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.
 A: 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
$$
S=2 \log(2)-1\approx 0.386294
$$
A: 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}
A: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)} = \sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)} = \sum_{n=1}^{\infty}\int_0^1 \int_0^1 \frac{x^{2n}y^{2n-2}}{n}\,dy \,dx $
$\displaystyle =\int_0^1 \int_0^1\sum_{n=1}^{\infty} \frac{x^{2n}y^{2n-2}}{n} \,dy \,dx = \int_0^1 \int_0^1 \frac{\log(1-x^2y^2)}{y^2}\,dy \,dx \\ \displaystyle = \int_0^1 2 x \tanh^{-1}(x)+\log(1-x^2)\,{dx} = \log(4)-1. $ 
A: 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
$$\sum_{n=1}^\infty \frac{1}{n(2n-1)(2n+1)}=\sum_{n=1}^\infty\left(\frac{1}{2n-1}-\frac{1}{2n}\right)+\sum_{n=1}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n}\right)=2\log(2)-1.$$
A: 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 $$\log (x+1) = \sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{k}.$$
A: Hint:
$$\frac{1}{n(4n^2-1)}=-\frac{1}{n}+\frac{1}{2n+1}+\frac{1}{2n-1}$$
