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

\begin{alignat}{2}
&(1)  \displaystyle\sum_{n=0}^{\infty} \frac{1}{(4n+3)(4n+4)(4n+5)}=\frac{3}{4}\log 2-\frac{1}{2}\\
&(2)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+3)(4n+4)(4n+5)}=\frac{1}{2}-\frac{1}{2\sqrt{2}}\coth^{-1}\sqrt{2} -\frac{1}{4}\log 2\\
&(3)  \displaystyle\sum_{n=1}^{\infty} \frac{1}{4n(4n+1)(4n+2)}=\frac{3}{4}-\frac{1}{2}\log 2-\frac{π}{8}\\
&(4)  \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{4n(4n+1)(4n+2)}=-\frac{3}{4}+\frac{2\sqrt{2}-1}{16}π+\frac{1}{2\sqrt{2}}\coth^{-1}\sqrt{2}+\frac{1}{8}\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)  ψ\left(\frac{1}{4}\right)=-γ-3 \log 2-\frac{π}{2}\\
&(B)  ψ\left(\frac{3}{4}\right)=-γ-3 \log 2+\frac{π}{2}\\
&(C)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{4n+1}=\frac{π+2 \coth^{-1}\sqrt{2}}{4\sqrt{2}}\\
&(D)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{4n+3}=\frac{π-2 \coth^{-1}\sqrt{2}}{4\sqrt{2}}\\
\end{alignat}


\begin{alignat}{2}
&(1)  \displaystyle\sum_{n=0}^{\infty} \frac{1}{(4n+3)(4n+4)(4n+5)}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}\left\{\frac{1}{(4n+3)(4n+4)}-\frac{1}{(4n+4)(4n+5)}\right\}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{4n+3}-\frac{1}{4n+4}-\frac{1}{4n+4}+\frac{1}{4n+5}\right)\\
&=\frac{1}{8}\left\{ψ\left(1\right)-ψ\left(\frac{3}{4}\right)+ψ\left(1\right)-ψ\left(\frac{5}{4}\right)\right\}\\
&=\frac{1}{8}\left\{-2γ-ψ\left(\frac{3}{4}\right)-ψ\left(\frac{1}{4}\right)-4\right\}\\
&=\frac{1}{8}\left\{-2γ-\left(-γ-3 \log 2+\frac{π}{2}\right)-\left(-γ-3 \log 2-\frac{π}{2}\right)-4\right\}\\
&=\frac{1}{8}\left(3 \log 2-\frac{π}{2}+3 \log 2+\frac{π}{2}-4\right)\\
&=\frac{1}{8}(6 \log 2-4)=\frac{3}{4}\log 2-\frac{1}{2}
\end{alignat}以上より$$\displaystyle\sum_{n=0}^{\infty} \frac{1}{(4n+3)(4n+4)(4n+5)}=\frac{3}{4}\log 2-\frac{1}{2}$$






\begin{alignat}{2}
&(2)  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+3)(4n+4)(4n+5)}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}(-1)^n\left\{\frac{1}{(4n+3)(4n+4)}-\frac{1}{(4n+4)(4n+5)}\right\}\\
&=\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}(-1)^n \left(\frac{1}{4n+3}-\frac{1}{4n+4}-\frac{1}{4n+4}+\frac{1}{4n+5}\right)\\
&=\frac{1}{2}+\frac{1}{2}\left\{\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{4n+3}-\frac{1}{2}\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}-\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{4n+1}\right\}\\
&=\frac{1}{2}+\frac{1}{2}\left(\frac{π-2 \coth^{-1}\sqrt{2}}{4\sqrt{2}}-\frac{1}{2}\log 2-\frac{π+2 \coth^{-1}\sqrt{2}}{4\sqrt{2}}\right)\\
&=\frac{1}{2}+\frac{1}{2}\left(-\frac{1}{\sqrt{2}}\coth^{-1}\sqrt{2}-\frac{1}{2}\log 2\right)=\frac{1}{2}-\frac{1}{2\sqrt{2}}\coth^{-1}\sqrt{2} -\frac{1}{4}\log 2
\end{alignat}以上より$$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+3)(4n+4)(4n+5)}=\frac{1}{2}-\frac{1}{2\sqrt{2}}\coth^{-1}\sqrt{2} -\frac{1}{4}\log 2$$







