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.

$\left(\frac{1}{an+b}\right)_{\substack{n\in\mathbb{N}\\an+b\neq0}}$ is not summable if $a\neq0$ (this has the same behavior as the harmonic series). Thus $\sum\frac{1}{an+b}$ diverges, and your $F(a,b)$ is not well-defined.
However you can sum up to some integer $N$, and use an asymptotic development of:
$$\sum_{n=1}^{N}\frac{1}{an+b}=\frac{1}{a}\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}$$
This can be done typically with a comparison with an integral.

UPD: Actually you just need to know an asymptotic estimation of the harmonic series:
$$H_N=\sum_{n=1}^N\frac{1}{n}=\log(n)+\gamma+o(1)$$
Because:
$$\sum_{n=1}^{N}\frac{1}{n+\lfloor{\frac{b}{a}}\rfloor+1}<\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}\le\sum_{n=1}^{N}\frac{1}{n+\lfloor\frac{b}{a}\rfloor}$$
Thus:
$$\sum_{n=1+\lfloor{\frac{b}{a}}\rfloor+1}^{N+\lfloor{\frac{b}{a}}\rfloor+1}\frac{1}{n}<\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}\le\sum_{n=1+\lfloor{\frac{b}{a}}\rfloor}^{N+\lfloor{\frac{b}{a}}\rfloor}\frac{1}{n}$$
i.e.
$$H_{N+\lfloor{\frac{b}{a}}\rfloor+1}-H_{1+\lfloor{\frac{b}{a}}\rfloor}<\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}\le H_{N+\lfloor\frac{b}{a}\rfloor}-H_{\lfloor\frac{b}{a}\rfloor}$$

Ok, let's write it.
$$\sum_{n=1}^{N}\frac{1}{an+b}=\frac{1}{a}\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}$$
According to our previous estimations: $\sum_{n=1}^{N}\frac{1}{an+b}=\frac{1}{a}\log(N)+O(1)$. But actually we need a $o(1)$ precision. So let $\epsilon_N=\sum_{n=1}^{N}\frac{1}{an+b}-\frac{1}{a}\log(N)$.
$$\epsilon_{N}-\epsilon_{N-1}=\frac{1}{aN+b}-\frac{1}{a}(\log(N)-\log(N-1))=\frac{1}{aN+b}+\frac{1}{a}\log\left(1-\frac{1}{N}\right)$$
So, using asymptotic estimations of $\log(1+x)$ and $\frac{1}{1+x}$ in $x=0$:
$$\epsilon_{N}-\epsilon_{N-1} = \frac{1}{aN}\left(1-\frac{b}{aN}+o\left(\frac{1}{N}\right)\right)+\frac{1}{a}\left(-\frac{1}{N}-\frac{1}{2N^2}+o\left(\frac{1}{N^2}\right)\right)
\sim \frac{C}{N^2}$$
with $C=-\frac{b}{a^2}-\frac{1}{2a}$
Summing equivalents there exists $C'$ s.t. $\epsilon_N = C' - \frac{C}{N}+o\left(\frac{1}{N}\right)$

Now we can solve the original problem, using the partial decomposition-form and adding our estimations.
