Find sum of series $\sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)}$ I need help with finding sum of this:
$$
\sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)}
$$
First, I tried to telescope it in some way, but it seems to be dead end. The only other idea I have is that this might have something to do with logarithm, but really I don't know how to proceed. Any hint would be greatly appreciated.

Let
$$ f(x)=\sum_{n=1}^\infty\frac{1}{n(4n^2-1)}x^{2n+1}. $$
Clearly $\sum_{n=1}^\infty\frac{1}{n(4n^2-1)}=f(1)$. Note
$$ f'(x)=\sum_{n=1}^\infty\frac{1}{n(2n-1)}x^{2n}, f''(x)=2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}, f'''(x)=2\sum_{n=1}^\infty x^{2n-2}=\frac{2}{1-x^2}$$
So
\begin{eqnarray}
f(1)&=&\int_0^1\int_0^x\int_0^y\frac{2}{1-z^2}dzdydx\\
&=&\int_0^1\int_z^1\int_x^1\frac{2}{1-z^2}dydxdz\\
&=&\int_0^1\frac{1-z}{1+z}dz\\
&=&2\log2-1.
\end{eqnarray}
