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.
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 $$\frac{1}{6}\log(2) + \frac{1}{12}\log(2) + \frac{1}{12} - \frac{1}{6}\log(4) = \frac{1}{12} - \frac{\log(2)}{12} $$
$\left(\frac{1}{an+b}\right)_{\substack{n\in\mathbb{N}\\an+b\neq0}}$ 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.
