How to show $\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$ How to show 
$$\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$$
 A: Stahl's approach basically works. A walk-through solution is also possible: Notice that the digamma function $\psi(z)$ satisfies
$$ \psi(z) = -\gamma+\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z}\right). $$
Consequently
\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{(24n+5)(24n+19)}
&= \frac{1}{14} \sum_{n=0}^{\infty} \left( \frac{1}{24n+5} - \frac{1}{24n+19} \right) \\
&= \frac{1}{14 \cdot 24} \left( \psi\left( \frac{19}{24} \right) - \psi\left( \frac{5}{24} \right) \right).
\end{align*}
Now applying the Euler's reflection formula shows that
$$ \psi(z) - \psi(1-z) = \frac{d}{dz}\log ( \Gamma(z)\Gamma(1-z) ) = \frac{d}{dz}\log\left( \frac{\pi}{\sin \pi z} \right) = -\pi \cot(\pi z). $$
Plugging $z = 19/24$ gives the same answer as Jack D'Aurizio. (Basically, one solution implies the other so both solutions are essentially equivalent.)
A: $$\sum_{n\geq 0}\frac{1}{(24n+12)^2-7^2}=\frac{1}{12^2}\sum_{n\geq 0}\frac{1}{(2n+1)^2-\left(\frac{7}{12}\right)^2}$$
but from the logarithmic derivatives of the Weierstrass product for the cosine function it follows that, for any $\alpha\in(0,1)$:
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^2-\alpha^2}=\frac{\pi}{4a}\tan\frac{\pi a}{2}$$
so we are just left with few trigonometric manipulations.
A: hint: $\displaystyle \sum_{n=0}^\infty\dfrac{1}{(24n+5)(24n+19)} = \dfrac{1}{14}\displaystyle \sum_{n=0}^\infty\left(\dfrac{1}{24n+5} - \dfrac{1}{24n+19}\right)= \dfrac{1}{14}\displaystyle \sum_{n=0}^\infty\left(\displaystyle \int_{0}^1\left(x^{24n+4} - x^{24n+18}\right)dx\right)= \dfrac{1}{14}\displaystyle \int_{0}^1\displaystyle \left(\sum_{n=0}^\infty\left(x^{24n+4} - x^{24n+18}\right)\right)dx$. Can you continue?
