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.

1.) Use partial fractions $a_n=\frac{1}{n(4n^2-1)}=\frac{1}{2 n-1}+\frac{1}{2 n+1}-\frac{1}{n}$
2.) Rewrite the sum as $S =\sum_{n=1}^{\infty}a_n\equiv \lim_{N\rightarrow \infty} \sum_{n=1}^N a_n$ 
3.) Use the series representation of the digamma function and their relation  to the harmonic numbers to obtain 
$$
S=\lim_{N\rightarrow \infty}\left(-H_N+H_{N-\frac{1}{2}}+\frac{1}{2 N+1}-1+2\log (2) \right)
$$
4.) It is clear that for $N\rightarrow \infty$ , $H_N\rightarrow H_{N-1/2}$ and $\frac{1}{2N+1}\rightarrow 0$ . 
The result therefore is
$$
S=2 \log(2)-1\approx 0.386294
$$
