How to prove $\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$? How to prove:
$$\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$$
Is it possible to convert it into a finite integral?

Let me try. 
Note that the $$RHS = 3\ln 2 + 2 = 2 + 3 \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = 2 + 3\sum_{k=1}^\infty \frac{1}{2k-1}-\frac{1}{2k}$$
We have $$\begin{eqnarray} \sum_{k=1}^\infty \frac{k-1}{2k(k+1)(2k+1)}  &=& \sum_{k=1}^\infty \frac{2k - (k+1)}{2k(k+1)(2k+1)} \\ &=&\sum_{k=1}^\infty \left(\frac{1}{(k+1)(2k+1)} - \frac{1}{2k(2k+1)}  \right)\\
&=& \sum_{k=1}^\infty \left(\frac{2}{2k+1} - \frac{1}{k+1} - \frac{1}{2k} + \frac{1}{2k+1}\right) \\ &=& \sum_{k=1}^\infty \left(\frac{3}{2k+1} - \frac{1}{k+1} - \frac{1}{2k}\right)\\
&=& 2 + 3\sum_{k=1}^\infty \frac{1}{2k-1}-\frac{1}{2k} \\&=& RHS.\end{eqnarray}$$
