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?
 A: 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 $$\begin{eqnarray} \sum_{k=1}^\infty \frac{k-1}{2k(k+1)(2k+1)}  &=& \sum_{k=1}^\infty \frac{2k - (k+1)}{2k(k+1)(2k+1)} \\ &=&\sum_{k=1}^\infty \left(\frac{1}{(k+1)(2k+1)} - \frac{1}{2k(2k+1)}  \right)\\
&=& \sum_{k=1}^\infty \left(\frac{2}{2k+1} - \frac{1}{k+1} - \frac{1}{2k} + \frac{1}{2k+1}\right) \\ &=& \sum_{k=1}^\infty \left(\frac{3}{2k+1} - \frac{1}{k+1} - \frac{1}{2k}\right)\\
&=& 2 + 3\sum_{k=1}^\infty \frac{1}{2k-1}-\frac{1}{2k} \\&=& RHS.\end{eqnarray}$$
A: $$
\begin{align}
\sum_{k=1}^\infty\frac{k-1}{2k(1+k)(1+2k)}
&=\sum_{k=1}^\infty\left(-\frac1{2k}-\frac1{k+1}+\frac3{2k+1}\right)\\
&=\color{#C00000}{\sum_{k=1}^\infty\left(\frac3{2k-1}-\frac3{2k}\right)}\\
&+\color{#00A000}{\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)}\\
&+\color{#0000F0}{\sum_{k=1}^\infty\left(\frac{3}{2k+1}-\frac3{2k-1}\right)}\\[3pt]
&=\color{#C00000}{3\log(2)}\color{#00A000}{+1}\color{#0000F0}{-3}\\[9pt]
&=\log(8)-2
\end{align}
$$
A: 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.
A: 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 $\log(n+1) = \log n + O(1/n)$.
