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?

Given how popular these seem to be becoming, it might be worth learning a nice fact to eliminate the need for a lot of cleverness:
$$ \sum_{k=1}^n \frac{1}{k} = \log n + \gamma + O(1/n) $$
where $\gamma$ is Euler's constant.
Partial fractions tells us
$$\frac{k-1}{2k(k+1)(2k+1)} = \frac{3}{2k+1} - \frac{1}{2k} - \frac{1}{k+1} $$
We can use the trick
$$ \begin{align}\sum_{k=1}^n \frac{1}{2k-1}
&= \left(\sum_{k=1}^{2n} \frac{1}{k}\right) - \frac{1}{2} \left( \sum_{k=1}^n \frac{1}{k} \right) 
\\&= (\log 2n + \gamma) - \frac{1}{2} (\log n + \gamma) + O(1/n)
\\&= \frac{1}{2}\log n + \frac{1}{2} \gamma + \log 2 + O(1/n)
\end{align}$$
and thus compute the partial sum of the three terms:
$$ \sum_{k=1}^n \frac{3}{2k+1} = 
\frac{3}{2}\log(n+1) + \frac{3}{2} \gamma + 3\log 2  - \frac{3}{1} + O(1/n)$$
$$ \sum_{k=1}^n \frac{1}{2k} = \frac{1}{2} \log(n) + \frac{1}{2} \gamma + O(1/n) $$
$$ \sum_{k=1}^n \frac{1}{k+1} = \log(n+1) + \gamma - \frac{1}{1} + O(1/n) $$
combine them, then take the limit. We can even simplify that by noting $\log(n+1) = \log n + O(1/n)$.
