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?

We can evaluate the series using partial fraction expansion.  To that end, we have 
$$\begin{align}
\sum_{k=1}^{\infty}\frac{k-1}{2k(k+1)(2k+1)}&=\frac12\sum_{k=1}^{\infty}\left(\frac{6}{2k+1}-\frac{3}{k+1}\right)+\frac12\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k}\right)\\\\
&=-\frac12+\frac32\sum_{k=1}^{\infty}\left(\frac{1}{k+1/2}-\frac{1}{k+1}\right)\end{align} \tag 1$$
Now, we recall from the definition of the digamma function that
$$\begin{align}
\sum_{k=1}^{\infty}\left(\frac{1}{k+1/2}-\frac{1}{k+1}\right)&=-1-\gamma-\psi^0(1/2)\\\\
&=2\log 2-1 \tag 2
\end{align}$$
where we used $\psi^0(1/2)=-\gamma-2\log 2$ to arrive at the right-hand side of $(2)$.  
Finally, substituting $(2)$ into $(1)$ yields
$$\begin{align}
\sum_{k=1}^{\infty}\frac{k-1}{2k(k+1)(2k+1)}&=-\frac12+\frac32\left(2\log 2-1\right)\\\\
&=3\log 2-2
\end{align}$$
which was to be shown!

NOTE:
Here we show another way to evaluate the sum $S$ given in $(2)$ by
$$S=\sum_{k=1}^{\infty}\left(\frac{1}{k+1/2}-\frac{1}{k+1}\right)$$
To that end, we write
$$\begin{align}
S&=\sum_{k=1}^{\infty}\left(\int_0^1 (x^{k-1/2}-x^k)\,dx\right)\\\\
&=\int_0^1\sum_{k=1}^{\infty}\left(x^{k-1/2}-x^k\right)\,dx\\\\
&=\int_0^1\frac{x^{1/2}-x}{1-x}\,dx\\\\
&=\left.\left(x-2x^{1/2}+2\log(1+x^{1/2})\right)\right|_0^1\\\\
&=2\log(2)-1
\end{align}$$
as expected!  
We remark that we tacitly used the Dominated Convergence Theorem to justify the interchange of the order of integration and summation.
