Who can solve it?

I see that this sum has a closed-form Wolframalpha $\frac{1}{3}\sum_{n=0}^\infty \prod_{k=0}^n\frac{3k+2}{3k+6}=\frac{2}{3}$ I record this on The product of fraction with arithmetic sequence

The result generated by WolframAlpha shows Gamma function as this series belongs to Hypergeometric function I add details for this method on The product of fraction with arithmetic sequence $\sum_{n=0}^\infty\prod_{k=0}^n\frac{ak+b}{ak+c} =\frac{b}{c}~_2F_1(1+\frac{b}{a},1;1+\frac{c}{a};1) =\frac{b\Gamma(1+\frac{c}{a})\Gamma(\frac{c-b}{a}-1)}{c\Gamma(\frac{c}{a})\Gamma(\frac{c-b}{a})} =\frac{b}{c-a-b}$

Is there any idea to start( tried V_p but I couldn't proceed)?

Note that If $a_n=\frac{2*5* ...*(3n-1)}{3*6*9*...*(3n+3)}$ and $s_n=a_1+...+a_n$ Then $s_n+(3n+2)a_n=\frac{2}{3}$

RagvaloD wrote: Note that If $a_n=\frac{2*5* ...*(3n-1)}{3*6*9*...*(3n+3)}$ and $s_n=a_1+...+a_n$ Then $s_n+(3n+2)a_n=\frac{2}{3}$ Good idea. I modify my proof as $(c-a-b)\sum_{n=0}^N\prod_{k=0}^n\frac{ak+b}{ak+c}+(a(N+1)+b)\prod_{k=0}^N\frac{ak+b}{ak+c}=b$