TITLE: Proving that $\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}=\frac{\pi}{3}$
QUESTION [12 upvotes]: The following problem(p.668, 7) is from Integrals and Series [ Интегралы и ряды, А.П. Прудников, Ю.А. Брычков, О.И. Маричев.] states that
$$\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}=\frac{\pi}{3}$$
How one can show that?

REPLY [4 votes]: \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}