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.

Let us use partial fraction decomposition and the identity
$$\frac{1}{n+1}=\int_0^1 x^n dx$$
to evaluate
$$\begin{align}S&=\frac{1}{2·3·4}+\frac{1}{4·5·6}+\frac{1}{6·7·8}+...\\
\\
&=\sum_{k=0}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)}=\sum_{k=0}^\infty \frac{1}{2} \left(\frac{1}{2k+2}-\frac{2}{2k+3}+\frac{1}{2k+4}\right)\\
\\
&=\frac{1}{2} \sum_{k=0}^\infty  \int_0^1 \left(x^{2k+1}-2x^{2k+2}+x^{2k+3}\right)dx=\frac{1}{2} \int_0^1   
(x-2x^2+x^3) \sum_{k=0}^\infty x^{2k}dx\\
\\
&=\frac{1}{2} \int_0^1   
\frac{x(1-x)^2}{1-x^{2}}dx=\frac{1}{2} \int_0^1   
\frac{x(1-x)}{1+x}dx\\
\\
&=\frac{1}{2} \int_0^1   
\frac{-x^2+x}{1+x}dx=\frac{1}{2} \int_0^1   
\frac{-x^2-x+2x+2-2}{1+x}dx\\
\\
&=\frac{1}{2} \int_0^1   \left(-x+2-\frac{2}{1+x}\right)dx=\frac{1}{2} \left(-\frac{x^2}{2}+2x-2\log(1+x)\right)|_0^1\\
\\
&=\frac{3}{4}-\log(2)\\
\end{align}$$
This series has all terms positive and thus is a direct proof of the inequality
$$\log(2)<\frac{3}{4}$$
Similar series and integrals may be found in Series and integrals for inequalities and approximations to $\log(n)$.
