Calculating the value of $\sum^\infty_{k=1}\frac{1}{k(9k^2-1)}$ My first thought is to split it up into:
$$\sum^\infty_{k=1}\;\frac{3}{2(3k-1)} + \frac{3}{2(3k+1)} - \frac{1}{k}$$
This is starting to look vaguely like some sort of rearrangement of the alternating harmonic series with some factors thrown in. The way I would evaluate one of those types of sums would be to exploit the fact that $\lim_{n \to \infty} H_n - \log n = \gamma$ exists to find a formula for the partial sum, then take the limit and the answer would just fall out.
Maybe I'm just being stupid, but I can't see exactly how to do it in this case. Alternative methods are also welcome.
 A: Note that for $0<r<1$ $$\frac{r^2}{2\cdot 2}-\frac{r^3}3+\frac{r^4}{2\cdot 4}+\frac{r^5}{2\cdot 5}-\frac{r^6}6+\frac{r^7}{2\cdot 7}+\ldots$$ is the real part of $$-\frac{r}2-\sum_{k=1}^{\infty}\frac{(r\zeta)^k}{k} =-\frac{r}2+\log(1-r\zeta)$$ where $\zeta=\exp(2\pi\textrm{i}/3)$. Taking $r\uparrow 1$ this has the limit $$-\frac12+\frac12\log(3)-\frac{\pi\textrm{i}}{6}.$$ So your series converges to $\tfrac32(\log(3)-1)$.
A: Let $q\ge2$ be an integer. Then set
$$f_q(a,b,c)=\sum_{n\ge1}a^{qn}b^{qn-1}c^{qn-2}=\frac{1}{bc^2}\sum_{n\ge1}(abc)^{qn}=\frac{a^{q}b^{q-1}c^{q-2}}{1-(abc)^q},
\qquad |abc|<1.$$
Then we may see
$$\sum_{n\ge1}\frac{1}{n(q^2n^2-1)}=q\int_0^1\int_0^1\int_0^1f_q(a,b,c)dadbdc=q\int_{(0,1)^3}f_q(a,b,c)dA.$$
Here, $dA=da\,db\,dc$.
Thus your sum is given as
$$S=\sum_{n\ge1}\frac{1}{n(9n^2-1)}=3\int_{(0,1)^3}\frac{a^3b^2c}{1-(abc)^3}dA.$$
I'm sure there is a closed form, but the triple integral is very tedious. Complex numbers may be involved.
An alternate form is
$$S=\int_0^1\int_0^x\int_0^u\frac{t}{1-t^3}dt\,du\,dx.$$
This 'simplifies' to
$$S=\int_0^1\left(\frac16J_1(x)-\frac{1}{\sqrt3}J_2(x)-\frac13J_3(x)+\frac{\pi}{2\sqrt3}x\right)dx,$$
where
$$\begin{align}
J_1(x)&=(x+2)\ln(1+x+x^2)+\sqrt3\arctan\tfrac{1}{\sqrt3}(2x+1)-2x-\frac\pi{2\sqrt3}\\
J_2(x)&=-\frac{\sqrt3}{4}\ln(1+x+x^2)+(x+\tfrac12)\arctan\tfrac{1}{\sqrt3}(2x+1)-\frac\pi{12}\\
J_3(x)&=(x-1)\ln(x-1)+\ln(-1)-x.
\end{align}$$
