Hi there, 


I was wondering if you guys could be able to find the sum of the following series: 


$ S = 1/((1\cdot2)^2) + 1/((3\cdot4)^2) + 1/((5\cdot6)^2) + ... + 1/(((2n-1)\cdot2n)^2) $, in which $\{n\to\infty}$ . 


This question came to mind when I was looking at this (<http://www.stat.purdue.edu/~dasgupta/publications/tr02-03.pdf>) paper by Professor Anirban DasGupta. In the last section, a couple of specific examples of his 'unified' method to find the sums of infinite series is pressented. In equation (34), he states that the following series:


$ 1/(1\cdot2) + 1/(3\cdot4) + 1/(5\cdot6) + ... 1/(2n\cdot(2n-1)) = log(2) $ (Note that $\{n\to\infty}$ again). I was wondering If it's possible to find the sum if the values of the denominators of the terms are squared.


Thanks in advance, 


Max Muller 



Evaluation of the following Series
Yes, it equals $\frac{\pi^2}{3}-3$. This follows from applying partial fractions and using $\zeta(2)=\frac{\pi^2}{6}$.


