Demostrando que $\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}=\frac{\pi}{3}$

Question

El siguiente problema(p.668, 7) es de las Integrales y Series [ Интегралы и ряды, А.П. Прудников, Ю.А. Брычков, О.И. Маричев.] los estados que

$$\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}=\frac{\pi}{3}$$

Cómo puede uno mostrar que?

Answer 1

\begin{align} \sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}&=-\sum_{k=0}^\infty \frac{1}{k+1/2}+\frac{2}{3}\sum_{k=0}^{\infty} \frac{1}{k+1/4}+\frac{1}{3} \sum_{k=0}^\infty \frac{1}{k+1} \\ &=\frac{1}{3}\sum_{k=0}^\infty \frac{1}{k+1}-\frac{1}{k+1/2}+\frac{2}{3} \sum_{k=0}^{\infty} \frac{1}{k+1/4}-\frac{1}{k+1/2} \\ &=-\frac{1}{6}\sum_{k=0}^{\infty} \frac{1}{(k+1)(k+1/2)}+\frac{1}{6}\sum_{k=0}^\infty \frac{1}{(k+1/4)(k+1/2)} \\ \text{Using Gauss's Digamma Theorem, }\\ &=-\frac{1}{6}\cdot \frac{\psi(1)-\psi(1/2)}{1/3-1/2}+ \frac{1}{6}\cdot \frac{\psi(1/4)-\psi(1/2)}{1/4-1/2} \\ &=-\frac{1}{6}\cdot \log 16+\frac{1}{6}\cdot 2(\pi+\log4)=\frac{\pi}{3} \end{align}