<s>[INST] 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.
 [/INST] I undeleted my answer and fixed it up since one of the other answers refers to it.

In the comments, Lucian mentioned the generalization $$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)(n+c)} = \frac{(b-c) \, \psi(a)+ (c-a) \, \psi(b)+(a-b) \, \psi(c)}{(a-b)(b-c)(c-a)} \tag{1}$$ where $\psi(x)$ is the digamma function;  $a,b,c >0$; $a \ne b$; $b \ne c$; and $a \ne c$.
We can prove $(1)$ by using the series representation $$\psi(x) = - \gamma + \sum_{n=0}^{\infty} \left(\frac{1}{n+1}-\frac{1}{n+x} \right) \, , \quad x>0.$$
Starting from the right side of $(1)$, we get
$$ \begin{align} &\frac{(b-c) \, \psi(a)+ (c-a) \, \psi(b)+(a-b) \, \psi(c)}{(a-b)(b-c)(c-a)} \\   &= \small\left[ (b-c)\left(-\gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+a} \right) \right)+ (c-a)\left(-  \gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+b} \right) \right)   + (a-b) \left( -\gamma +\sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+c} \right) \right) \right] /{\left((a-b)(b-c)(c-a) \right)} \\  &= \frac{1}{(a-b)(b-c)(c-a)}\sum_{n=0}^{\infty} \left( \frac{c-b}{n+a}+\frac{a-c}{n+b} + \frac{b-a}{n+c} \right) \\ &=\frac{1}{(a-b)(b-c)(c-a)} \sum_{n=0}^{\infty} \ \frac{bc^{2}-b^{2}c+a^{2}c-ac^{2}+ab^{2}-a^{2}b}{(n+a)(n+b)(n+c)} \\ &= \frac{1}{(a-b)(b-c)(c-a)}\sum_{n=0}^{\infty} \frac{(a-b)(ac-c^{2}-ab+bc)}{(n+a)(n+b)(n+c)} \\ &=\frac{1}{(a-b)(b-c)(c-a)}\sum_{n=0}^{\infty} \frac{(a-b)(b-c)(c-a)}{(n+a)(n+b)(n+c)} \\ &= \sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)(n+c)} \end{align}$$
</s>