<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] Hint. First observe that
$$
\frac{1}{2i(2i+1)(2i+2)}=\frac{1}{2}\left(\frac{1}{2i(2i+1)}-\frac{1}{(2i+1)(2i+2)}\right)=\frac{1}{2}\left(\frac{1}{2i}-\frac{2}{2i+1}+\frac{1}{2i+2}\right)
$$
Then
$$
\sum_{i=1}^n\frac{1}{2i(2i+1)(2i+2)}=\sum_{i=1}^n\left(\frac{1}{2i}-\frac{1}{2i+1}\right)-\frac{1}{4}+\frac{1}{2(2n+2)}\\=\sum_{i=1}^n\frac{1}{2i(2i+1)}-\frac{1}{4}+\frac{1}{2n(2n+2)}
$$
Clearly
$$
\sum_{i=1}^n\frac{1}{2i(2i+1)}=\int_0^1\int_0^x(t+t^3+t^5+\cdots+t^{2n-1})\,dt\,dx\longrightarrow\int_0^1\int_0^x\frac{t\,dt}{1-t^2}\,dx\\=\frac{1}{2}\int_0^1\int_0^x\left(\frac{1}{1-t}-\frac{1}{1+t}\right)dt\,dx=-\frac{1}{2}\int_0^1\left(\log(1-x)+\log(1+x)\right)\,dx=\cdots
$$
We have
$$
\int_0^1 \log(1-x)\,dx=\int_0^1 \log x\,dx=\left.x\log x-x\right|_0^1=-1, \\
\int_0^1\log(1+x)\,dx=\int_1^2\log x\,dx=\left.x\log x-x\right|_1^2=2\log 2-1.
$$
Hence
$$
\sum_{i=1}^\infty\frac{1}{2i(2i+1)}=1-\log 2
$$
and finally
$$
\sum_{i=1}^\infty\frac{1}{2i(2i+1)(2i+2)}=\frac{3}{4}-\log 2.
$$
</s>