Calculate the value of the series $\,\sum_{n=1}^\infty\frac{1}{2n(2n+1)(2n+2)}$ Calculate the infinite sum

$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$

I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum. 
Is there any easy way to calculate this?
Please someone help.

In the style suggested by Michael Rozenberg's comment (solution 2), let us take one and two terms out of Mercator series
$$\begin{align}
S_0=\log(2)&=\sum_{k=0}^\infty\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)\\
&=1-\sum_{k=0}^\infty \left(\frac{1}{2k+2}-\frac{1}{2k+3}\right)\\
&=1-\frac{1}{2}+\sum_{k=0}^\infty \left(\frac{1}{2k+3}-\frac{1}{2k+4}\right)\\
&=\frac{1}{2}+\sum_{k=0}^\infty \left(\frac{1}{2k+3}-\frac{1}{2k+4}\right)\\
\end{align}$$
to obtain
$$S_1 = \sum_{k=0}^\infty \left(\frac{1}{2k+2}-\frac{1}{2k+3}\right)=1-\log(2)$$
and 
$$S_2=\sum_{k=0}^\infty \left(\frac{1}{2k+3}-\frac{1}{2k+4}\right)=\log(2)-\frac{1}{2}$$
This reduces the evaluation to
$$\begin{align}S&=\sum_{k=0}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)}\\
\\
&=\frac{1}{2}\sum_{k=0}^\infty\left(\frac{1}{2k+2}-\frac{2}{2k+3}+\frac{1}{2k+4}\right)\\
\\
&=\frac{1}{2}\sum_{k=0}^\infty\left(\frac{1}{2k+2}-\frac{2}{2k+3}+\frac{1}{2k+4}\right)\\
\\
&=\frac{1}{2}\sum_{k=0}^\infty\left(\frac{1}{2k+2}-\frac{1}{2k+3}-\frac{1}{2k+3}+\frac{1}{2k+4}\right)\\
\\
&=\frac{S_1-S_2}{2}=\dfrac{1-\log(2)-\left(\log(2)-\dfrac{1}{2}\right)}{2}=\dfrac{\dfrac{3}{2}-2\log(2)}{2}\\
\\
&=\frac{3}{4}-\log(2)\\
\end{align}$$
the semidifference of the precomputed series.
