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!

  • 1
    $\begingroup$ Take a look at the properties of digamma function. $\endgroup$ – Math-fun Mar 19 '19 at 11:36
  • $\begingroup$ If you just want the answer, Wolfram Alpha gives $$\frac{\pi}{2\sqrt 3}+2\log 2-\frac32\log 3$$ But I don't know how that was calculated. $\endgroup$ – TonyK Mar 19 '19 at 11:37

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 $$\psi ^{(0)}\left(\frac{2}{3}\right)=-\gamma +\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}\qquad \text{and} \qquad \psi ^{(0)}\left(\frac{1}{2}\right)=-\gamma -2\log (2)$$


Using they fact that $$\psi(z+1)-\psi(s+1)=\sum_{n=1}^\infty \frac{1}{n+s}-\frac{1}{n+z}$$ Your sum is $$\psi(2/3)-\psi(1/2)$$ This simplifies to $$\frac{\pi}{2\sqrt{3}}-\frac{3\log 3}{2}+2\log 2$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.