Given : $$\frac{1}{1\cdot 2\cdot 3}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{7\cdot 8\cdot 9}+ ...$$
What is the sum of this series, how can one rewrite it to look simpler ?
EDIT :
Actually I found how to rewrite it :
$$\sum _{n=1}^{\infty }\:\frac{1}{\left(3n-2\right)\left(3n-1\right)3n}$$
Let's find the sum of the geometric series: $$\sum_{k=0}^\infty x^{3k}=\frac{1}{1-x^{3}}$$
For $0<x<1$. Now let's integrate this series by term three times:
$$\sum_{k=0}^\infty \frac{x^{3k+1}}{3k+1}=\int_0^x \frac{dt}{1-t^{3}}$$
$$\sum_{k=0}^\infty \frac{x^{3k+2}}{(3k+1)(3k+2)}=\int_0^x \int_0^y \frac{dt~ dy}{1-t^{3}}$$
$$\sum_{k=0}^\infty \frac{x^{3k+3}}{(3k+1)(3k+2)(3k+3)}=\int_0^x \int_0^y \int_0^z \frac{dt~ dy~ dz}{1-t^{3}}$$
Changing $k+1 \to k$ we have:
$$\sum_{k=1}^\infty \frac{x^{3k}}{(3k-2)(3k-1)3k}=\int_0^x \int_0^y \int_0^z \frac{dt~ dy~ dz}{1-t^{3}}$$
For $x=1$ we have:
$$\sum_{k=1}^\infty \frac{1}{(3k-2)(3k-1)3k}=\int_0^1 \int_0^y \int_0^z \frac{dt~ dy~ dz}{1-t^{3}}$$
Looks complicated, but if we use Cauchy formula for repeated integration, we simply get:
$$\int_0^1 \int_0^y \int_0^z \frac{dt~ dy~ dz}{1-t^{3}}=\frac{1}{2} \int_0^1 \frac{(1-t)^2dt}{1-t^{3}}=\frac{1}{2} \int_0^1 \frac{(1-t)dt}{1+t+t^2}$$
So the integral form of this series is:
$$\sum_{k=1}^\infty \frac{1}{(3k-2)(3k-1)3k}=\frac{1}{2} \int_0^1 \frac{(1-t)dt}{1+t+t^2}=\frac{1}{12} \left(\sqrt{3} \pi -3 \log (3)\right)$$
To solve the integral we just need to modify the functions so it looks like the definitions of the arctangent and the logarithm. It's not that hard, but I will add the full solution if the OP needs it.
A little hint:
$$1+t+t^2=\frac{3}{4}+\left(\frac{1}{2}+t \right)^2=\frac{3}{4} \left(1+ \frac{4}{3}\left(\frac{1}{2}+t \right)^2 \right)$$
Thus, we need a change of variable: $$u=\frac{2}{\sqrt{3}} \left(\frac{1}{2}+t \right)$$
