\left(\frac{\sqrt{3}}{9}π-\frac{1}{3}\log 2-\frac{2}{3}\log 2-\frac{\sqrt{3}}{9}π-\frac{1}{3}\log 2\right)+\frac{1}{2}=\frac{1}{2}-\frac{2}{3}\log 2
\end{alignat}以上より$$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(3n+2)(3n+3)(3n+4)}=\frac{1}{2}-\frac{2}{3}\log 2$$

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