<s>[INST] 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.
 [/INST] 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.
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