Finding the sum of $\sum_{n=1}^\infty {2 \over (4n-1)(4n+3)}$ I know this is very similar to this question, so I thought the answer should be $\frac \pi 4$ until I saw another question that is more like mine that doesn't have a good answer (or if it is I am having trouble understanding it), but it is 2 years old so I don't think it will be getting another answer soon. 
Anyway, I was able to prove that this sum converges by the direct comparison test to $\sum_{n=1}^\infty{\frac 1 {n^2}}$ (thanks to the help in my previous question), but now to find the sum.
I tried to use partial sums 
$${2 \over (4n-1)(4n+3)} = {2 \over 19(4n-1)} - {1 \over 2(4n+3)}.$$
If this is right I think I am stuck since $19(4n-1)$ is always odd, but $2(4n+3)$ is always even so telescoping won't work unless I am mistaken. 
 A: $$\frac{1/2}{4n-1} - \frac{1/2}{4n+3} = \frac{2}{(4n-1)(4n+3)}$$
Hence your sum is telescoping
$$\frac{1/2}{3} - \frac{1/2}{7} + \frac{1/2}{7} - \frac{1/2}{11} + \frac{1/2}{11} - \frac{1/2}{15} + \cdots$$
and it's clear that the sum is $\frac{1}{6}$.
A: There is a trick to find those fractions with little effort ( i will post here in a more general form )
If we have only two terms in the denominator
$$\frac{1}{(ak+b)(ak+b+a) }=\frac{1}{a}\frac{(ak+b+a) -(ak+b)}{(ak+b)(ak+b+a) }$$
$$=\frac{1}{a}\frac{1}{(ak+b) } -\frac{1}{sa}\frac{1}{ (ak+b+a) }. $$
A still more general case:
Calculate the sum
$$\sum^{n}_{k=0}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }. $$
Every time we have this kind of summation we can use the telescoping sum. To find the terms of telescoping sum  we can  add in the numerator the following difference:  $(ak+b+sa)$ the greatest term" minus thelowest term" $(ak+b)$, and to not alter the fraction divide by $\frac{1}{sa}$  (which is the resultant  difference  of those terms).
$$\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }=\frac{1}{sa}\frac{(ak+b+sa) -(ak+b)}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }$$
$$=\frac{1}{sa}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+(s-1)a) } -\frac{1}{sa}\frac{1}{(ak+b+a)\ldots (ak+b+sa) }. $$
Take $f(k)=\frac{1}{sa}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+(s-1)a) }.$
Then the sum is
$$\frac{-1}{sa}\sum^{n}_{k=0} \left(f(k+1) -f(k)\right) $$
by the telescopic sum it's
$$=\frac{-1}{sa}\left(f(n+1) -f(0)\right)=\frac{-1}{sa}\left(\frac{1}{(an+b)\ldots (an+b+(s-1)a) } -\frac{1}{(b)(b+a)\ldots (+b+(s-1)a) }\right). $$
