<s>[INST] Show that:$\sum\limits_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$ Show that
$$\sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$$
My try:
We split into partial decomposition 
$$n={A\over 2n-1}+{B\over 2n+1}+{C\over 4n-1}+{D\over 4n+1}$$
Setting $n={1\over 2}$, ${-1\over2}$ we have $A={1\over3}$ and $B={-1\over 3}$
Finding C and D is a bit tedious 
I wonder what is the closed form for
$$\sum_{n=1}^{\infty}{1\over an+b}=F(a,b)?$$
This way is not a good approach. Can anyone help me with a better approach to tackle this problem? Thank you.
 [/INST] Note that 
$$S = \sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}= \sum_{n=1}^\infty \frac{1}{12n}\left\{\frac{1}{4n^2-1}-\frac{1}{16n^2-1} \right\}$$
Hence
$$ \frac{1}{12n}\left\{\frac{1}{4n^2-1}-\frac{1}{16n^2-1} \right\}=   \frac{1}{24n}\left\{\frac{1}{2n-1}-\frac{1}{2n+1}+\frac{1}{4n+1}-\frac{1}{4n-1} \right\}$$
Using the digamma function we have 
$$S = \frac{1}{24}\left\{- \psi \left(\frac{3}{2}\right)-\psi \left(\frac{1}{2}\right)+\psi \left(\frac{5}{4}\right)+\psi \left(\frac{3}{4}\right) \right \}  = \frac{1}{12}(1-\log 2)$$
Since
$$\psi(x+1) = -\gamma + \sum \frac{x}{n(n+x)}$$
Note that 
$$\psi\left(\frac{1}{2} \right) = -\gamma -2\log(2)$$
$$\psi\left(\frac{1}{4} \right) = -\gamma -\frac{\pi}{2}-3\log(2)$$
$$\psi(1+x) = \psi(x)+\frac{1}{x}$$
$$\psi(1-x) = \psi(x)+\pi \cot(\pi x)$$
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