Interesting Infinite Sums

Here are some interesting values of infinite sums that I calculated:

\[\displaystyle \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)} = \frac{\pi}{3\sqrt{3}}\]

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+3)} = \frac{\pi}{8}$ (Leibniz's formula)

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(6n+1)(6n+5)} = \frac{\pi}{8\sqrt{3}}$

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(8n+1)(8n+7)} = \frac{\pi}{48}(\sqrt{2}+1)$

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(8n+3)(8n+5)} = \frac{\pi}{16}(\sqrt{2}-1)$

My method for calculation using the result:

$\displaystyle \sum_{r=1}^n \cos rx = \frac{\sin(n+\frac{1}{2})x-\sin\frac{1}{2}x}{2\sin\frac{1}{2}x}$ for $\sin\frac{1}{2}x\neq0$

Integrating both sides we get $\displaystyle \sum_{r=1}^n \frac{1}{r}\sin rx = \int \frac{\sin(n+\frac{1}{2})x-\sin\frac{1}{2}x}{2\sin\frac{1}{2}x} dx$

Simplifying the integral as making the appropriate substitution $u=\frac{1}{2}x$ we get the result:

$x + \displaystyle \sum_{r=1}^n \frac{1}{r}\sin 2rx = \int \frac{\sin (2n+1)x}{\sin x} dx$

Using the fact that $\lim_{n\to\infty} \int_{0}^{a} \frac{\sin (2n+1)x}{\sin x} dx = \frac{\pi}{2}$ for positive $a$ and $a\neq q\pi$ where $q$ is a positive integer, we can establish infinite sums such as:

$\frac{\pi}{6} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{r\pi}{3} = \frac{\pi}{2}$

$\frac{\pi}{8} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{r\pi}{4} = \frac{\pi}{2}$

$\frac{\pi}{3} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{2r\pi}{3} = \frac{\pi}{2}$

$\frac{\pi}{4} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{r\pi}{2} = \frac{\pi}{2}$