Telescoping property: $\sum_{k \geq 0}\frac{1}{(4k+1)(4k+3)}$

Question

I need to calculate the sum $$\sum_{k \geq 0}\frac{1}{(4k+1)(4k+3)}$$

I've made some attempts to transform this in a summation that I could apply the telescopic property, but I didnt have any success.

Thanks in advance!

Answer 1

First notice that

$\sum \frac{1}{(4k+1)(4k+3)} = \frac{1}{2} \sum \frac{2}{(4k+1)(4k+3)} = \frac{1}{2} \sum \frac{1}{(4k+1)} - \frac{1}{4k+3} =\frac{1}{2} \sum \frac{(-1)^{k}}{2k+1} = \frac{1}{2}\frac{\pi}{4} = \frac{\pi}{8}$

To know why $\sum \frac{(-1)^{k}}{2k+1} = \frac{\pi}{4} $ check here.