I would like to compute the value of the following sum
$$\sum_{ n = 1}^\infty \frac{1}{(2n-1)(3n-1)}$$
Clearly, it converges since $ \frac{1}{(2n-1)(3n-1)} = O(n^{-2})$. I tried to use the partial fraction decomposition to get :
$$\frac{1}{(2n-1)(3n-1)} = \frac{2}{2n-1}- \frac{3}{3n-1}$$
yet, it doesn't seem to lead anywhere since it's hard to see where the terms cancel out. So I don't really know what I could do in order to to attack this sum.
Thank you for your help!
Best Answer
As Math-fun commented, this is related to the digamma function $$S_p=\sum_{ n = 1}^p \frac{1}{(2n-1)(3n-1)}=2\sum_{ n = 1}^p\frac{1}{2n-1}-3\sum_{ n = 1}^p \frac{1}{3n-1}$$ $$S_p=\psi ^{(0)}\left(p+\frac{1}{2}\right)-\psi ^{(0)}\left(p+\frac{2}{3}\right)+\psi ^{(0)}\left(\frac{2}{3}\right)-\psi ^{(0)}\left(\frac{1}{2}\right)$$ Now, using the asymptotics $$S_p=\psi ^{(0)}\left(\frac{2}{3}\right)-\psi ^{(0)}\left(\frac{1}{2}\right)-\frac{1}{6 p}+O\left(\frac{1}{p^2}\right)$$ and $$\psi ^{(0)}\left(\frac{2}{3}\right)=-\gamma +\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}\qquad \text{and} \qquad \psi ^{(0)}\left(\frac{1}{2}\right)=-\gamma -2\log (2)$$