Find sum of series $\sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)}$

The easiest thing to do is to further decompose the decomposition; i.e., $$\frac{1}{n(4n^2 - 1)} = \frac{1}{2n-1} - \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n},$$ and look at the alternating harmonic series $$\log (x+1) = \sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{k}.$$

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

$$ S=2 \log(2)-1\approx 0.386294 $$

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}