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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!

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

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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 $$\frac{\pi}{2\sqrt{3}}-\frac{3\log 3}{2}+2\log 2$$

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