Originally Posted by
Sudharaka In your previous thread we have found that, $\displaystyle \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right)$
Here we have to find, $\displaystyle \sum_{r = 0}^{n}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{n}\frac{1}{4r^2-8r+3}$
Therefore taking the limit as n tend to infinity,
$\displaystyle \sum_{r = 0}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}$
Hope I did not confuse you with my last post, there I did not notice that the summation must be taken from r=0.