How find this sum $\sum\limits_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$ Find this sum

$$I=\sum_{n=0}^{\infty}\dfrac{1}{(3n+1)(3n+2)(3n+3)}$$

My try: let

$$f(x)=\sum_{n=0}^{\infty}\dfrac{x^{3n+3}}{(3n+1)(3n+2)(3n+3)},|x|\le 1$$

then we have
$$f^{(3)}(x)=\sum_{n=0}^{\infty}x^{3n}=\dfrac{1}{1-x^3}$$
then we find the $f(x)$,Following is very ugly(can you someone can post your follow solution,) have other simple methods? Thank you very much.

OK, so let's start with 
$$\frac{d^3}{dx^3} f(x) = \frac{1}{1-x^3}$$
where $f(0)=f'(0)=f''(0)=0$.  Note that we may use partial fractions and a little rearranging to find that
$$\frac{1}{1-x^3} = \frac13 \left [\frac{1}{1-x} + \frac{x+\frac12}{\left (x+\frac12\right)^2+\frac{3}{4}}+\frac{3}{2} \frac{1}{\left (x+\frac12\right)^2+\frac{3}{4}}\right ]$$
Integrating and using $f''(0)=0$ we get that
$$f''(x) = -\frac13 \log{(1-x)} + \frac16 \log{(1+x+x^2)} + \frac{1}{\sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}} - \frac{\pi}{6 \sqrt{3}}$$
From here the challenge will be managing the various terms rather than individual integrations, which are easily done through parts.  Note that, for example,
$$\int dx \, \log{(1+x+x^2)} = x \log{(1+x+x^2)} - \int dx \frac{2 x^2+x}{x^2+x+1}$$
You may use partial fractions on the latter integral and determine that
$$\int dx \frac{2 x^2+x}{x^2+x+1} = 2 x - \frac12 \log{(1+x+x^2)} - \sqrt{3} \arctan{\frac{2 x+1}{\sqrt{3}}} +C$$
Using a substitution ($u=(2 x+1)/\sqrt{3}$) and integration by parts, we similarly find that
$$\int dx \, \arctan{\frac{2 x+1}{\sqrt{3}}} = \frac{\sqrt{3}}{2}\left [\frac{2 x+1}{\sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}}  - \frac12 \log{(1+x+x^2)}\right ] + C$$
Omitting a lot of algebra, I find that, using $f'(0)=0$:
$$f'(x) = \frac13 (1-x) \log{(1-x)} - \frac16 (1-x) \log{(1+x+x^2)} + \frac{1+x}{\sqrt{3}}  \arctan{\frac{2 x+1}{\sqrt{3}}} - \frac{\pi}{6 \sqrt{3}} (1+x)$$
Now we integrate once more.  Again, the challenge is bookkeeping, as all the integrals may be done by parts.  Using $f(0)=0$, I get that
$$f(x) = -\frac16(1-x)^2 \log{(1-x)} + \frac{1}{12} (x^2-2 x-2) \log{(1+x+x^2)} + \frac{(x^2+2 x)}{2 \sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}}  - \frac{\pi}{12 \sqrt{3}} (x^2+2 x)$$
The sum is then 
$$f(1) = \frac{\sqrt{3}}{12} \pi - \frac14 \log{3} \approx 0.178$$
which checks out with, e.g., WA.