\begin{alignat}{2}
&(3)  \displaystyle\sum_{n=1}^{\infty} \frac{1}{4n(4n+1)(4n+2)}\\
&=\frac{1}{2}\displaystyle\sum_{n=1}^{\infty}\left\{\frac{1}{4n(4n+1)}-\frac{1}{(4n+1)(4n+2)}\right\}\\
&=\frac{1}{2}\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{4n}-\frac{1}{4n+1}-\frac{1}{4n+1}+\frac{1}{4n+2}\right)\\
&=\frac{3}{4}+\frac{1}{2}\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{4n+4}-\frac{1}{4n+1}-\frac{1}{4n+1}+\frac{1}{4n+2}\right)\\
&=\frac{3}{4}+\frac{1}{8}\left\{ψ\left(\frac{1}{4}\right)-ψ\left(1\right)+ψ\left(\frac{1}{4}\right)-ψ\left(\frac{1}{2}\right)\right\}\\
&=\frac{3}{4}+\frac{1}{8}\left\{2\left(-γ-3\log 2-\frac{π}{2}\right)+γ+γ+2 \log 2\right\}\\
&=\frac{3}{4}+\frac{1}{8}(-6 \log 2-π+2 \log 2)=\frac{3}{4}-\frac{1}{2}\log 2-\frac{π}{8}
\end{alignat}以上より$$\displaystyle\sum_{n=1}^{\infty} \frac{1}{4n(4n+1)(4n+2)}=\frac{3}{4}-\frac{1}{2}\log 2-\frac{π}{8}$$







\begin{alignat}{2}
&(4)  \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{4n(4n+1)(4n+2)}\\
&=\frac{1}{2}\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\left\{\frac{1}{4n(4n+1)}-\frac{1}{(4n+1)(4n+2)}\right\}\\
&=\frac{1}{2}\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1} \left(\frac{1}{4n}-\frac{1}{4n+1}-\frac{1}{4n+1}+\frac{1}{4n+2}\right)\\
&=-\frac{3}{4}+\frac{1}{2}\displaystyle\sum_{n=0}^{\infty}(-1)^{n-1} \left(-\frac{1}{4n+4}-\frac{1}{4n+1}-\frac{1}{4n+1}+\frac{1}{4n+2}\right)\\
&=-\frac{3}{4}+\frac{1}{2}\left\{\frac{1}{4}\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}+2\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}-\frac{1}{2}\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}\right\}\\
&=-\frac{3}{4}+\frac{1}{2}\left(\frac{1}{4}\log 2+2 \cdot \frac{π+2 \coth^{-1}\sqrt{2}}{4\sqrt{2}} -\frac{1}{2} \cdot \frac{π}{4} \right)\\
&=-\frac{3}{4}+\frac{1}{8}\log 2+\frac{π}{4\sqrt{2}}+\frac{1}{2\sqrt{2}}\coth^{-1}\sqrt{2}-\frac{π}{16}\\
&=-\frac{3}{4}+\frac{2\sqrt{2}-1}{16}π+\frac{1}{2\sqrt{2}}\coth^{-1}\sqrt{2}+\frac{1}{8}\log 2
\end{alignat}以上より$$\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{4n(4n+1)(4n+2)}=-\frac{3}{4}+\frac{2\sqrt{2}-1}{16}π+\frac{1}{2\sqrt{2}}\coth^{-1}\sqrt{2}+\frac{1}{8}\log 2$$

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