TITLE: Find sum of series $\sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)}$
QUESTION [12 upvotes]: 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.

REPLY [3 votes]: One way forward is to note that
$$\begin{align}
\sum_{n=1}^{N}\left(\frac{1}{2n-1}-\frac{1}{2n}\right)&=\sum_{n=1}^{N}\left(\frac{1}{2n-1}+\frac{1}{2n}\right)-\sum_{n=1}^{N}\frac1n \tag 1\\\\
&=\sum_{n=1}^{2N}\frac1n -\sum_{n=1}^{N}\frac1n \tag 2\\\\
&=\sum_{n=N+1}^{2N}\frac1n \\\\
&=\sum_{n=1}^{N}\frac{1}{n+N} \\\\
&=\frac1N \sum_{n=1}^{N}\frac{1}{1+n/N} \tag 3
\end{align}$$
In going from $(1)$ to $(2)$ we simply noted that the sum, $\sum\limits_{n=1}^{2N}\frac1n$, can be written in terms of sums of even and odd indexed terms.  
Now, we observe that limit of $(3)$ is the Riemann sum for the integral $$\int_0^1 \frac{1}{1+x}\,dx=\log(2).$$
Similarly, we see that 
$$\begin{align}
\sum_{n=1}^{N}\left(\frac{1}{2n+1}-\frac{1}{2n}\right)&=-1+\frac1N \sum_{n=1}^{N}\frac{1}{1+n/N}
\end{align}$$
is the Riemann sum for $$-1+\int_0^1\frac{1}{1+x}\,dx=-1+\log(2).$$
Putting all of this together, we recover the expected result
$$\sum_{n=1}^\infty \frac{1}{n(2n-1)(2n+1)}=\sum_{n=1}^\infty\left(\frac{1}{2n-1}-\frac{1}{2n}\right)+\sum_{n=1}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n}\right)=2\log(2)-1.$$