series: $\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$ 
We have the infinite series:$$\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$$

This is not my series: $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{4\cdot 5\cdot 6}+\cdots$ so I cannot use $\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$
My attempt:
I know that this type of series solved by making it telescoping series but here, I am unable find general term of the series. Thank you.

Let
$$ f(x)=\sum_{n=1}^\infty\frac1{n(n+1)(n+2)}x^{n+2}. $$
Then
$$ f'(x)=\sum_{n=1}^\infty\frac1{n(n+1)}x^{n+1},f''(x)=\sum_{n=1}^\infty\frac1{n}x^{n},f'''(x)=\sum_{n=1}^\infty x^{n-1}=\frac1{1-x}. $$
So
$$ f''(x)=-\ln(1-x)$$
and
$$ f(1)=-\int_0^1\int_0^x\ln(1-t)dtdx=\cdots$$
