Show that:$\sum\limits_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$

If we consider $$ f(x)=\frac{x}{(4x^2-1)(16x^2-1)} $$ we may compute its partial fraction decomposition through the residue theorem: $$ f(x) = \frac{1}{24}\left(\frac{1}{x-\tfrac{1}{2}}+\frac{1}{x+\tfrac{1}{2}}\right)-\frac{1}{24}\left(\frac{1}{x-\tfrac{1}{4}}+\frac{1}{x+\tfrac{1}{4}}\right)$$ and that leads to: $$\begin{eqnarray*} \sum_{n\geq 1}f(n) &=& \frac{1}{6}\sum_{n\geq 1}\left(\frac{1}{4n-2}+\frac{1}{4n+2}-\frac{1}{4n-1}-\frac{1}{4n+1}\right) \\&=&\frac{1}{6}\int_{0}^{1}\sum_{n\geq 1}\left(x^{4n-3}+x^{4n+1}-x^{4n-2}-x^{4n}\right)\,dx\\&=&\frac{1}{6}\int_{0}^{1}\frac{x(1-x)(1-x^3)}{1-x^4}\,dx\\&=&\frac{1}{6}\left(\int_{0}^{1}(1-x)\,dx-\int_{0}^{1}\frac{(1-x)^2}{1-x^4}\,dx\right)\\&=&\frac{1}{6}\left(\int_{0}^{1}(1-x)\,dx-\int_{0}^{1}\frac{dx}{x+1}+\frac{1}{2}\int_{0}^{1}\frac{2x}{x^2+1}\,dx\right)\\&=&\frac{1-\log 2}{12}.\end{eqnarray*}$$

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

$$\psi\left(\frac{1}{2} \right) = -\gamma -2\log(2)$$

$$\psi\left(\frac{1}{4} \right) = -\gamma -\frac{\pi}{2}-3\log(2)$$

$$\psi(1+x) = \psi(x)+\frac{1}{x}$$ $$\psi(1-x) = \psi(x)+\pi \cot(\pi x)$$

You can use the decomposition $$\frac{n}{(4n^2-1)(16n^2-1)}=\frac{1/12}{2n-1}+\frac{1/12}{2n+1}+\frac{-1/6}{4n+1}+\frac{-1/6}{4n-1}$$ Then let $H_N=\sum_{n=1}^N \frac{1}{n}=\log(N)+\gamma+o(1)$

You can compute $$\sum_{n=1}^N \frac{1}{2n-1}=\sum_{n=1}^{2N} \frac{1}{n}-\sum_{n=1}^N \frac{1}{2n}=H_{2N}-H_N/2$$

$$\sum_{n=1}^N \frac{1}{2n+1}=\sum_{n=1}^{N+1} \frac{1}{2n-1}-1=H_{2N+2}-H_{N+1}/2-1$$

And, $$\sum_{n=1}^N \frac{1}{4n-1} + \sum_{n=1}^N \frac{1}{4n+1} + \sum_{n=1}^N \frac{1}{4n} + \sum_{n=1}^N \frac{1}{4n-2} = \sum_{n=2}^{4N+1} \frac{1}{n} = H_{4N+1}-1 $$ Hence, $$\sum_{n=1}^N \frac{1}{4n-1} + \sum_{n=1}^N \frac{1}{4n+1} = H_{4N+1} - 1 - (1/2) \sum_{n=1}^N \frac{1}{2n-1} -(1/4)H_N$$

At this point you can compute your sum: $$\sum_{n=1}^N \frac{1/12}{2n-1}+\frac{1/12}{2n+1}+\frac{-1/6}{4n+1}+\frac{-1/6}{4n-1} = \frac{1}{12}H_{2N} - \frac{1}{24}H_N +\frac{1}{12} H_{2N+2} - \frac{1}{12} -\frac{1}{24}H_{N+1} -\frac{1}{6} \left(H_{4N+1} -1 -\frac{1}{2}H_{2N} + \frac{1}{4}H_N - \frac{1}{4} H_N \right)$$ You can use the development with $\gamma$, and you get: $$\frac{1}{6}\log(2N)+\frac{1}{6}\gamma -\frac{1}{24} \log(N) - \frac{1}{24}\gamma+\frac{1}{12}\log(2N+2) + \frac{1}{12}\gamma + \frac{1}{12} - \frac{1}{24}\log(N+1) - \frac{1}{24}\gamma - \frac{1}{6}\gamma - \frac{1}{6}\log(4N+1) + o(1)$$ You notice that all the $\gamma$ term vanish and you compute the equivalent when $N$ goes to $\infty$ (mainly replace $N+1$ by $N$), all the $\log(N)$ terms vanish too and you get $$\frac{1}{6}\log(2) + \frac{1}{12}\log(2) + \frac{1}{12} - \frac{1}{6}\log(4) = \frac{1}{12} - \frac{\log(2)}{12} $$